# String matching

String matching
You are encouraged to solve this task according to the task description, using any language you may know.

Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.

You may see other such operations in the Basic Data Operations category, or:

Integer Operations
Arithmetic | Comparison

Boolean Operations
Bitwise | Logical

String Operations
Concatenation | Interpolation | Comparison | Matching

Memory Operations

Given two strings, demonstrate the following three types of string matching:

1.   Determining if the first string starts with second string
2.   Determining if the first string contains the second string at any location
3.   Determining if the first string ends with the second string

Optional requirements:

1.   Print the location of the match for part 2
2.   Handle multiple occurrences of a string for part 2.

## 360 Assembly

*        String matching           04/04/2017
STRMATCH CSECT
USING STRMATCH,R15
XPRNT SS,L'SS
*
CLC SS(L'S1),S1
BNE NOT1
XPRNT =C'-- STARTS WITH',14
XPRNT S1,L'S1
NOT1 EQU *
*
CLC SS+L'SS-L'S2(L'S2),S2
BNE NOT2
XPRNT =C'-- ENDS WITH',12
XPRNT S2,L'S2
NOT2 EQU *
*
LA R0,L'SS-L'S3+1
LA R1,SS
LOOP CLC 0(L'S3,R1),S3
BNE NOT3
XPRNT =C'-- CONTAINS',11
XPRNT S3,L'S3
NOT3 LA R1,1(R1)
BCT R0,LOOP
*
BR R14
SS DC CL6'ABCDEF'
S1 DC CL2'AB'
S2 DC CL2'EF'
S3 DC CL2'CD'
PG DC CL80' '
YREGS
END STRMATCH
Output:
ABCDEF
-- STARTS WITH
AB
-- ENDS WITH
EF
-- CONTAINS
CD

procedure Match_Strings is
S1 : constant String := "abcd";
S2 : constant String := "abab";
S3 : constant String := "ab";
begin
if S1'Length >= S3'Length and then S1 (S1'First..S1'First + S3'Length - 1) = S3 then
Put_Line (''' & S1 & "' starts with '" & S3 & ''');
end if;
if S2'Length >= S3'Length and then S2 (S2'Last - S3'Length + 1..S2'Last) = S3 then
Put_Line (''' & S2 & "' ends with '" & S3 & ''');
end if;
Put_Line (''' & S3 & "' first appears in '" & S1 & "' at" & Integer'Image (Index (S1, S3)));
Put_Line
( ''' & S3 & "' appears in '" & S2 & ''' &
Integer'Image (Ada.Strings.Fixed.Count (S2, S3)) & " times"
);
end Match_Strings;

Output:
'abcd' starts with 'ab'
'abab' ends with 'ab'
'ab' first appears in 'abcd' at 1
'ab' appears in 'abab' 2 times

## Aime

text t;
data b;

b = "Bangkok";

t = "Bang";

o_form("starts with, embeds, ends with \"~\": ~, ~, ~\n", t, b.seek(t) == 0,
b.seek(t) != -1,
b.seek(t) != -1 && b.seek(t) + ~t == ~b);

t = "ok";

o_form("starts with, embeds, ends with \"~\": ~, ~, ~\n", t, b.seek(t) == 0,
b.seek(t) != -1,
b.seek(t) != -1 && b.seek(t) + ~t == ~b);

t = "Summer";

o_form("starts with, embeds, ends with \"~\": ~, ~, ~\n", t, b.seek(t) == 0,
b.seek(t) != -1,
b.seek(t) != -1 && b.seek(t) + ~t == ~b);
Output:
starts with, embeds, ends with "Bang": 1, 1, 0
starts with, embeds, ends with "ok": 0, 1, 1
starts with, embeds, ends with "Summer": 0, 0, 0

## ALGOL 68

Translation of: python
Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
# define some appropriate OPerators #
PRIO STARTSWITH = 5, ENDSWITH = 5;
OP STARTSWITH = (STRING str, prefix)BOOL: # assuming LWB = 1 #
IF UPB str < UPB prefix THEN FALSE ELSE str[:UPB prefix]=prefix FI;
OP ENDSWITH = (STRING str, suffix)BOOL: # assuming LWB = 1 #
IF UPB str < UPB suffix THEN FALSE ELSE str[UPB str-UPB suffix+1:]=suffix FI;

INT loc, loc2;

print((
"abcd" STARTSWITH "ab", # returns TRUE #
"abcd" ENDSWITH "zn", # returns FALSE #
string in string("bb",loc,"abab"), # returns FALSE #
string in string("ab",loc,"abab"), # returns TRUE #
(string in string("bb",loc,"abab")|loc|-1), # returns -1 #
(string in string("ab",loc,"abab")|loc|-1), # returns +1 #
(string in string("ab",loc2,"abab"[loc+1:])|loc+loc2|-1) # returns +3 #
))
Output:
TFFT         -1         +1         +3

## AppleScript

set stringA to "I felt happy because I saw the others were happy and because I knew I should feel happy, but I wasn’t really happy."

set string1 to "I felt happy"
set string2 to "I should feel happy"
set string3 to "I wasn't really happy"

-- Determining if the first string starts with second string
stringA starts with string1 --> true

-- Determining if the first string contains the second string at any location
stringA contains string2 --> true

-- Determining if the first string ends with the second string
stringA ends with string3 --> false

-- Print the location of the match for part 2
offset of string2 in stringA --> 69

AppleScript doesn't have a builtin means of matching multiple occurrences of a substring, however one can redefine the existing offset command to add this functionality:

-- Handle multiple occurrences of a string for part 2
on offset of needle in haystack
local needle, haystack

if the needle is not in the haystack then return {}
set my text item delimiters to the needle
script
property N : needle's length
property t : {1 - N} & haystack's text items
end script

tell the result
repeat with i from 2 to (its t's length) - 1
set x to item i of its t
set y to item (i - 1) of its t
set item i of its t to (its N) + (x's length) + y
end repeat

items 2 thru -2 of its t
end tell
end offset

offset of "happy" in stringA --> {8, 44, 83, 110}

or, defining an offsets function in terms of a more general findIndices:

-- offsets :: String -> String -> [Int]
on offsets(needle, haystack)
script match
property mx : length of haystack
property d : (length of needle) - 1
on |λ|(x, i, xs)
set z to d + i
mx ≥ z and needle = text i thru z of xs
end |λ|
end script

findIndices(match, haystack)
end offsets

-- TEST ---------------------------------------------------
on run
set txt to "I felt happy because I saw the others " & ¬
"were happy and because I knew I should " & ¬
"feel happy, but I wasn’t really happy."

offsets("happy", txt)

--> {8, 44, 83, 110}
end run

-- GENERIC -------------------------------------------------

-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
set lng to length of xs
set acc to {}
tell mReturn(f)
repeat with i from 1 to lng
set acc to acc & (|λ|(item i of xs, i, xs))
end repeat
end tell
return acc
end concatMap

-- findIndices :: (a -> Bool) -> [a] -> [Int]
-- findIndices :: (String -> Bool) -> String -> [Int]
on findIndices(p, xs)
script go
property f : mReturn(p)
on |λ|(x, i, xs)
if f's |λ|(x, i, xs) then
{i}
else
{}
end if
end |λ|
end script
concatMap(go, xs)
end findIndices

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
Output:
{8, 44, 83, 110}

## ARM Assembly

Works with: as version Raspberry Pi

/* ARM assembly Raspberry PI */
/* program strMatching.s */

/* Constantes */
.equ STDOUT, 1 @ Linux output console
.equ EXIT, 1 @ Linux syscall
.equ WRITE, 4 @ Linux syscall

/* Initialized data */
.data
szMessFound: .asciz "String found. \n"
szString: .asciz "abcdefghijklmnopqrstuvwxyz"
szString2: .asciz "abc"
szStringStart: .asciz "abcd"
szStringEnd: .asciz "xyz"
szStringStart2: .asciz "abcd"
szStringEnd2: .asciz "xabc"
szCarriageReturn: .asciz "\n"

/* UnInitialized data */
.bss

/* code section */
.text
.global main
main:

bl searchStringDeb @ Determining if the first string starts with second string
cmp r0,#0
ble 1f
bl affichageMess
b 2f
1:
bl affichageMess
2:
bl searchStringFin @ Determining if the first string ends with the second string
cmp r0,#0
ble 3f
bl affichageMess
b 4f
3:
bl affichageMess
4:

bl searchStringDeb @
cmp r0,#0
ble 5f
bl affichageMess
b 6f
5:
bl affichageMess
6:
bl searchStringFin
cmp r0,#0
ble 7f
bl affichageMess
b 8f
7:
bl affichageMess
8:
bl searchSubString @ Determining if the first string contains the second string at any location
cmp r0,#0
ble 9f
bl affichageMess
b 10f
9:
ldr r0,iAdrszMessNotFound @ display substring result
bl affichageMess
10:

100: @ standard end of the program
mov r0, #0 @ return code
mov r7, #EXIT @ request to exit program
svc 0 @ perform system call
/******************************************************************/
/* search substring at begin of input string */
/******************************************************************/
/* r0 contains the address of the input string */
/* r1 contains the address of substring */
/* r0 returns 1 if find or 0 if not or -1 if error */
searchStringDeb:
push {r1-r4,lr} @ save registers
mov r3,#0 @ counter byte string
ldrb r4,[r1,r3] @ load first byte of substring
cmp r4,#0 @ empty string ?
moveq r0,#-1 @ error
beq 100f
1:
ldrb r2,[r0,r3] @ load byte string input
cmp r2,#0 @ zero final ?
moveq r0,#0 @ not find
beq 100f
cmp r4,r2 @ bytes equals ?
movne r0,#0 @ no not find
bne 100f
ldrb r4,[r1,r3] @ and load next byte of substring
cmp r4,#0 @ zero final ?
bne 1b @ no -> loop
mov r0,#1 @ yes is ok
100:
pop {r1-r4,lr} @ restaur registers
bx lr @ return

/******************************************************************/
/* search substring at end of input string */
/******************************************************************/
/* r0 contains the address of the input string */
/* r1 contains the address of substring */
/* r0 returns 1 if find or 0 if not or -1 if error */
searchStringFin:
push {r1-r5,lr} @ save registers
mov r3,#0 @ counter byte string
@ search the last character of substring
1:
ldrb r4,[r1,r3] @ load byte of substring
cmp r4,#0 @ zero final ?
addne r3,#1 @ no increment counter
bne 1b @ and loop
cmp r3,#0 @ empty string ?
moveq r0,#-1 @ error
beq 100f
sub r3,#1 @ index of last byte
ldrb r4,[r1,r3] @ load last byte of substring
@ search the last character of string
mov r2,#0 @ index last character
2:
ldrb r5,[r0,r2] @ load first byte of substring
cmp r5,#0 @ zero final ?
addne r2,#1 @ no -> increment counter
bne 2b @ and loop
cmp r2,#0 @ empty input string ?
beq 100f
sub r2,#1 @ index last character
3:
ldrb r5,[r0,r2] @ load byte string input
cmp r4,r5 @ bytes equals ?
bne 100f
subs r3,#1 @ decrement counter
movlt r0,#1 @ if zero -> ok found
blt 100f
subs r2,#1 @ decrement counter input string
blt 100f
ldrb r4,[r1,r3] @ load previous byte of substring
b 3b @ and loop

100:
pop {r1-r5,lr} @ restaur registers
bx lr @ return

/******************************************************************/
/* search a substring in the string */
/******************************************************************/
/* r0 contains the address of the input string */
/* r1 contains the address of substring */
/* r0 returns index of substring in string or -1 if not found */
searchSubString:
push {r1-r6,lr} @ save registers
mov r2,#0 @ counter byte input string
mov r3,#0 @ counter byte string
mov r6,#-1 @ index found
ldrb r4,[r1,r3]
1:
ldrb r5,[r0,r2] @ load byte string
cmp r5,#0 @ zero final ?
moveq r0,#-1 @ yes returns error
beq 100f
cmp r5,r4 @ compare character
beq 2f
mov r6,#-1 @ no equals - > raz index
mov r3,#0 @ and raz counter byte
add r2,#1 @ and increment counter byte
b 1b @ and loop
2: @ characters equals
cmp r6,#-1 @ first characters equals ?
moveq r6,r2 @ yes -> index begin in r6
add r3,#1 @ increment counter substring
ldrb r4,[r1,r3] @ and load next byte
cmp r4,#0 @ zero final ?
beq 3f @ yes -> end search
add r2,#1 @ else increment counter string
b 1b @ and loop
3:
mov r0,r6
100:
pop {r1-r6,lr} @ restaur registers
bx lr

/******************************************************************/
/* display text with size calculation */
/******************************************************************/
/* r0 contains the address of the message */
affichageMess:
push {r0,r1,r2,r7,lr} @ save registers
mov r2,#0 @ counter length */
1: @ loop length calculation
ldrb r1,[r0,r2] @ read octet start position + index
cmp r1,#0 @ if 0 its over
bne 1b @ and loop
@ so here r2 contains the length of the message
mov r1,r0 @ address message in r1
mov r0,#STDOUT @ code to write to the standard output Linux
mov r7, #WRITE @ code call system "write"
svc #0 @ call system
pop {r0,r1,r2,r7,lr} @ restaur registers
bx lr @ return

## AutoHotkey

String1 = abcd
String2 = abab

If (SubStr(String1,1,StrLen(String2)) = String2)
MsgBox, "%String1%" starts with "%String2%".
IfInString, String1, %String2%
{
Position := InStr(String1,String2)
StringReplace, String1, String1, %String2%, %String2%, UseErrorLevel
MsgBox, "%String1%" contains "%String2%" at position %Position%`, and appears %ErrorLevel% times.
}
StringRight, TempVar, String1, StrLen(String2)
If TempVar = %String2%
MsgBox, "%String1%" ends with "%String2%".

