Copy a string

This task is about copying a string.

Task

Where it is relevant, distinguish between copying the contents of a string versus making an additional reference to an existing string.

Other tasks related to string operations:

Metrics

Counting

Remove/replace

Anagrams/Derangements/shuffling

Find/Search/Determine

Formatting

Song lyrics/poems/Mad Libs/phrases

Tokenize

Sequences

11l

<lang 11l>V src = ‘hello’ V dst = copy(src)</lang>

360 Assembly

To copy a string, we use an MVC (Move Character). To make a reference to a string, we use a LA (Load Address). <lang 360asm>* Duplicate a string

        MVC    A,=CL64'Hello'     a='Hello'
        MVC    B,A                b=a          memory copy
        MVC    A,=CL64'Goodbye'   a='Goodbye'
        XPRNT  A,L'A              print a
        XPRNT  B,L'B              print b
        ...

Make reference to a string a string

        MVC    A,=CL64'Hi!'       a='Hi!'
        LA     R1,A               r1=@a        set pointer
        ST     R1,REFA            refa=@a      store pointer
        XPRNT  A,L'A              print a
        XPRNT  0(R1),L'A          print %refa
        ...

A DS CL64 a B DS CL64 b REFA DS A @a</lang>

68000 Assembly

Making a reference to an existing string is simple. Just load its memory location into an address register. <lang 68000devpac>myString: DC.B "HELLO WORLD",0 EVEN

LEA myString,A3</lang>

Copying a string involves a little more work: <lang 68000devpac>StringRam equ $100000

myString: DC.B "HELLO WORLD",0 EVEN

LEA myString,A3 LEA StringRam,A4

CopyString: MOVE.B (A3)+,D0 MOVE.B D0,(A4)+ ;we could have skipped D0 but this makes it easier to check for the terminator. BEQ Terminated BRA CopyString

Terminated: ;the null terminator is already stored along with the string itself, so we are done.

program ends here.</lang>

AArch64 Assembly

Works with: as version Raspberry Pi 3B version Buster 64 bits

<lang AArch64 Assembly> /* ARM assembly AARCH64 Raspberry PI 3B */ /* program copystr64.s */

/*******************************************/ /* Constantes file */ /*******************************************/ /* for this file see task include a file in language AArch64 assembly*/ .include "../includeConstantesARM64.inc" /*******************************************/ /* Initialized data */ /*******************************************/ .data szString: .asciz "ABCDEFGHIJKLMNOPQRSTUVWXYZ\n" /*******************************************/ /* UnInitialized data */ /*******************************************/ .bss .align 4 qPtString: .skip 8 szString1: .skip 80 /*******************************************/ /* code section */ /*******************************************/ .text .global main main: // entry of program

                        // display start string 
   ldr x0,qAdrszString
   bl affichageMess
                        // copy pointer string
   ldr x0,qAdrszString
   ldr x1,qAdriPtString
   str x0,[x1]
                        // control
   ldr x1,qAdriPtString
   ldr x0,[x1]
   bl affichageMess
                        // copy string
   ldr x0,qAdrszString    
   ldr x1,qAdrszString1

1:

   ldrb w2,[x0],1       // read one byte and increment pointer one byte
   strb w2,[x1],1       // store one byte and increment pointer one byte
   cmp x2,#0            // end of string ?
   bne 1b               // no -> loop 
                        // control
   ldr x0,qAdrszString1
   bl affichageMess

100: // standard end of the program */

   mov x0,0             // return code
   mov x8,EXIT          // request to exit program
   svc 0                // perform the system call

qAdrszString: .quad szString qAdriPtString: .quad qPtString qAdrszString1: .quad szString1 /********************************************************/ /* File Include fonctions */ /********************************************************/ /* for this file see task include a file in language AArch64 assembly */ .include "../includeARM64.inc" </lang>

ABAP

<lang ABAP>data: lv_string1 type string value 'Test',

     lv_string2 type string.