## AutoIt

\$string1 = "arduinoardblobard"
\$string2 = "ard"

; == Determining if the first string starts with second string
If StringLeft(\$string1, StringLen(\$string2)) = \$string2 Then
ConsoleWrite("1st string starts with 2nd string." & @CRLF)
Else
ConsoleWrite("1st string does'nt starts with 2nd string." & @CRLF)
EndIf

; == Determining if the first string contains the second string at any location
; == Print the location of the match for part 2
; == Handle multiple occurrences of a string for part 2
\$start = 1
\$count = 0
\$pos = StringInStr(\$string1, \$string2)
While \$pos
\$count += 1
ConsoleWrite("1st string contains 2nd string at position: " & \$pos & @CRLF)
\$pos = StringInStr(\$string1, \$string2, 0, 1, \$start + \$pos + StringLen(\$string2))
WEnd
If \$count = 0 Then ConsoleWrite("1st string does'nt contain 2nd string." & @CRLF)

; == Determining if the first string ends with the second string
If StringRight(\$string1, StringLen(\$string2)) = \$string2 Then
ConsoleWrite("1st string ends with 2nd string." & @CRLF)
Else
ConsoleWrite("1st string does'nt ends with 2nd string." & @CRLF)
EndIf

## AWK

#!/usr/bin/awk -f
{
if (\$1 ~ "^"\$2) {
print \$1" begins with "\$2;
} else {
print \$1" does not begin with "\$2;
}

if (\$1 ~ \$2) {
print \$1" contains "\$2;
} else {
print \$1" does not contain "\$2;
}

if (\$1 ~ \$2"\$") {
print \$1" ends with "\$2;
} else {
print \$1" does not end with "\$2;
}
}

## BASIC

Works with: QBasic
first\$ = "qwertyuiop"

'Determining if the first string starts with second string
second\$ = "qwerty"
IF LEFT\$(first\$, LEN(second\$)) = second\$ THEN
PRINT "'"; first\$; "' starts with '"; second\$; "'"
ELSE
END IF

'Determining if the first string contains the second string at any location
'Print the location of the match for part 2
second\$ = "wert"
x = INSTR(first\$, second\$)
IF x THEN
PRINT "'"; first\$; "' contains '"; second\$; "' at position "; x
ELSE
PRINT "'"; first\$; "' does not contain '"; second\$; "'"
END IF

' Determining if the first string ends with the second string
second\$ = "random garbage"
IF RIGHT\$(first\$, LEN(second\$)) = second\$ THEN
PRINT "'"; first\$; "' ends with '"; second\$; "'"
ELSE
PRINT "'"; first\$; "' does not end with '"; second\$; "'"
END IF

Output:
'qwertyuiop' starts with 'qwerty'
'qwertyuiop' contains 'wert' at position  2
'qwertyuiop' does not end with 'random garbage'

### Applesoft BASIC

10 A\$ = "THIS, THAT, AND THE OTHER THING"
20 S\$ = "TH"
30 DEF FN S(P) = MID\$(A\$, P, LEN(S\$)) = S\$
40 PRINT A\$ : PRINT

110 S\$(1) = "STARTS"
120 S\$(0) = "DOES NOT START"
130 PRINT S\$(FN S(1))" WITH "S\$ : PRINT

210 R\$ = "" : FOR I = 1 TO LEN(A\$) - LEN(S\$) : IF FN S(I) THEN R\$ = R\$ + STR\$(I) + " "
220 NEXT I
230 IF LEN(R\$) = 0 THEN PRINT "DOES NOT CONTAIN "S\$
240 IF LEN(R\$) THEN PRINT "CONTAINS "S\$" LOCATED AT POSITION "R\$
250 PRINT

310 E\$(1) = "ENDS"
320 E\$(0) = "DOES NOT END"
330 PRINT E\$(FN S(LEN(A\$) - LEN(S\$) + 1))" WITH "S\$

## Batch File

::NOTE #1: This implementation might crash, or might not work properly if
::you put some of the CMD special characters (ex. %,!, etc) inside the strings.
::
::NOTE #2: The comparisons here are case-SENSITIVE.
::NOTE #3: Spaces in strings are considered.

@echo off
setlocal enabledelayedexpansion

::The main things...

set "str1=qwertyuiop"
set "str2=qwerty"
call :str2_lngth
call :matchbegin

set "str1=qweiuoiocghiioyiocxiisfguiioiuygvd"
set "str2=io"
call :str2_lngth
call :matchcontain

set "str1=blablabla"
set "str2=bbla"
call :str2_lngth
call :matchend

echo.
pause
exit /b 0
::/The main things.

::The functions...

:matchbegin
echo.
if "!str1:~0,%length%!"=="!str2!" (
echo "%str1%" begins with "%str2%".
) else (
echo "%str1%" does not begin with "%str2%".
)
goto :EOF

:matchcontain
echo.
set curr=0&set exist=0
:scanchrloop
if "!str1:~%curr%,%length%!"=="" (
if !exist!==0 echo "%str1%" does not contain "%str2%".
goto :EOF
)
if "!str1:~%curr%,%length%!"=="!str2!" (
echo "%str1%" contains "%str2%". ^(in Position %curr%^)
set exist=1
)
set /a curr+=1&goto scanchrloop

:matchend
echo.
if "!str1:~-%length%!"=="!str2!" (
echo "%str1%" ends with "%str2%".
) else (
echo "%str1%" does not end with "%str2%".
)
goto :EOF

:str2_lngth
set length=0
:loop
if "!str2:~%length%,1!"=="" goto :EOF
set /a length+=1
goto loop
::/The functions.
Output:
"qwertyuiop" begins with "qwerty".

"qweiuoiocghiioyiocxiisfguiioiuygvd" contains "io". (in Position 6)
"qweiuoiocghiioyiocxiisfguiioiuygvd" contains "io". (in Position 12)
"qweiuoiocghiioyiocxiisfguiioiuygvd" contains "io". (in Position 15)
"qweiuoiocghiioyiocxiisfguiioiuygvd" contains "io". (in Position 26)

"blablabla" does not end with "bbla".

Press any key to continue . . .

## BBC BASIC

first\$ = "The fox jumps over the dog"

FOR test% = 1 TO 3
starts% = FN_first_starts_with_second(first\$, second\$)
IF starts% PRINT """" first\$ """ starts with """ second\$ """"
ends% = FN_first_ends_with_second(first\$, second\$)
IF ends% PRINT """" first\$ """ ends with """ second\$ """"
where% = FN_first_contains_second_where(first\$, second\$)
IF where% PRINT """" first\$ """ contains """ second\$ """ at position " ; where%
howmany% = FN_first_contains_second_howmany(first\$, second\$)
IF howmany% PRINT """" first\$ """ contains """ second\$ """ " ; howmany% " time(s)"
NEXT
DATA "The", "he", "dog"
END

DEF FN_first_starts_with_second(A\$, B\$)
= B\$ = LEFT\$(A\$, LEN(B\$))

DEF FN_first_ends_with_second(A\$, B\$)
= B\$ = RIGHT\$(A\$, LEN(B\$))

DEF FN_first_contains_second_where(A\$, B\$)
= INSTR(A\$, B\$)

DEF FN_first_contains_second_howmany(A\$, B\$)
LOCAL I%, N% : I% = 0
REPEAT
I% = INSTR(A\$, B\$, I%+1)
IF I% THEN N% += 1
UNTIL I% = 0
= N%

Output:
"The fox jumps over the dog" starts with "The"
"The fox jumps over the dog" contains "The" at position 1
"The fox jumps over the dog" contains "The" 1 time(s)
"The fox jumps over the dog" contains "he" at position 2
"The fox jumps over the dog" contains "he" 2 time(s)
"The fox jumps over the dog" ends with "dog"
"The fox jumps over the dog" contains "dog" at position 24
"The fox jumps over the dog" contains "dog" 1 time(s)

## Bracmat

Bracmat does pattern matching in expressions subject:pattern and in strings @(subject:pattern). The (sub)pattern ? is a wild card.

( (sentence="I want a number such that that number will be even.")
& out\$(@(!sentence:I ?) & "sentence starts with 'I'" | "sentence does not start with 'I'")
& out\$(@(!sentence:? such ?) & "sentence contains 'such'" | "sentence does not contain 'such'")
& out\$(@(!sentence:? "even.") & "sentence ends with 'even.'" | "sentence does not end with 'even.'")
& 0:?N
& ( @(!sentence:? be (? & !N+1:?N & ~))
| out\$str\$("sentence contains " !N " occurrences of 'be'")
)
)

In the last line, Bracmat is forced by the always failing node ~ to backtrack until all occurrences of 'be' are found. Thereafter the pattern match expression fails. The interesting part is the side effect: while backtracking, the accumulator N keeps track of how many are found.

Output:
sentence starts with 'I'
sentence contains 'such'
sentence ends with 'even.'
sentence contains 3 occurrences of 'be'

## C

Case sensitive matching:

#include <string.h>
#include <stdio.h>

int startsWith(const char* container, const char* target)
{
size_t clen = strlen(container), tlen = strlen(target);
if (clen < tlen)
return 0;
return strncmp(container, target, tlen) == 0;
}

int endsWith(const char* container, const char* target)
{
size_t clen = strlen(container), tlen = strlen(target);
if (clen < tlen)
return 0;
return strncmp(container + clen - tlen, target, tlen) == 0;
}

int doesContain(const char* container, const char* target)
{
return strstr(container, target) != 0;
}

int main(void)
{
printf("Starts with Test ( Hello,Hell ) : %d\n", startsWith("Hello","Hell"));
printf("Ends with Test ( Code,ode ) : %d\n", endsWith("Code","ode"));

return 0;
}
Output:
Starts with Test ( Hello,Hell ) : 1
Ends with Test ( Code,ode ) : 1
Contains Test ( Google,msn ) : 0

Code without using string library to demonstrate how char strings are just pointers:

#include <stdio.h>

/* returns 0 if no match, 1 if matched, -1 if matched and at end */
int s_cmp(const char *a, const char *b)
{
char c1 = 0, c2 = 0;
while (c1 == c2) {
c1 = *(a++);
if ('\0' == (c2 = *(b++)))
return c1 == '\0' ? -1 : 1;
}
return 0;
}

/* returns times matched */
int s_match(const char *a, const char *b)
{
int i = 0, count = 0;
printf("matching `%s' with `%s':\n", a, b);

while (a[i] != '\0') {
switch (s_cmp(a + i, b)) {
case -1:
printf("matched: pos %d (at end)\n\n", i);
return ++count;
case 1:
printf("matched: pos %d\n", i);
++count;
break;
}
i++;
}
printf("end match\n\n");
return count;
}

int main()
{
s_match("A Short String", "ort S");
s_match("aBaBaBaBa", "aBa");
s_match("something random", "Rand");

return 0;
}
Output:
matching `A Short String' with `ort S':
matched: pos 4
end match

matching `aBaBaBaBa' with `aBa':
matched: pos 0
matched: pos 2
matched: pos 4
matched: pos 6 (at end)

matching `something random' with `Rand':
end match

## C++

#include <string>
using namespace std;

string s1="abcd";
string s2="abab";
string s3="ab";
//Beginning
s1.compare(0,s3.size(),s3)==0;
//End
s1.compare(s1.size()-s3.size(),s3.size(),s3)==0;
//Anywhere
s1.find(s2)//returns string::npos
int loc=s2.find(s3)//returns 0
loc=s2.find(s3,loc+1)//returns 2

## C#

Works with: Mono version 2.6

class Program
{
public static void Main (string[] args)
{
var value = "abcd".StartsWith("ab");
value = "abcd".EndsWith("zn"); //returns false
value = "abab".Contains("bb"); //returns false
value = "abab".Contains("ab"); //returns true
int loc = "abab".IndexOf("bb"); //returns -1
loc = "abab".IndexOf("ab"); //returns 0
loc = "abab".IndexOf("ab",loc+1); //returns 2
}
}

## Clojure

Translation of: Java
(def evals '((. "abcd" startsWith "ab")
(. "abcd" endsWith "zn")
(. "abab" contains "bb")
(. "abab" contains "ab")
(. "abab" indexOf "bb")
(let [loc (. "abab" indexOf "ab")]
(. "abab" indexOf "ab" (dec loc)))))

user> (for [i evals] [i (eval i)])
([(. "abcd" startsWith "ab") true] [(. "abcd" endsWith "zn") false] [(. "abab" contains "bb") false] [(. "abab" contains "ab") true] [(. "abab" indexOf "bb") -1] [(let [loc (. "abab" indexOf "ab")] (. "abab" indexOf "ab" (dec loc))) 0])

## CoffeeScript

This example uses string slices, but a better implementation might use indexOf for slightly better performance.

matchAt = (s, frag, i) ->
s[i...i+frag.length] == frag

startsWith = (s, frag) ->
matchAt s, frag, 0

endsWith = (s, frag) ->
matchAt s, frag, s.length - frag.length

matchLocations = (s, frag) ->
(i for i in [0..s.length - frag.length] when matchAt s, frag, i)

console.log startsWith "tacoloco", "taco" # true
console.log startsWith "taco", "tacoloco" # false
console.log startsWith "tacoloco", "talk" # false
console.log endsWith "tacoloco", "loco" # true
console.log endsWith "loco", "tacoloco" # false
console.log endsWith "tacoloco", "yoco" # false
console.log matchLocations "bababab", "bab" # [0,2,4]
console.log matchLocations "xxx", "x" # [0,1,2]

## Common Lisp

(defun starts-with-p (str1 str2)
"Determine whether `str1` starts with `str2`"
(let ((p (search str2 str1)))
(and p (= 0 p))))