lv_string2 = lv_string1.</lang>

Inline Declaration

Works with: ABAP version 7.4 Or above only

<lang ABAP>DATA(string1) = |Test|. DATA(string2) = string1.</lang>

ActionScript

Strings are immutable in ActionScript, and can safely be assigned with the assignment operator, much as they can in Java.[1] <lang ActionScript>var str1:String = "Hello"; var str2:String = str1;</lang>

Ada

Ada provides three different kinds of strings. The String type is a fixed length string. The Bounded_String type is a string with variable length up to a specified maximum size. The Unbounded_String type is a variable length string with no specified maximum size. The Bounded_String type behaves a lot like C strings, while the Unbounded_String type behaves a lot like the C++ String class.

Fixed Length String Copying.

<lang ada>Src : String := "Hello"; Dest : String := Src;</lang> Ada provides the ability to manipulate slices of strings. <lang ada>Src : String := "Rosetta Stone"; Dest : String := Src(1..7); -- Assigns "Rosetta" to Dest Dest2 : String := Src(9..13); -- Assigns "Stone" to Dest2</lang>

Bounded Length String Copying

<lang ada>-- Instantiate the generic package Ada.Strings.Bounded.Generic_Bounded_Length with a maximum length of 80 characters package Flexible_String is new Ada.Strings.Bounded.Generic_Bounded_Length(80); use Flexible_String;

Src : Bounded_String := To_Bounded_String("Hello"); Dest : Bounded_String := Src;</lang> Ada Bounded_String type provides a number of functions for dealing with slices.

Unbounded Length String Copying

<lang ada>-- The package Ada.Strings.Unbounded contains the definition of the Unbounded_String type and all its methods Src : Unbounded_String := To_Unbounded_String("Hello"); Dest : Unbounded_String := Src;</lang>

Aime

The intrinsic text type is immediate, immutable and cannot be referred more than once.

Copying an intrinsic string: <lang aime>text s, t; t = "Rosetta"; s = t;</lang> Data of the non intrinsic byte array type can be referred more than once. Copying a binary array of bytes: <lang aime>data s, t;

Copy -t- into -s-

b_copy(s, t);

Set -s- as a reference of the object -t- is pointing

b_set(s, t);

or:

s = t; </lang>

ALGOL 68

In ALGOL 68 strings are simply flexible length arrays of CHAR;

<lang algol68>(

 STRING src:="Hello", dest;
 dest:=src

)</lang>

ALGOL W

<lang algolw>begin

   % strings are (fixed length) values in algol W. Assignment makes a copy   %
   string(10) a, copyOfA;
   a := "some text";
   copyOfA := a;
   % assignment to a will not change copyOfA                                 %
   a := "new value";
   write( a, copyOfA )

end.</lang>

Output:

new value some text

Apex

In Apex, Strings are a primitive data type <lang apex>String original = 'Test'; String cloned = original; //"original == cloned" is true

cloned += ' more'; //"original == cloned" is false</lang>

AppleScript

<lang AppleScript>set src to "Hello" set dst to src</lang>

ARM Assembly

Works with: as version Raspberry Pi

<lang ARM Assembly> /* ARM assembly Raspberry PI */ /* program copystr.s */

/* Constantes */ .equ STDOUT, 1 @ Linux output console .equ EXIT, 1 @ Linux syscall .equ WRITE, 4 @ Linux syscall /* Initialized data */ .data szString: .asciz "ABCDEFGHIJKLMNOPQRSTUVWXYZ\n"

/* UnInitialized data */ .bss .align 4 iPtString: .skip 4 szString1: .skip 80

/* code section */ .text .global main main: /* entry of program */

   push {fp,lr}    /* saves 2 registers */

   @ display start string 
   ldr r0,iAdrszString
   bl affichageMess
   @ copy pointer string
   ldr r0,iAdrszString
   ldr r1,iAdriPtString
   str r0,[r1]
   @ control
   ldr r1,iAdriPtString
   ldr r0,[r1]
   bl affichageMess
   @ copy string
   ldr r0,iAdrszString    
   ldr r1,iAdrszString1