(print (starts-with-p "foobar" "foo")) ; T
(print (starts-with-p "foobar" "bar")) ; NIL

(defun ends-with-p (str1 str2)
"Determine whether `str1` ends with `str2`"
(let ((p (mismatch str2 str1 :from-end T)))
(or (not p) (= 0 p))))

(print (ends-with-p "foobar" "foo")) ; NIL
(print (ends-with-p "foobar" "bar")) ; T

(defun containsp (str1 str2)
"Determine whether `str1` contains `str2`.
Instead of just returning T, return a list of starting locations
for every occurence of `str2` in `str1`"

(unless (string-equal str2 "")
(loop for p = (search str2 str1) then (search str2 str1 :start2 (1+ p))
while p
collect p)))

(print (containsp "foobar" "oba")) ; (2)
(print (containsp "ababaBa" "ba")) ; (1 3)
(print (containsp "foobar" "x")) ; NIL

## Component Pascal

BlackBox Component Builder

MODULE StringMatch;
IMPORT StdLog,Strings;
CONST
strSize = 1024;
patSize = 256;

TYPE
Matcher* = POINTER TO LIMITED RECORD
str: ARRAY strSize OF CHAR;
pat: ARRAY patSize OF CHAR;
pos: INTEGER
END;

PROCEDURE NewMatcher*(IN str: ARRAY OF CHAR): Matcher;
VAR
m: Matcher;
BEGIN
NEW(m);m.str := str\$;m.pos:= 0;
RETURN m
END NewMatcher;

PROCEDURE (m: Matcher) Match*(IN pat: ARRAY OF CHAR): INTEGER,NEW;
VAR
pos: INTEGER;
BEGIN
m.pat := pat\$;
pos := m.pos;
Strings.Find(m.str,m.pat,pos,m.pos);
RETURN m.pos
END Match;

PROCEDURE (m: Matcher) Next*(): INTEGER, NEW;
VAR
pos: INTEGER;
BEGIN
pos := m.pos + LEN(m.pat\$);
Strings.Find(m.str,m.pat,pos,m.pos);
RETURN m.pos;
END Next;

(* Some Helper functions based on Strings module *)
PROCEDURE StartsWith(IN str: ARRAY OF CHAR;IN pat: ARRAY OF CHAR): BOOLEAN;
VAR
pos: INTEGER;
BEGIN
Strings.Find(str,pat,0,pos);
RETURN pos = 0
END StartsWith;

PROCEDURE Contains(IN str: ARRAY OF CHAR;IN pat: ARRAY OF CHAR; OUT pos: INTEGER): BOOLEAN;
BEGIN
Strings.Find(str,pat,0,pos);
RETURN pos >= 0
END Contains;

PROCEDURE EndsWith(IN str: ARRAY OF CHAR;IN pat: ARRAY OF CHAR): BOOLEAN;
VAR
pos: INTEGER;
BEGIN
Strings.Find(str,pat,0,pos);
RETURN pos + LEN(pat\$) = LEN(str\$)
END EndsWith;

PROCEDURE Do*;
CONST
aStr = "abcdefghijklmnopqrstuvwxyz";
VAR
pat: ARRAY 128 OF CHAR;
res: BOOLEAN;
at: INTEGER;
m: Matcher;
BEGIN
(* StartsWith *)
pat := "abc";
StdLog.String(aStr + " startsWith " + pat + " :>");StdLog.Bool(StartsWith(aStr,pat));StdLog.Ln;
pat := "cba";
StdLog.String(aStr + " startsWith " + pat + " :>");StdLog.Bool(StartsWith(aStr,pat));StdLog.Ln;
pat := "def";
StdLog.String(aStr + " startsWith " + pat + " :>");StdLog.Bool(StartsWith(aStr,pat));StdLog.Ln;
StdLog.Ln;
(* Contains *)
pat := 'def';
StdLog.String(aStr + " contains " + pat + " :>");StdLog.Bool(Contains(aStr,pat,at));
StdLog.String(" at: ");StdLog.Int(at);StdLog.Ln;
pat := 'efd';
StdLog.String(aStr + " contains " + pat + " :>");StdLog.Bool(Contains(aStr,pat,at));
StdLog.String(" at: ");StdLog.Int(at);StdLog.Ln;
pat := 'abc';
StdLog.String(aStr + " contains " + pat + " :>");StdLog.Bool(Contains(aStr,pat,at));
StdLog.String(" at: ");StdLog.Int(at);StdLog.Ln;
pat := 'xyz';
StdLog.String(aStr + " contains " + pat + " :>");StdLog.Bool(Contains(aStr,pat,at));
StdLog.String(" at: ");StdLog.Int(at);StdLog.Ln;
StdLog.Ln;
(* EndsWith *)
pat := 'xyz';
StdLog.String(aStr + " endsWith " + pat + " :>");StdLog.Bool(EndsWith(aStr,pat));StdLog.Ln;
pat := 'zyx';
StdLog.String(aStr + " endsWith " + pat + " :>");StdLog.Bool(EndsWith(aStr,pat));StdLog.Ln;
pat := 'abc';
StdLog.String(aStr + " endsWith " + pat + " :>");StdLog.Bool(EndsWith(aStr,pat));StdLog.Ln;
pat:= 'def';
StdLog.String(aStr + " endsWith " + pat + " :>");StdLog.Bool(EndsWith(aStr,pat));StdLog.Ln;
StdLog.Ln;

m := NewMatcher("abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz");
StdLog.String("Matching 'abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz' against 'abc':> ");
StdLog.Ln;
StdLog.String("Match at: ");StdLog.Int(m.Match("abc"));StdLog.Ln;
StdLog.String("Match at: ");StdLog.Int(m.Next());StdLog.Ln;
StdLog.String("Match at: ");StdLog.Int(m.Next());StdLog.Ln
END Do;
END StringMatch.

Execute: ^Q StringMatching.Do

Output:
abcdefghijklmnopqrstuvwxyz startsWith abc :> \$TRUE
abcdefghijklmnopqrstuvwxyz startsWith cba :> \$FALSE
abcdefghijklmnopqrstuvwxyz startsWith def :> \$FALSE

abcdefghijklmnopqrstuvwxyz contains def :> \$TRUE at:  3
abcdefghijklmnopqrstuvwxyz contains efd :> \$FALSE at:  -1
abcdefghijklmnopqrstuvwxyz contains abc :> \$TRUE at:  0
abcdefghijklmnopqrstuvwxyz contains xyz :> \$TRUE at:  23

abcdefghijklmnopqrstuvwxyz endsWith xyz :> \$TRUE
abcdefghijklmnopqrstuvwxyz endsWith zyx :> \$FALSE
abcdefghijklmnopqrstuvwxyz endsWith abc :> \$FALSE
abcdefghijklmnopqrstuvwxyz endsWith def :> \$FALSE

Matching 'abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz' against 'abc':>
Match at:  0
Match at:  26
Match at:  -1

## D

void main() {
import std.stdio;
import std.algorithm: startsWith, endsWith, find, countUntil;

"abcd".startsWith("ab").writeln; // true
"abcd".endsWith("zn").writeln; // false
"abab".find("bb").writeln; // empty array (no match)
"abcd".find("bc").writeln; // "bcd" (substring start
// at match)
"abab".countUntil("bb").writeln; // -1 (no match)
"abab".countUntil("ba").writeln; // 1 (index of 1st match)

// std.algorithm.startsWith also works on arrays and ranges:
[1, 2, 3].countUntil(3).writeln; // 2
[1, 2, 3].countUntil([2, 3]).writeln; // 1
}
Output:
true
false

bcd
-1
1
2
1

## DCL

\$ first_string = p1
\$ length_of_first_string = f\$length( first_string )
\$ second_string = p2
\$ length_of_second_string = f\$length( second_string )
\$ offset = f\$locate( second_string, first_string )
\$ if offset .eq. 0
\$ then
\$ write sys\$output "first string starts with second string"
\$ else
\$ endif
\$ if offset .ne. length_of_first_string
\$ then
\$ write sys\$output "first string contains the second string at location ", offset
\$ else
\$ write sys\$output "first string does not contain the second string at any location"
\$ endif
\$ temp = f\$extract( length_of_first_string - length_of_second_string, length_of_second_string, first_string )
\$ if second_string .eqs. temp
\$ then
\$ write sys\$output "first string ends with the second string"
\$ else
\$ write sys\$output "first string does not end with the second string"
\$ endif
Output:
\$ @string_matching efabcdef ef
first string starts with second string
first string contains the second string at location 0
first string ends with the second string
\$ @string_matching efabcdef ab
first string contains the second string at location 2
first string does not end with the second string
\$ @string_matching efabcdef def
first string contains the second string at location 5
first string ends with the second string
\$ @string_matching efabcdef defx
first string does not contain the second string at any location
first string does not end with the second string

## Delphi

program CharacterMatching;

{\$APPTYPE CONSOLE}

uses StrUtils;

begin
WriteLn(AnsiStartsText('ab', 'abcd')); // True
WriteLn(AnsiEndsText('zn', 'abcd')); // False
WriteLn(AnsiContainsText('abcd', 'bb')); // False
Writeln(AnsiContainsText('abcd', 'ab')); // True
WriteLn(Pos('ab', 'abcd')); // 1
end.

## Dyalect

var value = "abcd".startsWith("ab")
value = "abcd".endsWith("zn") //returns false
value = "abab".contains("bb") //returns false
value = "abab".contains("ab") //returns true
var loc = "abab".indexOf("bb") //returns -1
loc = "abab".indexOf("ab") //returns 0

## E

def f(string1, string2) {
println(string1.startsWith(string2))

var index := 0
while ((index := string1.startOf(string2, index)) != -1) {
println(`at \$index`)
index += 1
}

println(string1.endsWith(string2))
}

## EchoLisp

(string-suffix? "nette" "Antoinette") → #t
(string-prefix? "Simon" "Simon & Garfunkel") → #t

(string-match "Antoinette" "net") → #t ;; contains
(string-index "net" "Antoinette")5 ;; substring location

## Elena

ELENA 4.x :

import extensions;

public program()
{
var s := "abcd";

console.printLine(s," starts with ab: ",s.startingWith:"ab");
console.printLine(s," starts with cd: ",s.startingWith:"cd");

console.printLine(s," ends with ab: ",s.endingWith:"ab");
console.printLine(s," ends with cd: ",s.endingWith:"cd");

console.printLine(s," contains ab: ",s.containing:"ab");
console.printLine(s," contains bc: ",s.containing:"bc");
console.printLine(s," contains cd: ",s.containing:"cd");
console.printLine(s," contains az: ",s.containing:"az");

console.printLine(s," index of az: ",s.indexOf(0, "az"));
console.printLine(s," index of cd: ",s.indexOf(0, "cd"));
console.printLine(s," index of bc: ",s.indexOf(0, "bc"));
console.printLine(s," index of ab: ",s.indexOf(0, "ab"));

}

## Elixir

The String module has functions that cover the base requirements.

s1 = "abcd"
s3 = "ab"

String.starts_with?(s1, s3)
# => true
String.starts_with?(s2, s3)
# => false

String.contains?(s1, s3)
# => true
String.contains?(s2, s3)
# => true

String.ends_with?(s1, s3)
# => false
String.ends_with?(s2, s3)
# => true

# Optional requirements:
Regex.run(~r/#{s3}/, s1, return: :index)
# => [{0, 2}]
Regex.run(~r/#{s3}/, s2, return: :index)
# => [{2, 2}]

Regex.scan(~r/#{s3}/, "abcabc", return: :index)
# => [[{0, 2}], [{3, 2}]]

## Emacs Lisp

(defun match (word str)
(progn

(setq regex (format "^%s.*\$" word) )

(if (string-match regex str)
(insert (format "%s found in beginning of: %s\n" word str) )
(insert (format "%s not found in beginning of: %s\n" word str) ))

(setq pos (string-match word str) )

(if pos
(insert (format "%s found at position %d in: %s\n" word pos str) )

(setq regex (format "^.*%s\$" word) )

(if (string-match regex str)
(insert (format "%s found in end of: %s\n" word str) )
(insert (format "%s not found in end of: %s\n" word str) ))))

(setq string "before center after")

(progn
(match "center" string)
(insert "\n")
(match "before" string)
(insert "\n")
(match "after" string) )

Output:

center found at position 7 in: before center after

before found in beginning of: before center after
before found at position 0 in: before center after

after found at position 14 in: before center after
after found in end of: before center after

## Erlang

-module(character_matching).
-export([starts_with/2,ends_with/2,contains/2]).

%% Both starts_with and ends_with are mappings to 'lists:prefix/2' and
%% 'lists:suffix/2', respectively.

starts_with(S1,S2) ->
lists:prefix(S2,S1).

ends_with(S1,S2) ->
lists:suffix(S2,S1).

contains(S1,S2) ->
contains(S1,S2,1,[]).

%% While S1 is at least as long as S2 we check if S2 is its prefix,
%% storing the result if it is. When S1 is shorter than S2, we return
%% the result. NB: this will work on any arbitrary list in erlang
%% since it makes no distinction between strings and lists.
contains([_|T]=S1,S2,N,Acc) when length(S2) =< length(S1) ->
case starts_with(S1,S2) of
true ->
contains(T,S2,N+1,[N|Acc]);
false ->
contains(T,S2,N+1,Acc)
end;
contains(_S1,_S2,_N,Acc) ->
Acc.