1:

   ldrb r2,[r0],#1   @ read one byte and increment pointer one byte
   strb r2,[r1],#1   @ store one byte and increment pointer one byte
   cmp r2,#0          @ end of string ?
   bne 1b            @ no -> loop 
   @ control
   ldr r0,iAdrszString1
   bl affichageMess

100: /* standard end of the program */

   mov r0, #0                  @ return code
   pop {fp,lr}                 @restaur 2 registers
   mov r7, #EXIT              @ request to exit program
   swi 0                       @ perform the system call

iAdrszString: .int szString iAdriPtString: .int iPtString iAdrszString1: .int szString1

/******************************************************************/ /* display text with size calculation */ /******************************************************************/ /* r0 contains the address of the message */ affichageMess:

   push {fp,lr}    			/* save  registres */ 
   push {r0,r1,r2,r7}    		/* save others 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 */
   addne r2,r2,#1   			/* else add 1 in the length */
   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" */
   swi #0                      /* call systeme */
   pop {r0,r1,r2,r7}     		/* restaur others registers */
   pop {fp,lr}    				/* restaur des  2 registres */ 
   bx lr	        			/* return  */

</lang>

Arturo

<lang rebol>a: new "Hello" b: a ; reference the same string

changing one string in-place
will change both strings

'b ++ "World" print b print a

c: "Hello" d: new c ; make a copy of the older string

changing one string in-place
will change only the string in question

'd ++ "World" print d print c</lang>

Output:

HelloWorld
HelloWorld
HelloWorld
Hello

AutoHotkey

<lang autohotkey>src := "Hello" dst := src</lang>

AutoIt

<lang autoit>$Src= "Hello" $dest = $Src</lang>

AWK

<lang awk>BEGIN {

 a = "a string"
 b = a
 sub(/a/, "X", a) # modify a
 print b  # b is a copy, not a reference to...

}</lang>

Axe

<lang axe>Lbl STRCPY r₁→S While {r₂}

{r₂}→{r₁}
r₁++
r₂++

End 0→{r₁} S Return</lang>

Babel

To copy a string in Babel is the same as copying any other object. Use the cp operator to make a deep-copy.

<lang babel>babel> "Hello, world\n" dup cp dup 0 "Y" 0 1 move8 babel> << << Yello, world Hello, world </lang>

BASIC

Works with: QuickBasic version 4.5

Works with: PB version 7.1

 src$ = "Hello"
 dst$ = src$

Applesoft BASIC

<lang ApplesoftBasic>100 DEF FN P(A) = PEEK (A) + PEEK(A + 1) * 256 : FOR I = FN P(105) TO FN P(107) - 1 STEP 7 : ON PEEK(I + 1) < 128 OR PEEK(I) > 127 GOTO 130 : ON LEFT$(P$, 1) <> CHR$(PEEK(I)) GOTO 130 110 IF LEN(P$) > 1 THEN ON PEEK(I + 1) = 128 GOTO 130 : IF MID$(P$, 2, 1) <> CHR$(PEEK(I + 1) - 128) GOTO 130 120 POKE I + 4, P / 256 : POKE I + 3, P - PEEK(I + 4) * 256 : RETURN 130 NEXT I : STOP</lang>

<lang ApplesoftBasic>S$ = "HELLO" : REM S$ IS THE ORIGINAL STRING C$ = S$ : REM C$ IS THE COPY</lang>

<lang ApplesoftBasic>P$ = "S" : P = 53637 : GOSUB 100"POINT STRING S AT SOMETHING ELSE ?S$ ?C$</lang>

BaCon

Strings by value or by reference

Strings can be stored by value or by reference. By value means that a copy of the original string is stored in a variable. This happens automatically when when a string variable name ends with the '$' symbol.

Sometimes it may be necessary to refer to a string by reference. In such a case, simply declare a variable name as STRING but omit the '$' at the end. Such a variable will point to the same memory location as the original string. The following examples should show the difference between by value and by reference.