## Euphoria

sequence first, second
integer x

first = "qwertyuiop"

-- Determining if the first string starts with second string
second = "qwerty"
if match(second, first) = 1 then
printf(1, "'%s' starts with '%s'\n", {first, second})
else
end if

-- Determining if the first string contains the second string at any location
-- Print the location of the match for part 2
second = "wert"
x = match(second, first)
if x then
printf(1, "'%s' contains '%s' at position %d\n", {first, second, x})
else
printf(1, "'%s' does not contain '%s'\n", {first, second})
end if

-- Determining if the first string ends with the second string
second = "uio"
if length(second)<=length(first) and match_from(second, first, length(first)-length(second)+1) then
printf(1, "'%s' ends with '%s'\n", {first, second})
else
printf(1, "'%s' does not end with '%s'\n", {first, second})
end if
Output:
'qwertyuiop' starts with 'qwerty'
'qwertyuiop' contains 'wert' at position 2
'qwertyuiop' does not end with 'uio'

## F#

[<EntryPoint>]
let main args =

let text = "一二三四五六七八九十"
let starts = "一二"
let ends = "九十"
let contains = "五六"
let notContains = "百"

printfn "text = %A" text
printfn "starts with %A: %A" starts (text.StartsWith(starts))
printfn "starts with %A: %A" ends (text.StartsWith(ends))
printfn "ends with %A: %A" ends (text.EndsWith(ends))
printfn "ends with %A: %A" starts (text.EndsWith(starts))
printfn "contains %A: %A" contains (text.Contains(contains))
printfn "contains %A: %A" notContains (text.Contains(notContains))
printfn "substring %A begins at position %d (zero-based)" contains (text.IndexOf(contains))
let text2 = text + text
printfn "text = %A" text2
Seq.unfold (fun (n : int) ->
let idx = text2.IndexOf(contains, n)
if idx < 0 then None else Some (idx, idx+1)) 0
|> Seq.iter (printfn "substring %A begins at position %d (zero-based)" contains)
0
Output:
text = "一二三四五六七八九十"
starts with "一二": true
starts with "九十": false
ends with "九十": true
ends with "一二": false
contains "五六": true
contains "百": false
substring "五六" begins at position 4 (zero-based)
text = "一二三四五六七八九十一二三四五六七八九十"
substring "五六" begins at position 4 (zero-based)
substring "五六" begins at position 14 (zero-based)

## Factor

Does cheesecake contain sec at any location?

"sec" "cheesecake" subseq?   ! t

Does cheesecake end with cake?

"cheesecake" "cake" tail?   ! t

Where in cheesecake is the leftmost sec?

"sec" "cheesecake" subseq-start   ! 4

Where in Mississippi are all occurrences of iss?

USE: regexp
"Mississippi" "iss" <regexp> all-matching-slices [ from>> ] map  ! { 1 4 }

## Falcon

'VBA/Python programmer's approach. I'm just a junior Falconeer but this code seems falconic

/* created by Aykayayciti Earl Lamont Montgomery
April 9th, 2018 */

s1 = "Naig Ialocin Olracnaig"
s2 = "Naig"

var = s1.startsWith(s2) ? s1 + " starts with " + s2 : s1 + " does not start with " + s2
> var

s2 = "loc"
var = s2 in s1 ? @ "\$s1 contains \$s2" : @ "\$s1 does not contain \$s2"
> var

> s1.endsWith(s2) ? @ "s1 ends with \$s2" : @ "\$s1 does not end with \$s2"

Output:
Naig Ialocin Olracnaig starts with Naig
Naig Ialocin Olracnaig contains loc
Naig Ialocin Olracnaig does not end with loc
[Finished in 1.2s]

## Fantom

Fantom provides several self-explanatory string-matching methods:

• startsWith
• endsWith
• contains
• index (takes an optional index, for the start position)
• indexIgnoreCase (like above, ignoring case for ASCII characters)
• indexr (start search from end of string, with an optional index)
• indexrIgnoreCase (like above, ignoring case for ASCII characters)

class Main
{
public static Void main ()
{
string := "Fantom Language"
echo ("String is: " + string)
echo ("does string contain 'age'? " + string.contains("age"))
echo ("does string contain 'page'? " + string.contains("page"))
echo ("does string end with 'Fantom'? " + string.endsWith("Fantom"))
echo ("does string end with 'Language'? " + string.endsWith("Language"))

echo ("Location of 'age' is: " + string.index("age"))
posn := string.index ("an")
echo ("First location of 'an' is: " + posn)
posn = string.index ("an", posn+1)
echo ("Second location of 'an' is: " + posn)
posn = string.index ("an", posn+1)
if (posn == null) echo ("No third location of 'an'")
}
}

Output:
String is: Fantom Language
does string contain 'age'? true
does string contain 'page'? false
does string end with 'Fantom'? false
does string end with 'Language'? true
Location of 'age' is: 12
First location of 'an' is: 1
Second location of 'an' is: 8
No third location of 'an'

## FBSL

#APPTYPE CONSOLE

DIM s = "roko, mat jane do"

IF LEFT(s, 4) = "roko" THEN PRINT STRENC(s), " starts with ", STRENC("roko")
IF INSTR(s, "mat") THEN PRINT STRENC(s), " contains ", STRENC("mat"), " at ", INSTR
IF RIGHT(s, 2) = "do" THEN PRINT STRENC(s), " ends with ", STRENC("do")
PRINT STRENC(s), " contains ", STRENC("o"), " at the following locations:", InstrEx(s, "o")

PAUSE

SUB InstrEx(mane, match)
INSTR = 0
WHILE INSTR(mane, match, INSTR + 1): PRINT " ", INSTR;: WEND
END SUB

Output:
"roko, mat jane do" starts with "roko"
"roko, mat jane do" contains "mat" at 7
"roko, mat jane do" ends with "do"
"roko, mat jane do" contains "o" at the following locations: 2 4 17

Press any key to continue...

## Forth

: starts-with ( a l a2 l2 -- ? )
tuck 2>r min 2r> compare 0= ;
: ends-with ( a l a2 l2 -- ? )
tuck 2>r negate over + 0 max /string 2r> compare 0= ;
\ use SEARCH ( a l a2 l2 -- a3 l3 ? ) for contains

## Fortran

Fortran does not offer a string type, but since F77 it has been possible to use a CHARACTER variable, of some specified size, whose size may be accessed via the LEN function. When passed as a parameter, a secret additional parameter specifies its size and so string-like usage is possible. Character matching is case sensitive, and, trailing spaces are ignored so that "xx" and "xx " are deemed equal. The function INDEX(text,target) determines the first index in text where target matches, and returns zero if there is no such match. Unfortunately, the function does not allow the specification of a starting position for a search, as to find any second and further matches. One must specify something like INDEX(text(5:),target) to start with position five, and then deal with the resulting offsets needed to relate the result to positions within the parameter. On the other hand, since there is no "length" conjoined to the text such substring selections can be made without copying the text to a work area, unlike the copy(text,start,stop) equivalent of Pascal for example. Some Fortran compilers do offer a starting point, and also an option to search backwards from the end, but these facilities are not guaranteed. Similarly, INDEX is only made available for CHARACTER searching, even though it could easily be generalised to other types.

A second problem is presented by the possibility that a logical expression such as L.LT.0 .OR. etc. will always or might possibly or in certain constructions but not others be fully evaluated, which is to say that the etc will be evaluated even though L < 0 is true so that the result is determined. And in this case, evaluating the etc will cause trouble because the indexing won't work! To be safe, therefore, a rather lame two-stage test is required - though optimising compilers might well shift code around anyway.

In the case of STARTS, these annoyances can be left to the INDEX function rather than comparing the start of A against B. At the cost of it searching the whole of A if B is not at the start. Otherwise, it would be the mirror of ENDS.

SUBROUTINE STARTS(A,B) !Text A starts with text B?
CHARACTER*(*) A,B
IF (INDEX(A,B).EQ.1) THEN !Searches A to find B.
WRITE (6,*) ">",A,"< starts with >",B,"<"
ELSE
END IF
END SUBROUTINE STARTS

SUBROUTINE HAS(A,B) !Text B appears somewhere in text A?
CHARACTER*(*) A,B
INTEGER L
L = INDEX(A,B) !The first position in A where B matches.
IF (L.LE.0) THEN
WRITE (6,*) ">",A,"< does not contain >",B,"<"
ELSE
WRITE (6,*) ">",A,"< contains a >",B,"<, offset",L
END IF
END SUBROUTINE HAS

SUBROUTINE ENDS(A,B) !Text A ends with text B.
CHARACTER*(*) A,B
INTEGER L
L = LEN(A) - LEN(B) !Find the tail end of A that B might match.
IF (L.LT.0) THEN !Dare not use an OR, because of full evaluation risks.
WRITE (6,*) ">",A,"< is too short to end with >",B,"<" !Might as well have a special message.
ELSE IF (A(L + 1:L + LEN(B)).NE.B) THEN !Otherwise, it is safe to look.
WRITE (6,*) ">",A,"< does not end with >",B,"<"
ELSE
WRITE (6,*) ">",A,"< ends with >",B,"<"
END IF
END SUBROUTINE ENDS

CALL STARTS("This","is")
CALL STARTS("Theory","The")
CALL HAS("Bananas","an")
CALL ENDS("Banana","an")
CALL ENDS("Banana","na")
CALL ENDS("Brief","Much longer")
END

Output: text strings are bounded by >etc.< in case of leading or trailing spaces.

>Theory< starts with >The<
>Bananas< contains a >an<, offset           2
>Banana< does not end with >an<
>Banana< ends with >na<
>Brief< is too short to end with >Much longer<

Similar program using modern Fortran style

!-----------------------------------------------------------------------
!Main program string_matching
!-----------------------------------------------------------------------
program string_matching
implicit none
character(len=*), parameter :: fmt= '(I0)'
write(*,fmt) starts("this","is")
write(*,fmt) starts("theory","the")
write(*,fmt) has("bananas","an")
write(*,fmt) ends("banana","an")
write(*,fmt) ends("banana","na")
write(*,fmt) ends("brief","much longer")

contains
! Determining if the first string starts with second string
implicit none
character(len=*), intent(in) :: string1
character(len=*), intent(in) :: string2
if(len(string2)>len(string1)) return
end function starts
! Determining if the first string contains the second string at any location
implicit none
character(len=*), intent(in) :: string1
character(len=*), intent(in) :: string2
character(len=:),allocatable :: temp
character(len=*), parameter :: fmt= '(A6,X,I0)'
if(len(string2)>len(string1)) return
! Print the location of the match for part 2
! Handle multiple occurrences of a string for part 2.
! deallocate(temp)
temp = string1(add+1:) ! auto reallocation
enddo
end function has
! Determining if the first string ends with the second string
implicit none
character(len=*), intent(in) :: string1
character(len=*), intent(in) :: string2
if(len(string2)>len(string1)) return
end function ends
end program string_matching

Output: false = 0, true = 1 ( + multiple occurrences if applicable)

0
1
at  2
at  4
1
0
1
0

In recent standards of Fortran strings as intrinsic first-class type and many intrinsic procedures for strings manipulation have been added.

## FreeBASIC

' FB 1.05.0 Win64

Dim As String s1 = "abracadabra"
Dim As String s2 = "abra"
Print "First string  : "; s1
Print "Second string : "; s2
Print
Print "First string begins with second string : "; CBool(s2 = Left(s1, Len(s2)))
Dim As Integer i1 = Instr(s1, s2)
Dim As Integer i2
Print "First string contains second string  : ";
If i1 Then
Print "at index"; i1;
i2 = Instr(i1 + Len(s2), s1, s2)
If i2 Then
Print " and at index"; i2
Else
Print
End If
Else
Print "false";
End If
Print "First string ends with second string  : "; CBool(s2 = Right(s1, Len(s2)))
Print
Print "Press any key to quit"
Sleep
Output:
Second string : abra

First string begins with second string : true
First string contains second string    : at index 1 and at index 8
First string ends with second string   : true

## Gambas

Public Sub Main()
Dim sString1 As String = "Hello world"
Dim sString2 As String = "Hello"

Print sString1 Begins Left(sString2, 5) 'Determine if the first string starts with second string
If InStr(sString1, sString2) Then Print "True" Else Print "False" 'Determine if the first string contains the second string at any location
Print sString1 Ends Left(sString2, 5) 'Determine if the first string ends with the second string

End

Output:

True
True
False

## GML

Translation of: BASIC
#define charMatch
{
first = "qwertyuiop";

// Determining if the first string starts with second string
second = "qwerty";
if (string_pos(second, first) > 0) {
show_message("'" + first + "' starts with '" + second + "'");
} else {
show_message("'" + first + "' does not start with '" + second + "'");
}

second = "wert"
// Determining if the first string contains the second string at any location
// Print the location of the match for part 2
if (string_pos(second, first) > 0) {
show_message("'" + first + "' contains '" + second + "' at position " + string(x));
} else {
show_message("'" + first + "' does not contain '" + second + "'");
}
// Handle multiple occurrences of a string for part 2.
x = string_count(second, first);
show_message("'" + first + "' contains " + string(x) + " instances of '" + second + "'");

// Determining if the first string ends with the second string
second = "random garbage"
temp = string_copy(first,
(string_length(first) - string_length(second)) + 1,
string_length(second));
if (temp == second) {
show_message("'" + first + "' ends with '" + second + "'");
} else {
show_message("'" + first + "' does not end with '" + second + "'");
}
}
Output:
(in message boxes, 1 per line):
'qwertyuiop' starts with 'qwerty'
'qwertyuiop' contains 'wert' at position 0
'qwertyuiop' contains 1 instances of 'wert'
'qwertyuiop' does not end with 'random garbage'

## Go

package main

import (
"fmt"
"strings"
)

func match(first, second string) {
fmt.Printf("1. %s starts with %s: %t\n",
first, second, strings.HasPrefix(first, second))
i := strings.Index(first, second)
fmt.Printf("2. %s contains %s: %t,\n", first, second, i >= 0)
if i >= 0 {
fmt.Printf("2.1. at location %d,\n", i)
for start := i+1;; {
if i = strings.Index(first[start:], second); i < 0 {
break
}
fmt.Printf("2.2. at location %d,\n", start+i)
start += i+1
}
fmt.Println("2.2. and that's all")
}
fmt.Printf("3. %s ends with %s: %t\n",
first, second, strings.HasSuffix(first, second))
}

func main() {
}
Output:
1. abracadabra starts with abr: true
2.1. at location 0,
2.2. at location 7,
2.2. and that's all
3. abracadabra ends with abr: false

## Groovy

Examples:

assert "abcd".startsWith("ab")
assert ! "abcd".startsWith("zn")
assert "abcd".endsWith("cd")
assert ! "abcd".endsWith("zn")
assert "abab".contains("ba")
assert ! "abab".contains("bb")

assert "abab".indexOf("ab") == 0

def indicesOf = { string, substring ->
if (!string) { return [] }
def indices = [-1]
while (true) {
indices << string.indexOf(substring, indices.last()+1)
if (indices.last() == -1) break
}
indices[1..<(indices.size()-1)]
}
assert indicesOf("abab", "ab") == [0, 2]
assert indicesOf("abab", "ba") == [1]
assert indicesOf("abab", "xy") == []

All assertions pass, so there is no output.

> import Data.List
> "abc" `isPrefixOf` "abcdefg"
True
> "efg" `isSuffixOf` "abcdefg"
True
> "bcd" `isInfixOf` "abcdefg"
True
> "abc" `isInfixOf` "abcdefg" -- Prefixes and suffixes are also infixes
True
> let infixes a b = findIndices (isPrefixOf a) \$ tails b
> infixes "ab" "abcdefabqqab"
[0,6,10]

## Icon and Unicon

procedure main()

write("Matching s2 :=",image(s2 := "ab")," within s1:= ",image(s1 := "abcdabab"))

write("Test #1 beginning ",if match(s2,s1) then "matches " else "failed")
writes("Test #2 all matches at positions [")
every writes(" ",find(s2,s1)|"]\n")
write("Test #3 ending ", if s1[0-:*s2] == s2 then "matches" else "fails")

end
Output:
Matching s2 :="ab" within s1:= "abcdabab"
Test #1 beginning matches
Test #2 all matches at positions [ 1 5 7 ]
Test #3 ending matches

## J

startswith=: ] -: ({.~ #)
contains=: +./@:E.~
endswith=: ] -: ({.~ [email protected]#)

Example use:

'abcd' startswith 'ab'
1
'abcd' startswith 'cd'
0
'abcd' endswith 'ab'
0
'abcd' endswith 'cd'
1
'abcd' contains 'bb'
0
'abcd' contains 'ab'
1
'abcd' contains 'bc'
1
'abab' contains 'ab'
1
'abab' [email protected]~ 'ab' NB. find starting indicies
0 2

Note that these verbs contain no constraints restricting them to sequences of characters and so also apply to arrays of type other than character:

0 1 2 3 startswith 0 1               NB. integer
1
4.2 5.1 1.3 9 3 contains 1.3 4.2 NB. floating point
0
4.2 5.1 1.3 4.2 9 3 contains 1.3 4.2
1

## Java

"abcd".startsWith("ab") //returns true
"abcd".endsWith("zn") //returns false
"abab".contains("bb") //returns false
"abab".contains("ab") //returns true
int loc = "abab".indexOf("bb") //returns -1
loc = "abab".indexOf("ab") //returns 0
loc = "abab".indexOf("ab",loc+1) //returns 2

// -----------------------------------------------------------// public class JavaApplication6 {

public static void main(String[] args) {
String strOne = "complexity";
String strTwo = "udacity";
//
stringMatch(strOne, strTwo);
}
public static void stringMatch(String one, String two) {
boolean match = false;
if (one.charAt(0) == two.charAt(0)) {
System.out.println(match = true);   // returns true
} else {
System.out.println(match);       // returns false
}
for (int i = 0; i < two.length(); i++) {
int temp = i;
for (int x = 0; x < one.length(); x++) {
if (two.charAt(temp) == one.charAt(x)) {
System.out.println(match = true);    //returns true
i = two.length();
}
}
}
int num1 = one.length() - 1;
int num2 = two.length() - 1;
if (one.charAt(num1) == two.charAt(num2)) {
System.out.println(match = true);
} else {
System.out.println(match = false);
}
}

}

## JavaScript

var stringA = "tacoloco"
, stringB = "co"
, q1, q2, q2multi, m
, q2matches = []

// stringA starts with stringB
q1 = stringA.substring(0, stringB.length) == stringB

// stringA contains stringB
q2 = stringA.indexOf(stringB)

// multiple matches
q2multi = new RegExp(stringB,'g')

while(m = q2multi.exec(stringA)){
q2matches.push(m.index)
}

// stringA ends with stringB
q3 = stringA.substr(-stringB.length) == stringB

console.log("1: Does '"+stringA+"' start with '"+stringB+"'? " + ( q1 ? "Yes." : "No."))
console.log("2: Is '"+stringB+"' contained in '"+stringA+"'? " + (~q2 ? "Yes, at index "+q2+"." : "No."))
if (~q2 && q2matches.length > 1){
console.log(" In fact, it happens "+q2matches.length+" times within '"+stringA+"', at index"+(q2matches.length > 1 ? "es" : "")+" "+q2matches.join(', ')+".")
}
console.log("3: Does '"+stringA+"' end with '"+stringB+"'? " + ( q3 ? "Yes." : "No."))
Output:
2: Is 'co' contained in 'tacoloco'? Yes, at index 2.
In fact, it happens 2 times within 'tacoloco', at indexes 2, 6.
3: Does 'tacoloco' end with 'co'? Yes.

## jq

Using the builtins of jq 1.4 and later:

# startswith/1 is boolean:
"abc" | startswith("ab")
#=> true
# index/1 returns the index or null,
# so the jq test "if index(_) then ...." can be used
# without any type conversion.

"abcd" | index( "bc")
#=> 1
# endswith/1 is also boolean:
"abc" | endswith("bc")
#=> true

Using the regex functions available in jq 1.5:

"abc" | test( "^ab")

"abcd" | test("bc")

"abcd" | test("cd\$")

### Multiple Occurrences

To determine all the indices of one string in another:

# In jq 1.4 or later:
jq -n '"abcdabcd" | indices("bc")'
[
1,
5
]

In jq 1.5, the regex function match/1 can also be used:

\$ jq -n '"abcdabcd" | match("bc"; "g") | .offset'
1
5

## Julia

startswith("abcd","ab") #returns true
findfirst("ab", "abcd") #returns 1:2, indices range where string was found
endswith("abcd","zn") #returns false
match(r"ab","abcd") != Nothing #returns true where 1st arg is regex string
for r in eachmatch(r"ab","abab")
println(r.offset)
end #returns 1, then 3 matching the two starting indices where the substring was found

## K

startswith: {:[0<#p:_ss[x;y];~*p;0]}
endswith: {0=(-#y)+(#x)-*_ss[x;y]}
contains: {0<#_ss[x;y]}

Example:

startswith["abcd";"ab"]
1
startswith["abcd";"bc"]
0
endswith["abcd";"cd"]
1
endswith["abcd";"bc"]
0
contains["abcdef";"cde"]
1
contains["abcdef";"bdef"]
0
_ss["abcdabceabc";"abc"] / location of matches
0 4 8

## Kotlin

// version 1.0.6

fun main(args: Array<String>) {
val s2 = "abra"
println("\$s1 begins with \$s2 : \${s1.startsWith(s2)}")
println("\$s1 ends with \$s2  : \${s1.endsWith(s2)}")
val b = s2 in s1
print("\$s1 contains \$s2  : \$b")
if (b) println(" at locations \${s1.indexOf(s2) + 1} and \${s1.lastIndexOf(s2) + 1}")
else println()
}
Output:
abracadabra begins with abra : true
abracadabra ends with abra   : true
abracadabra contains abra    : true at locations 1 and 8

## LabVIEW

These images solve the task's requirements in order.
This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.

## Lasso

local(
a = 'a quick brown peanut jumped over a quick brown fox',
b = 'a quick brown'
)

//Determining if the first string starts with second string
#a->beginswith(#b) // true

//Determining if the first string contains the second string at any location
#a >> #b // true
#a->contains(#b) // true

//Determining if the first string ends with the second string
#a->endswith(#b) // false

## Lang5

: 2array  2 compress ; : bi*  '_ set dip _ execute ;  : [email protected]  dup bi* ;
: comb "" split ;  : concat "" join ;  : dip swap '_ set execute _ ;
: first 0 extract swap drop ; : flip comb reverse concat ;

: contains
swap 'comb [email protected] length do # create a matrix.
1 - "dup 1 1 compress rotate" dip dup 0 == if break then
loop drop length compress swap drop
"length 1 -" [email protected] rot 0 0 "'2array dip" '2array bi* swap 2array slice reverse
: concat.(*) concat ;
'concat "'concat. apply" bi* eq 1 1 compress index collapse
length if expand drop else drop 0 then ; # r: position.
: end-with 'flip [email protected] start-with ;
: start-with 'comb [email protected] length rot swap iota subscript 'concat [email protected] eq ;

"rosettacode" "rosetta" start-with . # 1
"rosettacode" "taco" contains . # 5
"rosettacode" "ocat" contains . # 0
"rosettacode" "edoc" end-with . # 0
"rosettacode" "code" contains . # 7

## Liberty BASIC

'1---Determining if the first string starts with second string
st1\$="first string"
st2\$="first"
if left\$(st1\$,len(st2\$))=st2\$ then
print "First string starts with second string."
end if

'2---Determining if the first string contains the second string at any location
'2.1---Print the location of the match for part 2
st1\$="Mississippi"
st2\$="i"
if instr(st1\$,st2\$) then
print "First string contains second string."
print "Second string is at location ";instr(st1\$,st2\$)
end if

'2.2---Handle multiple occurrences of a string for part 2.
pos=instr(st1\$,st2\$)
while pos
count=count+1
pos=instr(st1\$,st2\$,pos+len(st2\$))
wend
print "Number of times second string appears in first string is ";count

'3---Determining if the first string ends with the second string
st1\$="first string"
st2\$="string"
if right\$(st1\$,len(st2\$))=st2\$ then
print "First string ends with second string."
end if

## Lingo

a = "Hello world!"
b = "Hello"

-- Determining if the first string starts with second string
put a starts b
-- 1

-- Determining if the first string contains the second string at any location
put a contains b
-- 1

-- Determining if the first string ends with the second string
put a.char[a.length-b.length+1..a.length] = b
-- 0

b = "world!"
put a.char[a.length-b.length+1..a.length] = b
-- 1

-- Print the location of the match for part 2
put offset(b, a)
-- 7

## Logo

to starts.with? :sub :thing
if empty? :sub [output "true]
if empty? :thing [output "false]
if not equal? first :sub first :thing [output "false]
output starts.with? butfirst :sub butfirst :thing
end

to ends.with? :sub :thing
if empty? :sub [output "true]
if empty? :thing [output "false]
if not equal? last :sub last :thing [output "false]
output ends.with? butlast :sub butlast :thing
end

show starts.with? "dog "doghouse  ; true
show ends.with? "house "doghouse  ; true
show substring? "gho "doghouse  ; true (built-in)

## Lua

s1 = "string"
s2 = "str"
s3 = "ing"
s4 = "xyz"

print( "s1 starts with s2: ", string.find( s1, s2 ) == 1 )
print( "s1 starts with s3: ", string.find( s1, s3 ) == 1, "\n" )

print( "s1 contains s3: ", string.find( s1, s3 ) ~= nil )
print( "s1 contains s3: ", string.find( s1, s4 ) ~= nil, "\n" )

print( "s1 ends with s2: ", select( 2, string.find( s1, s2 ) ) == string.len( s1 ) )
print( "s1 ends with s3: ", select( 2, string.find( s1, s3 ) ) == string.len( s1 ) )

## M2000 Interpreter

Module StringMatch {
A\$="Hello World"
Print A\$ ~ "Hello*"
Print A\$ ~ "*llo*"
p=Instr(A\$, "llo")
Print p=3
\\ Handle multiple occurance for "o"
p=Instr(A\$, "o")
While p > 0 {
Print "position:";p;{ for "o"}
p=Instr(A\$, "o", p+1)
}
Print A\$ ~ "*orld"
}
{{out}}
<pre>
True
True
True
position:5 for "o"
position:8 for "o"
True
</pre>
StringMatch

## Maple

These facilities are all to be found in the StringTools package in Maple.

> with( StringTools ): # bind package exports at the top-level
> s := "dzrIemaWWIMidXYZwGiqkOOn":
> s[1..4]; # pick a prefix
"dzrI"

> IsPrefix( s[ 1 .. 4 ], s ); # check it
true

> s[ -4 .. -1 ]; # pick a suffix
"kOOn"

> IsSuffix( s[ -4 .. -1 ], s ); # check it
true

> p := Search( "XYZ", s ); # find a substring
p := 14

> s[ p .. p + 2 ]; # check
"XYZ"

> SearchAll( [ "WWI", "XYZ" ], s ); # search for multiple patterns
[8, 1], [14, 2]

> to 3 do s := cat( s, s ) end: length( s ); # build a longer string by repeated doubling
192

> p := SearchAll( "XYZ", s ); # find all occurrences
p := 14, 38, 62, 86, 110, 134, 158, 182

> {seq}( s[ i .. i + 2 ], i = p ); # check them
{"XYZ"}

The StringTools package also contains facilities for regular expression matching, but for fixed string patterns, the Search and SearchAll tools are much faster.