When using string variables by value:

<lang freebasic>a$ = "I am here" b$ = a$ a$ = "Hello world..." PRINT a$, b$</lang>

This will print "Hello world...I am here". The variables point to their individual memory areas so they contain different strings. Now consider the following code:

<lang freebasic>a$ = "Hello world..." LOCAL b TYPE STRING b = a$ a$ = "Goodbye..." PRINT a$, b</lang>

This will print "Goodbye...Goodbye..." because the variable 'b' points to the same memory area as 'a$'.

Commodore BASIC

<lang basic>10 A$ = "HELLO" 20 REM COPY CONTENTS OF A$ TO B$ 30 B$ = A$ 40 REM CHANGE CONTENTS OF A$ 50 A$ = "HI" 60 REM DISPLAY CONTENTS 70 PRINT A$, B$</lang> Commodore BASIC can't do pointers or 'reference to'

Sinclair ZX81 BASIC

Creating a new reference to an existing string is not possible, or at least not easy. (You could probably do it with PEEKs and POKEs.) This program demonstrates that an assignment statement copies a string, by showing that the two strings can afterwards be independently modified. <lang basic>10 LET A$="BECAUSE I DO NOT HOPE TO TURN AGAIN" 20 LET B$=A$ 30 LET A$=A$( TO 21) 40 PRINT B$ 50 PRINT A$ 60 LET B$=A$+B$(22 TO 29) 70 PRINT B$</lang>

Output:

BECAUSE I DO NOT HOPE TO TURN AGAIN
BECAUSE I DO NOT HOPE
BECAUSE I DO NOT HOPE TO TURN

Batch File

Since the only variables are environment variables, creating a string copy is fairly straightforward: <lang dos>set src=Hello set dst=%src%</lang>

BBC BASIC

Works with: BBC BASIC for Windows

<lang bbcbasic> source$ = "Hello, world!"

     REM Copy the contents of a string:
     copy$ = source$
     PRINT copy$
     
     REM Make an additional reference to a string:
     !^same$ = !^source$
     ?(^same$+4) = ?(^source$+4)
     ?(^same$+5) = ?(^source$+5)
     PRINT same$</lang>

Bracmat

Because in Bracmat strings are unalterable, you never want to copy a string. Still, you will obtain a copy of a string by overflowing the reference counter of the string. (Currently, reference counters on strings and on most operators are 10 bits wide. The = operator has a much wider 'inexhaustible' reference counter, because it anchors alterable objects.) Still, you won't be able to test whether you got the original or a copy other than by looking at overall memory usage of the Bracmat program at the OS-level or by closely timing comparison operations. You obtain a new reference to a string or a copy of the string by simple assignment using the = or the : operator: <lang bracmat>abcdef:?a; !a:?b;

c=abcdef; !c:?d;

!a:!b { variables a and b are the same and probably referencing the same string } !a:!d { variables a and d are also the same but not referencing the same string } </lang>

C

<lang c>#include <stdlib.h> /* exit(), free() */

include <stdio.h> /* fputs(), perror(), printf() */
include <string.h>

int main() { size_t len; char src[] = "Hello"; char dst1[80], dst2[80]; char *dst3, *ref;

/* * Option 1. Use strcpy() from <string.h>. * * DANGER! strcpy() can overflow the destination buffer. * strcpy() is only safe if the source string is shorter than * the destination buffer. We know that "Hello" (6 characters * with the final '\0') easily fits in dst1 (80 characters). */ strcpy(dst1, src);

/* * Option 2. Use strlen() and memcpy() from <string.h>, to copy * strlen(src) + 1 bytes including the final '\0'. */ len = strlen(src); if (len >= sizeof dst2) { fputs("The buffer is too small!\n", stderr); exit(1); } memcpy(dst2, src, len + 1);

/* * Option 3. Use strdup() from <string.h>, to allocate a copy. */ dst3 = strdup(src); if (dst3 == NULL) { /* Failed to allocate memory! */ perror("strdup"); exit(1); }

/* Create another reference to the source string. */ ref = src;