## Mathematica

StartWith[x_, y_] := MemberQ[Flatten[StringPosition[x, y]], 1]
EndWith[x_, y_] := MemberQ[Flatten[StringPosition[x, y]], StringLength[x]]
StartWith["XYZaaabXYZaaaaXYZXYZ", "XYZ"]
EndWith["XYZaaabXYZaaaaXYZXYZ", "XYZ"]
StringPosition["XYZaaabXYZaaaaXYZXYZ", "XYZ"]
Output:
True
True
{{1,3},{8,10},{15,17},{18,20}}

## MATLAB / Octave

% 1. Determining if the first string starts with second string
strcmp(str1,str2,length(str2))
% 2. Determining if the first string contains the second string at any location
~isempty(strfind(s1,s2))
% 3. Determining if the first string ends with the second string
( (length(str1)>=length(str2)) && strcmp(str1(end+[1-length(str2):0]),str2) )

% 1. Print the location of the match for part 2
disp(strfind(s1,s2))
% 2. Handle multiple occurrences of a string for part 2.
ix = strfind(s1,s2); % ix is a vector containing the starting positions of s2 within s1

## NetRexx

/* NetRexx */
options replace format comments java crossref savelog symbols

lipsum = ''
x_ = 0
x_ = x_ + 1; lipsum[0] = x_; lipsum[x_] = 'Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua.'
x_ = x_ + 1; lipsum[0] = x_; lipsum[x_] = lipsum[1].reverse

srch = ''
srch[1] = 'Lorem ipsum dolor sit amet'
srch[3] = 'dolore magna aliqua.'

loop j_ = 1 to lipsum[0]
x1 = lipsum[j_].pos(srch[1])
x2 = lipsum[j_].pos(srch[2])
x3 = lipsum[j_].lastpos(srch[3])

report(x1 = 1, lipsum[j_], srch[1], 'Test string starts with search string:', 'Test string does not start with search string:')
report(x2 > 0, lipsum[j_], srch[2], 'Search string located in test string at position:' x2, 'Search string not found within test string:')
report(x3 \= srch[3].length, lipsum[j_], srch[3], 'Test string ends with search string:', 'Test string does not start with search string:')
end j_

many = ''
x_ = 0
x_ = x_ + 1; many[0] = x_; many[x_] = 'How many times does "many times" occur in this string?'
x_ = x_ + 1; many[0] = x_; many[x_] = 'How often does "many times" occur in this string?'
x_ = x_ + 1; many[0] = x_; many[x_] = 'How often does it occur in this string?'
srch[4] = 'many times'

loop j_ = 1 to many[0]
o_ = 0
k_ = 0
loop label dups until o_ = 0
o_ = many[j_].pos(srch[4], o_ + 1)
if o_ \= 0 then k_ = k_ + 1
end dups
report(k_ > 0, many[j_], srch[4], 'Number of times search string occurs:' k_, 'Search string not found')
end j_

method report(state = boolean, ts, ss, testSuccess, testFailure) public static
if state then say testSuccess
else say testFailure
say ' Test string:' ts
say ' Search string:' ss
say

return

## NewLISP

(setq str "abcdefbcghi")

;; test if str starts with "ab"
(starts-with str "ab")

;; find "bc" inside str
(find "bc" str)

;; test if str ends with "ghi"
(ends-with str "ghi")

;; find all positions of pattern inside str
(define (find-all-pos pattern str)
(let ((idx -1) (pos '()))
(while (setq idx (find pattern str 0 (+ idx 1)))
(push idx pos -1))))

(find-all-pos "bc" str)

## Nim

import strutils

var s: string = "The quick brown fox"
if startsWith(s, "The quick"):
echo("Starts with: The quick")
if endsWith(s, "brown Fox"):
echo("Ends with: brown fox")
var pos = find(s, " brown ") # -1 if not found
if contains(s, " brown "): # showing the contains() proc, but could use if pos!=-1:
echo('"' & " brown " & '"' & " is located at position: " & \$pos)
Output:
Starts with: The quick
Ends with: brown fox
" brown " is located at position: 9

## Objective-C

[@"abcd" hasPrefix:@"ab"] //returns true
[@"abcd" hasSuffix:@"zn"] //returns false
int loc = [@"abab" rangeOfString:@"bb"].location //returns -1
loc = [@"abab" rangeOfString:@"ab"].location //returns 0
loc = [@"abab" rangeOfString:@"ab" options:0 range:NSMakeRange(loc+1, [@"abab" length]-(loc+1))].location //returns 2

## Objeck

bundle Default {
class Matching {
function : Main(args : System.String[]) ~ Nil {
"abcd"->StartsWith("ab")->PrintLine(); # returns true
"abcd"->EndsWith("zn")->PrintLine(); # returns false
("abab"->Find("bb") <> -1)->PrintLine(); # returns false
("abab"->Find("ab") <> -1)->PrintLine(); # returns true
loc := "abab"->Find("bb"); # returns -1
loc := "abab"->Find("ab"); # returns 0
loc := "abab"->Find("ab", loc + 1); # returns 2
}
}
}

## OCaml

let match1 s1 s2 =
let len1 = String.length s1
and len2 = String.length s2 in
if len1 < len2 then false else
let sub = String.sub s1 0 len2 in
(sub = s2)

testing in the top-level:

# match1 "Hello" "Hello World!" ;;
- : bool = false
# match1 "Hello World!" "Hello" ;;
- : bool = true
let match2 s1 s2 =
let len1 = String.length s1
and len2 = String.length s2 in
if len1 < len2 then false else
let rec aux i =
if i < 0 then false else
let sub = String.sub s1 i len2 in
if (sub = s2) then true else aux (pred i)
in
aux (len1 - len2)
# match2 "It's raining, Hello World!" "umbrella" ;;
- : bool = false
# match2 "It's raining, Hello World!" "Hello" ;;
- : bool = true
let match3 s1 s2 =
let len1 = String.length s1
and len2 = String.length s2 in
if len1 < len2 then false else
let sub = String.sub s1 (len1 - len2) len2 in
(sub = s2)
# match3 "Hello World" "Hello" ;;
- : bool = false
# match3 "Hello World" "World" ;;
- : bool = true
let match2_loc s1 s2 =
let len1 = String.length s1
and len2 = String.length s2 in
if len1 < len2 then (false, -1) else
let rec aux i =
if i < 0 then (false, -1) else
let sub = String.sub s1 i len2 in
if (sub = s2) then (true, i) else aux (pred i)
in
aux (len1 - len2)
# match2_loc "The sun's shining, Hello World!" "raining" ;;
- : bool * int = (false, -1)
# match2_loc "The sun's shining, Hello World!" "shining" ;;
- : bool * int = (true, 10)
let match2_num s1 s2 =
let len1 = String.length s1
and len2 = String.length s2 in
if len1 < len2 then (false, 0) else
let rec aux i n =
if i < 0 then (n <> 0, n) else
let sub = String.sub s1 i len2 in
if (sub = s2)
then aux (pred i) (succ n)
else aux (pred i) (n)
in
aux (len1 - len2) 0
# match2_num "This cloud looks like a camel, \
that other cloud looks like a llama" "stone" ;;
- : bool * int = (false, 0)
# match2_num "This cloud looks like a camel, \
that other cloud looks like a llama" "cloud" ;;
- : bool * int = (true, 2)

## Oforth

: stringMatching(s1, s2)
| i |
s2 isAllAt(s1, 1) ifTrue: [ System.Out s1 << " begins with " << s2 << cr ]
s2 isAllAt(s1, s1 size s2 size - 1 + ) ifTrue: [ System.Out s1 << " ends with " << s2 << cr ]

s1 indexOfAll(s2) ->i
i ifNotNull: [ System.Out s1 << " includes " << s2 << " at position : " << i << cr ]

"\nAll positions : " println
1 ->i
while (s1 indexOfAllFrom(s2, i) dup ->i notNull) [
System.Out s1 << " includes " << s2 << " at position : " << i << cr
i s2 size + ->i
] ;
Output:
> "arduinoardblobard", "ard" stringMatching
arduinoardblobard begins with ard
arduinoardblobard ends with ard
arduinoardblobard includes ard at position : 1

All positions :
arduinoardblobard includes ard at position : 1
arduinoardblobard includes ard at position : 8
arduinoardblobard includes ard at position : 15

## OxygenBasic

string s="sdfkjhgsdfkdfgkbopefioqwurti487sdfkrglkjfs9wrtgjglsdfkdkjcnmmb.,msfjflkjsdfk"

string f="sdfk"

string cr=chr(13)+chr(10),tab=chr(9)

string pr="FIND STRING LOCATIONS" cr cr

sys a=0, b=1, count=0, ls=len(s), lf=len(f)

do
a=instr b,s,f
if a=0 then exit do
count++
if a=1 then pr+="Begins with keyword" cr
pr+=count tab a cr
if a=ls-lf+1 then pr+="Ends with keyword at " a cr
b=a+1
end do

pr+=cr "Total matches: " count cr

print pr

'FIND STRING LOCATIONS
'
'Begins with keyword
'1 1
'2 8
'3 32
'4 51
'5 73
'Ends with keyword at 73
'
'Total matches: 5

## PARI/GP

This meets the first but not the second of the optional requirements. Note that GP treats any nonzero value as true so the location found by contains() can be ignore if not needed.

startsWith(string, prefix)={
string=Vec(string);
prefix=Vec(prefix);
if(#prefix > #string, return(0));
for(i=1,#prefix,if(prefix[i]!=string[i], return(0)));
1
};
contains(string, inner)={
my(good);
string=Vec(string);
inner=Vec(inner);
for(i=0,#string-#inner,
good=1;
for(j=1,#inner,
if(inner[j]!=string[i+j], good=0; break)
);
if(good, return(i+1))
);
0
};
endsWith(string, suffix)={
string=Vec(string);
suffix=Vec(suffix);
if(#suffix > #string, return(0));
for(i=1,#suffix,if(prefix[i]!=string[i+#string-#suffix], return(0)));
1
};

## Perl

Using regexes:

\$str1 =~ /^\Q\$str2\E/  # true if \$str1 starts with \$str2
\$str1 =~ /\Q\$str2\E/ # true if \$str1 contains \$str2
\$str1 =~ /\Q\$str2\E\$/ # true if \$str1 ends with \$str2

Using index:

index(\$str1, \$str2) == 0                               # true if \$str1 starts with \$str2
index(\$str1, \$str2) != -1 # true if \$str1 contains \$str2
rindex(\$str1, \$str2) == length(\$str1) - length(\$str2) # true if \$str1 ends with \$str2

Using substr:

substr(\$str1, 0, length(\$str2)) eq \$str2  # true if \$str1 starts with \$str2
substr(\$str1, - length(\$str2)) eq \$str2 # true if \$str1 ends with \$str2

Bonus task (printing all positions where \$str2 appears in \$str1):

print \$-[0], "\n" while \$str1 =~ /\Q\$str2\E/g;  # using a regex
my \$i = -1; print \$i, "\n" while (\$i = index \$str1, \$str2, \$i + 1) != -1;  # using index

## Perl 6

Using string methods:

\$haystack.starts-with(\$needle)  # True if \$haystack starts with \$needle
\$haystack.contains(\$needle) # True if \$haystack contains \$needle
\$haystack.ends-with(\$needle) # True if \$haystack ends with \$needle

Using regexes:

so \$haystack ~~ /^ \$needle  /  # True if \$haystack starts with \$needle
so \$haystack ~~ / \$needle / # True if \$haystack contains \$needle
so \$haystack ~~ / \$needle \$/ # True if \$haystack ends with \$needle

Using substr:

substr(\$haystack, 0, \$needle.chars) eq \$needle  # True if \$haystack starts with \$needle
substr(\$haystack, *-\$needle.chars) eq \$needle # True if \$haystack ends with \$needle

\$haystack.match(\$needle, :g)».from;  # List of all positions where \$needle appears in \$haystack

## Phix

constant word = "the",                                      -- (also try this with "th"/"he")
sentence = "the last thing the man said was the"
-- sentence = "thelastthingthemansaidwasthe" -- (practically the same results)

-- A common, but potentially inefficient idiom for checking for a substring at the start is:
if match(word,sentence)=1 then
?"yes(1)"
end if
-- A more efficient method is to test the appropriate slice
if length(sentence)>=length(word)
and sentence[1..length(word)]=word then
?"yes(2)"
end if
-- Which is almost identical to checking for a word at the end
if length(sentence)>=length(word)
and sentence[-length(word)..-1]=word then
?"yes(3)"
end if
-- Or sometimes you will see this, a tiny bit more efficient:
if length(sentence)>=length(word)
and match(word,sentence,length(sentence)-length(word)+1) then
?"yes(4)"
end if
-- Finding all occurences is a snap:
integer r = match(word,sentence)
while r!=0 do
?r
r = match(word,sentence,r+1)
end while
Output:
"yes(1)"
"yes(2)"
"yes(3)"
"yes(4)"
1
16
33

## PHP

<?php
/**********************************************************************************
* This program gets needle and haystack from the caller (chm.html) (see below)
* and checks for occurrences of the needle in the haystack
* 02.05.2013 Walter Pachl
**********************************************************************************/

\$haystack = \$_POST['haystack']; if (\$haystack=='') {\$haystack='no haystack given';}
\$needle = \$_POST['needle']; if (\$needle=='') {\$needle='no needle given';}

function rexxpos(\$h,\$n) {
\$pos = strpos(\$h,\$n);
if (\$pos === false) { \$pos=-1; }
else { \$pos=\$pos+1; }
return (\$pos);
}

\$pos=rexxpos(\$haystack,\$needle);
\$tx1 = "";
else { \$n=1; } // found once (so far)
// Special cases
if (\$pos==1){ \$tx1="needle found to be the start of the haystack"; }
if (\$pos==strlen(\$haystack)-strlen(\$needle)+1)
{ \$tx1="needle found at end of haystack"; }

if (\$n>0) { // look for other occurrences
\$pl=\$pos; // list of positions
\$p=\$pos; //
\$x="*************************************";
\$h=\$haystack;
while (\$p>0) {
\$h=substr(\$x,0,\$p).substr(\$h,\$p);
\$p=rexxpos(\$h,\$needle);
if ( \$p>0 ) { \$n=\$n+1; \$pl=\$pl.",&nbsp;".\$p; }
}
if (\$n==1) { \$txt="needle found once in haystack, position: \$pl."; }
else if (\$n==2) { \$txt="needle found twice in haystack, position(s): \$pl."; }
else { \$txt="needle found \$n times in haystack, position(s): \$pl."; }
}
?>
<html>
<title>Character Matching</title>
<meta name="author" content="Walter Pachl">
<meta name="date" content="02.05.2013">
<style>
p { font: 120% courier; }
</style>
<body>
<p><strong>Haystack:&nbsp;'<?php echo "\$haystack" ?>'</strong></p>
<p><strong>Needle:&nbsp;&nbsp;&nbsp;'<?php echo "\$needle" ?>'</strong></p>
<p><strong><?php echo "\$txt" ?></strong></p>
<!-- special message: -->
<p style="color: red";><strong><?php echo "\$tx1" ?></strong></p>
</body>
</html>
<?php
<!DOCTYPE html>
<!--
/************************************************************************
* Here we prompt the user for a haystack and a needle
* We then invoke program chmx.php
* to check for occurrences of the needle in the haystack
* 02.05.2013 Walter Pachl
************************************************************************/

-->
<html>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Character matching</title>
<body>
<form id="test" name="test" method="post" action="chmx.php">
<h1>Character matching</h1>
<p>Given two strings, demonstrate the following 3 types of matchings:
<ol style="margin-top:2; margin-bottom:2;">
<li>Determining if the first string starts with second string
<li>Determining if the first string contains the second string at any location
<li>Determining if the first string ends with the second string
</ol>
<p>Optional requirements:
<ol style="margin-top:2; margin-bottom:2;">
<li>Print the location of the match(es) for part 2
<li>Handle multiple occurrences of a string for part 2.
</ol>
<p style="margin-top:5; margin-bottom:3;">
<font face="Courier"><strong>Haystack:</strong>
<strong><input type="text" name="haystack" size="80"></strong></font></p>
<p style="margin-top:5; margin-bottom:3;">
<font face="Courier"><strong>Needle:&nbsp;&nbsp;</strong>
<strong><input type="text" name="needle" size="80"></strong></font></p>
<p>Press <input name="Submit" type="submit" class="erfolg" value="CHECK"/>
to invoke chmx.php.</p>
</form>
</body>
</html>

## PicoLisp

: (pre? "ab" "abcd")
-> "abcd"
: (pre? "xy" "abcd")
-> NIL

: (sub? "bc" "abcd")
-> "abcd"
: (sub? "xy" "abcd")
-> NIL

: (tail (chop "cd") (chop "abcd"))
-> ("c" "d")
: (tail (chop "xy") (chop "abcd"))
-> NIL

(de positions (Pat Str)
(setq Pat (chop Pat))
(make
(for ((I . L) (chop Str) L (cdr L))

: (positions "bc" "abcdabcd")
-> (2 6)

## PL/I

/* Let s be one string, t be the other that might exist within s. */
/* 8-1-2011 */
k = index(s, t);
if k = 0 then put skip edit (t, ' is nowhere in sight') (a);
else if k = 1 then
put skip edit (t, ' starts at the beginning of ', s) (a);
else if k+length(t)-1 = length(s) then
put skip edit (t, ' is at the end of ', s) (a);
else put skip edit (t, ' is within ', s) (a);

if k > 0 then put skip edit (t, ' starts at position ', k) (a);

Optional extra:

/* Handle multiple occurrences. */
n = 1;
do forever;
k = index(s, t, n);
if k = 0 then
do;
if n = 1 then put skip list (t, ' is nowhere in sight');
stop;
end;
else if k = 1 then
put skip edit ('<', t, '> starts at the beginning of ', s) (a);
else if k+length(t)-1 = length(s) then
put skip edit ('<', t, '> is at the end of ', s) (a);
else put skip edit ('<', t, '> is within ', s) (a);
n = k + length(t);

if k > 0 then
put skip edit ('<', t, '> starts at position ', trim(k)) (a);
else stop;
end;

## PureBasic

Procedure StartsWith(String1\$, String2\$)
Protected Result
If FindString(String1\$, String2\$, 1) =1 ; E.g Found in possition 1
Result =CountString(String1\$, String2\$)
EndIf
ProcedureReturn Result
EndProcedure

Procedure EndsWith(String1\$, String2\$)
Protected Result, dl=Len(String1\$)-Len(String2\$)
If dl>=0 And Right(String1\$, Len(String2\$))=String2\$
Result =CountString(String1\$, String2\$)
EndIf
ProcedureReturn Result
EndProcedure

And a verification

If OpenConsole()
PrintN(Str(StartsWith("Rosettacode", "Rosetta"))) ; = 1
PrintN(Str(StartsWith("Rosettacode", "code"))) ; = 0
PrintN(Str(StartsWith("eleutherodactylus cruralis", "e"))) ; = 3
PrintN(Str(EndsWith ("Rosettacode", "Rosetta"))) ; = 0
PrintN(Str(EndsWith ("Rosettacode", "code"))) ; = 1
PrintN(Str(EndsWith ("Rosettacode", "e"))) ; = 2

Print(#CRLF\$ + #CRLF\$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf

An alternate and more complete solution:

Procedure startsWith(string1\$, string2\$)
;returns one if string1\$ starts with string2\$, otherwise returns zero
If FindString(string1\$, string2\$, 1) = 1
ProcedureReturn 1
EndIf
ProcedureReturn 0
EndProcedure

Procedure contains(string1\$, string2\$, location = 0)
;returns the location of the next occurrence of string2\$ in string1\$ starting from location,
;or zero if no remaining occurrences of string2\$ are found in string1\$
ProcedureReturn FindString(string1\$, string2\$, location + 1)
EndProcedure

Procedure endsWith(string1\$, string2\$)
;returns one if string1\$ ends with string2\$, otherwise returns zero
Protected ls = Len(string2\$)
If Len(string1\$) - ls >= 0 And Right(string1\$, ls) = string2\$
ProcedureReturn 1
EndIf
ProcedureReturn 0
EndProcedure

If OpenConsole()
PrintN(Str(startsWith("RosettaCode", "Rosetta"))) ; = 1, true
PrintN(Str(startsWith("RosettaCode", "Code"))) ; = 0, false

PrintN("")
PrintN(Str(contains("RosettaCode", "luck"))) ; = 0, no occurrences
Define location
Repeat
location = contains("eleutherodactylus cruralis", "e", location)
PrintN(Str(location)) ;display each occurrence: 1, 3, 7, & 0 (no more occurrences)
Until location = 0

PrintN("")
PrintN(Str(endsWith ("RosettaCode", "Rosetta"))) ; = 0, false
PrintN(Str(endsWith ("RosettaCode", "Code"))) ; = 1, true

Print(#CRLF\$ + #CRLF\$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf
Output:
1
0

0
1
3
7
0

0
1

## PowerShell

"spicywiener".StartsWith("spicy")
"spicywiener".Contains("icy")
"spicywiener".EndsWith("wiener")
"spicywiener".IndexOf("icy")
[regex]::Matches("spicywiener", "i").count

Output:
True
True
True
2
2

## Python

"abcd".startswith("ab") #returns True
"abcd".endswith("zn") #returns False
"bb" in "abab" #returns False
"ab" in "abab" #returns True
loc = "abab".find("bb") #returns -1
loc = "abab".find("ab") #returns 0
loc = "abab".find("ab",loc+1) #returns 2

## Racket

#lang racket
(require srfi/13)
(string-prefix? "ab" "abcd")
(string-suffix? "cd" "abcd")
(string-contains "abab" "bb")
(string-contains "abab" "ba")

Output:
#t
#t
#f
1

## Retro

: startsWith? ( \$1 \$2 - f )
withLength &swap dip 0 swap ^strings'getSubset compare ;

"abcdefghijkl" "abcde" startsWith?
"abcdefghijkl" "bcd" startsWith?

"abcdefghijkl" "bcd" ^strings'search
"abcdefghijkl" "zmq" ^strings'search

: endsWith? ( \$1 \$2 - f )
swap withLength + over getLength - compare ;

"abcdefghijkl" "ijkl" endsWith?
"abcdefghijkl" "abc" endsWith?

## REXX

Extra coding was added to take care of using plurals in the last output message.

/*REXX program  demonstrates  some  basic   character string   testing  (for matching). */
parse arg A B; LB=length(B) /*obtain A and B from the command line.*/
say 'string A = ' A /*display string A to the terminal.*/
say 'string B = ' B /* " " B " " " */
say copies('░', 70)
if left(A, LB)==B then say 'string A starts with string B'
say /* [↓] another method using COMPARE BIF*/
/*╔══════════════════════════════════════════════════════════════════════════╗
║ if compare(A,B)==LB then say 'string A starts with string B' ║
╚══════════════════════════════════════════════════════════════════════════╝*/

p=pos(B, A)
if p==0 then say "string A doesn't contain string B"
else say 'string A contains string B (starting in position' p")"
say
if right(A, LB)==b then say 'string A ends with string B'
else say "string A doesn't end with string B"
say
\$=; p=0; do until p==0; p=pos(B, A, p+1)
if p\==0 then \$=\$',' p
end /*until*/
\$=space(strip(\$, 'L', ",")) /*elide extra blanks and leading comma.*/
#=words(\$) /*obtain number of words in \$ string.*/
if #==0 then say "string A doesn't contain string B"
else say 'string A contains string B ' # " time"left('s', #>1),
"(at position"left('s', #>1) \$")" /*stick a fork in it, we're done*/
output   when the following is specified (the five Marx brothers):   Chico_Harpo_Groucho_Zeppo_Gummo p
string  A  =  Chico_Harpo_Groucho_Zeppo_Gummo
string  B  =  p
░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░

string  A  contains string  B  (starting in position 10)

string  A  doesn't end with string B

string  A  contains string  B  3  times (at positions 10, 23, 24)
output   when the following is specified:   Chico_Harpo_Groucho_Zeppo_Gummo Z
string  A  =  Chico_Harpo_Groucho_Zeppo_Gummo
string  B  =  Z
░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░

string  A  contains string  B  (starting in position 21)

string  A  doesn't end with string  B

string  A  contains string  B  1  time (at position 21)
output   when the following is specified:   Chico_Harpo_Groucho_Zeppo_Gummo Chi
string  A  =  Chico_Harpo_Groucho_Zeppo_Gummo
string  B  =  Chi
░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░
string  A  starts with string  B

string  A  contains string B  (starting in position 1)

string  A  doesn't end with string  B

string  A  contains string  B  1  time (at position 1)
output   when the following is specified:   Chico_Harpo_Groucho_Zeppo_Gummo mmo
string  A  =  Chico_Harpo_Groucho_Zeppo_Gummo
string  B  =  mmo
░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░

string  A  contains string  B  (starting in position 29)

string  A  ends with string  B

string  A  contains string  B  1  time (at position 29)

## Ring

aString = "Welcome to the Ring Programming Language"
bString = "Ring"
bStringIndex = substr(aString,bString)
if bStringIndex > 0 see "" + bStringIndex + " : " + bString ok

## Ruby

p 'abcd'.start_with?('ab')  #returns true
p 'abcd'.end_with?('ab') #returns false
p 'abab'.include?('bb') #returns false
p 'abab'.include?('ab') #returns true
p 'abab'['bb'] #returns nil
p 'abab'['ab'] #returns "ab"
p 'abab'.index('bb') #returns nil
p 'abab'.index('ab') #returns 0
p 'abab'.index('ab', 1) #returns 2
p 'abab'.rindex('ab') #returns 2

## Run BASIC

s1\$ = "abc def ghi klmnop"
s2\$ = "abc" ' begins with
s3\$ = "ef" ' is in the string
s4\$ = "nop" ' ends with

sn2\$ = "abcx" ' not begins with
sn3\$ = "efx" ' not in the string
sn4\$ = "nopx" ' not ends with

if left\$(s1\$,len(s2\$)) <> s2\$ then a\$ = "Not "
print "String:";s1\$;" does ";a\$;"begin with:";s2\$

if instr(s1\$,s3\$) = 0 then a\$ = "Not "
print "String:";s1\$;" does ";a\$;"contain:";s3\$

if mid\$(s1\$,len(s1\$) + 1 - len(s4\$),len(s4\$)) <> s4\$ then a\$ = "Not "
print "String:";s1\$;" does ";a\$;"end with:";s4\$

' ----------- not -----------------------------
print
if left\$(s1\$,len(sn2\$)) <> sn2\$ then a\$ = "Not "
print "String:";s1\$;" does ";a\$;"begin with:";sn2\$

if instr(s1\$,sn3\$) = 0 then a\$ = "Not "
print "String:";s1\$;" does ";a\$;"contain:";sn3\$

if mid\$(s1\$,len(s1\$) + 1 - len(sn4\$),len(sn4\$)) <> sn4\$ then a\$ = "Not "
print "String:";s1\$;" does ";a\$;"end with:";sn4\$
Output:
String:abc def ghi klmnop does begin with:abc
String:abc def ghi klmnop does contain:ef
String:abc def ghi klmnop does end with:nop

String:abc def ghi klmnop does Not begin with:abcx
String:abc def ghi klmnop does Not contain:efx
String:abc def ghi klmnop does Not end with:nopx

## Rust

fn print_match(possible_match: Option<usize>) {
match possible_match {
Some(match_pos) => println!("Found match at pos {}", match_pos),
None => println!("Did not find any matches")
}
}

fn main() {
let s1 = "abcd";
let s2 = "abab";
let s3 = "ab";

// Determining if the first string starts with second string
assert!(s1.starts_with(s3));
// Determining if the first string contains the second string at any location
assert!(s1.contains(s3));
// Print the location of the match
print_match(s1.find(s3)); // Found match at pos 0
print_match(s1.find(s2)); // Did not find any matches
// Determining if the first string ends with the second string
assert!(s2.ends_with(s3));
}

fn main(){
let hello = String::from("Hello world");
println!(" Start with \"he\" {} \n Ends with \"rd\" {}\n Contains \"wi\" {}",
hello.starts_with("He"),
hello.ends_with("ld"),
hello.contains("wi"));
}
Output:
Ends with "ld" true
Contains "wi" false

## Scala

"abcd".startsWith("ab") //returns true
"abcd".endsWith("zn") //returns false
"abab".contains("bb") //returns false
"abab".contains("ab") //returns true

var loc="abab".indexOf("bb") //returns -1
loc = "abab".indexOf("ab") //returns 0
loc = "abab".indexOf("ab", loc+1) //returns 2

## Seed7

\$ include "seed7_05.s7i";

const proc: main is func
local
var integer: position is 0;
begin
writeln(startsWith("abcd", "ab")); # write TRUE
writeln(endsWith("abcd", "zn")); # write FALSE
writeln(pos("abab", "bb") <> 0); # write FALSE
writeln(pos("abab", "ab") <> 0); # write TRUE
writeln(pos("abab", "bb")); # write 0
position := pos("abab", "ab");
writeln(position); # position is 1
position := pos("abab", "ab", succ(position));
writeln(position); # position is 3
end func;
Output:
TRUE
FALSE
FALSE
TRUE
0
1
3

## Sidef

var first = "abc-abcdef-abcd";
var second = "abc";

say first.begins_with(second); #=> true
say first.contains(second); #=> true
say first.ends_with(second); #=> false

# Get and print the location of the match
say first.index(second); #=> 0

# Find multiple occurrences of a string
var pos = -1;
while (pos = first.index(second, pos+1) != -1) {
say "Match at pos: #{pos}";
}

## Smalltalk

a startsWith: b
a includesSubCollection: b
a endsWith: b
a indexOfSubCollection: b
a indexOfSubCollection: b startingAt: pos

## SNOBOL4

s1 = 'abcdabefgab'
s2 = 'ab'
s3 = 'xy'
OUTPUT = ?(s1 ? POS(0) s2) "1. " s2 " begins " s1
OUTPUT = ?(s1 ? POS(0) s3) "1. " s3 " begins " s1  ;# fails

n = 0
again s1 POS(n) ARB s2 @a  :F(p3)
OUTPUT = "2. " s2 " found at position "
+ a - SIZE(s2) " in " s1
n = a  :(again)

p3 OUTPUT = ?(s1 ? s2 RPOS(0)) "3. " s2 " ends " s1
END
Output:
1. ab begins abcdabefgab
2. ab found at position 0 in abcdabefgab
2. ab found at position 4 in abcdabefgab
2. ab found at position 9 in abcdabefgab
3. ab ends abcdabefgab

## Standard ML

String.isPrefix "ab" "abcd"; (* returns true *)
String.isSuffix "zn" "abcd"; (* returns false *)
String.isSubstring "bb" "abab"; (* returns false *)
String.isSubstring "ab" "abab"; (* returns true *)
#2 (Substring.base (#2 (Substring.position "bb" (Substring.full "abab")))); (* returns 4 *)
val loc = #2 (Substring.base (#2 (Substring.position "ab" (Substring.full "abab")))); (* returns 0 *)
val loc' = #2 (Substring.base (#2 (Substring.position "ab" (Substring.extract ("abab", loc+1, NONE))))); (* returns 2 *)

## Swift

var str = "Hello, playground"
str.hasPrefix("Hell") //True
str.hasPrefix("hell") //False

str.containsString("llo") //True
str.containsString("xxoo") //False

str.hasSuffix("playground") //True
str.hasSuffix("world") //False

## Tcl

In this code, we are looking in various ways for the string in the variable needle in the string in the variable haystack.

set isPrefix    [string equal -length [string length \$needle] \$haystack \$needle]
set isContained [expr {[string first \$needle \$haystack] >= 0}]
set isSuffix [string equal \$needle [string range \$haystack end-[expr {[string length \$needle]-1}] end]]

Of course, in the cases where the needle is a glob-safe string (i.e., doesn't have any of the characters “*?[\” in), this can be written far more conveniently:

set isPrefix    [string match  \$needle* \$haystack]
set isContained [string match *\$needle* \$haystack]
set isSuffix [string match *\$needle \$haystack]

Another powerful technique is to use the regular expression engine in literal string mode:

set isContained [regexp ***=\$needle \$haystack]

This can be extended by getting the regexp to return the locations of the matches, enabling the other forms of match to be done:

set matchLocations [regexp -indices -all -inline ***=\$needle \$haystack]
# Each match location is a pair, being the index into the string where the needle started
# to match and the index where the needle finished matching

set isContained [expr {[llength \$matchLocations] > 0}]
set isPrefix [expr {[lindex \$matchLocations 0 0] == 0}]
set isSuffix [expr {[lindex \$matchLocations end 1] == [string length \$haystack]-1}]
set firstMatchStart [lindex \$matchLocations 0 0]
puts "Found \"\$needle\" in \"\$haystack\" at \$firstMatchStart"
foreach location \$matchLocations {
puts "needle matched at index [lindex \$location 0]"
}

## TUSCRIPT

\$\$ MODE TUSCRIPT

IF (string1.sw.string2) THEN
PRINT string1," starts with ",string2
ELSE
PRINT string1," not starts with ",string2
ENDIF
SET beg=STRING (string1,string2,0,0,0,end)
IF (beg!=0) THEN
PRINT string1," contains ",string2
PRINT " starting in position ",beg
PRINT " ending in position ",end
ELSE
PRINT string1," not contains ",string2
ENDIF

IF (string1.ew.string2) THEN
PRINT string1," ends with ",string2
ELSE
PRINT string1," not ends with ",string2
ENDIF

Output:
string1 >Rosetta Code
string2 >Code
Rosetta Code not starts with Code
Rosetta Code contains        Code
starting in position 9
ending   in position 13
Rosetta Code ends with       Code

## TXR

### TXR Lisp

(tree-case *args*
((big small)
(cond
((< (length big) (length small))
(put-line `@big is shorter than @small`))
((str= big small)
(put-line `@big and @small are equal`))
((starts-with small big)
(put-line `@small is a prefix of @big`))
((ends-with small big)
(put-line `@small is a suffix of @big`))
(t (iflet ((pos (search-str big small)))
(put-line `@small occurs in @big at position @pos`)
(put-line `@small does not occur in @big`)))))
(otherwise
(put-line `usage: @(ldiff *full-args* *args*) <bigstring> <smallstring>`)))
Output:
\$ txr cmatch2.tl x
usage: txr cmatch2.tl <bigstring> <smallstring>
\$ txr cmatch2.tl x y z
usage: txr cmatch2.tl <bigstring> <smallstring>
\$ txr cmatch2.tl catalog cat
cat is a prefix of catalog
\$ txr cmatch2.tl catalog log
log is a suffix of catalog
\$ txr cmatch2.tl catalog at
at occurs in catalog at position 1
\$ txr cmatch2.tl catalog catalogue
catalog is shorter than catalogue
\$ txr cmatch2.tl catalog catalog
catalog and catalog are equal
\$ txr cmatch2.tl catalog dog
dog does not occur in catalog

### Pattern Language

@line
@(cases)
@ line
@ (output)
second line is the same as first line
@ (end)
@(or)
@ (skip)@line
@ (output)
first line is a suffix of the second line
@ (end)
@(or)
@ [email protected](skip)
@ (output)
first line is a suffix of the second line
@ (end)
@(or)
@ [email protected]@(skip)
@ (output)
first line is embedded in the second line at position @(length prefix)
@ (end)
@(or)
@ (output)
@ (end)
@(end)
Output:
\$ txr cmatch.txr -
123
01234
first line is embedded in the second line at position 1
\$ txr cmatch.txr -
123
0123
first line is a suffix of the second line

## VBA

Translation of: Phix
Public Sub string_matching()
word = "the" '-- (also try this with "th"/"he")
sentence = "the last thing the man said was the"
'-- sentence = "thelastthingthemansaidwasthe" '-- (practically the same results)

'-- A common, but potentially inefficient idiom for checking for a substring at the start is:
If InStr(1, sentence, word) = 1 Then
Debug.Print "yes(1)"
End If
'-- A more efficient method is to test the appropriate slice
If Len(sentence) >= Len(word) _
And Mid(sentence, 1, Len(word)) = word Then
Debug.Print "yes(2)"
End If
'-- Which is almost identical to checking for a word at the end
If Len(sentence) >= Len(word) _
And Mid(sentence, Len(sentence) - Len(word) + 1, Len(word)) = word Then
Debug.Print "yes(3)"
End If
'-- Or sometimes you will see this, a tiny bit more efficient:
If Len(sentence) >= Len(word) _
And InStr(Len(sentence) - Len(word) + 1, sentence, word) Then
Debug.Print "yes(4)"
End If
'-- Finding all occurences is a snap:
r = InStr(1, sentence, word)
Do While r <> 0
Debug.Print r
r = InStr(r + 1, sentence, word)
Loop
End Sub
Output:
yes(1)
yes(2)
yes(3)
yes(4)
1
16
33

## VBScript

Function StartsWith(s1,s2)
StartsWith = False
If Left(s1,Len(s2)) = s2 Then
StartsWith = True
End If
End Function

Function Contains(s1,s2)
Contains = False
If InStr(1,s1,s2) Then
Contains = True & " at positions "
j = 1
Do Until InStr(j,s1,s2) = False
Contains = Contains & InStr(j,s1,s2) & ", "
If j = 1 Then
If Len(s2) = 1 Then
j = j + InStr(j,s1,s2)
Else
j = j + (InStr(j,s1,s2) + (Len(s2) - 1))
End If
Else
If Len(s2) = 1 Then
j = j + ((InStr(j,s1,s2) - j) + 1)
Else
j = j + ((InStr(j,s1,s2) - j) + (Len(s2) - 1))
End If
End If
Loop
End If
End Function

Function EndsWith(s1,s2)
EndsWith = False
If Right(s1,Len(s2)) = s2 Then
EndsWith = True
End If
End Function

WScript.StdOut.Write "Starts with test, 'foo' in 'foobar': " & StartsWith("foobar","foo")
WScript.StdOut.WriteLine
WScript.StdOut.Write "Contains test, 'o' in 'fooooobar': " & Contains("fooooobar","o")
WScript.StdOut.WriteLine
WScript.StdOut.Write "Ends with test, 'bar' in 'foobar': " & EndsWith("foobar","bar")
Output:
Starts with test, 'foo' in 'foobar': True
Contains test, 'o' in 'fooooobar': True at positions 2, 3, 4, 5, 6,
Ends with test, 'bar' in 'foobar': True

## Visual Basic

Works with: Visual Basic version VB6 Standard

works the same as in VBA, see String_matching#VBA

## XPL0

include c:\cxpl\codes;  \intrinsic 'code' declarations
string 0; \use zero-terminated strings

func StrLen(A); \Return number of characters in a string
char A;
int I;
for I:= 0 to -1>>1-1 do
if A(I) = 0 then return I;

func StrFind(A, B); \Search for string B in string A
\Returns index of first occurrence of string B in A, or -1 if B is not found
char A, B; \strings to be compared
int LA, LB, I, J;
[LA:= StrLen(A);
LB:= StrLen(B);
for I:= 0 to LA-LB do
[for J:= 0 to LB-1 do
if B(J) # A(J+I) then J:= LB+1;
if J = LB then return I; \found
];
return -1;
];

char Str; int I, J;
[Str:= "pack my box with";
\ 0123456789012345
Text(0, if StrFind(Str, "pack") = 0 then "yes" else "no"); CrLf(0); \1.
Text(0, if StrFind(Str, "ack") = 0 then "yes" else "no"); CrLf(0);
I:= StrFind(Str, "x w");
Text(0, if I >= 0 then "yes" else "no"); \2.
Text(0, ", at offset "); IntOut(0, I); CrLf(0);
I:= 0; J:= 0; \offsets of space characters
loop [I:= StrFind(Str+J, " ");
if I < 0 then quit;
IntOut(0, I+J); ChOut(0, ^ );
J:= J+I+1;
];
CrLf(0);
Text(0, if StrFind(Str, "X w") >= 0 then "yes" else "no"); CrLf(0);
Text(0, if StrFind(Str, "with") = StrLen(Str)-StrLen("with") then "yes" else "no"); CrLf(0); \3.
Text(0, if StrFind(Str, "x w" ) = StrLen(Str)-StrLen("x w" ) then "yes" else "no"); CrLf(0);
]
Output:
yes
no
yes, at offset 10
4 7 11
no
yes
no

## zkl

fcn f(text,p){ if(text.find(p)==0)println("Yep") else println("Nope") }
f("foobar","foo") //--> Yep
f("foobar","bar") //--> Nope
fcn f(text,p){ if(Void!=(n:=text.find(p)))println("Contained @",n) else println("Nope") }
f("foobar","ob") //--> Contained @2
f("foobar","food") //--> Nope
fcn f(text,p){
if( Void!=(n:=text.rfind(p)) and n+p.len()==text.len() )
println("tail gunner") else println("Nope")
}
f("foobar","r"); f("foobar","ar"); //--> tail gunners
f("foobar","ob"); //--> Nope
f("foobarfoobar","bar"); //--> tail gunner