/* Modify the source string, not its copies. */ memset(src, '-', 5);

printf(" src: %s\n", src); /* src: ----- */ printf("dst1: %s\n", dst1); /* dst1: Hello */ printf("dst2: %s\n", dst2); /* dst2: Hello */ printf("dst3: %s\n", dst3); /* dst3: Hello */ printf(" ref: %s\n", ref); /* ref: ----- */

/* Free memory from strdup(). */ free(dst3);

return 0; }</lang>

Library: BSD libc

<lang c>#include <stdlib.h> /* exit() */

include <stdio.h> /* fputs(), printf() */
include <string.h>

int main() { char src[] = "Hello"; char dst[80];

/* Use strlcpy() from <string.h>. */ if (strlcpy(dst, src, sizeof dst) >= sizeof dst) { fputs("The buffer is too small!\n", stderr); exit(1); }

memset(src, '-', 5); printf("src: %s\n", src); /* src: ----- */ printf("dst: %s\n", dst); /* dst: Hello */

return 0; }</lang>

C#

<lang csharp>string src = "Hello"; string dst = src;</lang>

C++

<lang cpp>#include <iostream>

include <string>

int main( ) {

  std::string original ("This is the original");
  std::string my_copy = original;
  std::cout << "This is the copy: " << my_copy << std::endl;
  original = "Now we change the original! ";
  std::cout << "my_copy still is " << my_copy << std::endl;

}</lang>

Clojure

<lang clojure>(let [s "hello"

     s1 s]
 (println s s1))</lang>

COBOL

Translation of: C#

<lang cobol>MOVE "Hello" TO src MOVE src TO dst</lang>

ColdFusion

In ColdFusion, only complex data types (structs, objects, etc.) are passed by reference. Hence, any string copy operations are by value.

Common Lisp

<lang lisp>(let* ((s1 "Hello") ; s1 is a variable containing a string

      (s1-ref s1)             ; another variable with the same value
      (s2     (copy-seq s1))) ; s2 has a distinct string object with the same contents
 (assert (eq s1 s1-ref))      ; same object
 (assert (not (eq s1 s2)))    ; different object
 (assert (equal s1 s2))       ; same contents
                             
 (fill s2 #\!)                ; overwrite s2
 (princ s1)                  
 (princ s2))                  ; will print "Hello!!!!!"</lang>

Component Pascal

<lang oberon2> VAR str1: ARRAY 128 OF CHAR; str2: ARRAY 32 OF CHAR; str3: ARRAY 25 OF CHAR; </lang> ... <lang oberon2> str1 := "abcdefghijklmnopqrstuvwxyz";

       str3 := str1; (* don't compile, incompatible assignement *)
       str3 := str1$; (* runtime error, string too long *)
       str2 := str1$; (* OK *)

</lang>

Computer/zero Assembly

Assuming a string to be a zero-terminated array of bytes, this program takes a string beginning at address src and makes a copy of it beginning at address dest. As an example, we copy the string "Rosetta". <lang czasm>ldsrc: LDA src stdest: STA dest

       BRZ  done  ; 0-terminated

       LDA  ldsrc
       ADD  one
       STA  ldsrc

       LDA  stdest
       ADD  one
       STA  stdest

       JMP  ldsrc

done: STP

one: 1

src: 82 ; ASCII

dest:</lang>

Crystal

<lang crystal>s1 = "Hello" s2 = s1</lang>

D

<lang d>void main() {

   string src = "This is a string";

   // copy contents:
   auto dest1 = src.idup;

   // copy contents to mutable char array
   auto dest2 = src.dup;

   // copy just the fat reference of the string
   auto dest3 = src;

}</lang>

dc

<lang dc>[a string] # push "a string" on the main stack d # duplicate the top value f # show the current contents of the main stack</lang>

Output:

a string
a string

Delphi

Delphi strings are reference counted with copy on write semantics.

<lang Delphi>program CopyString;

{$APPTYPE CONSOLE}

var

 s1: string;
 s2: string;

begin

 s1 := 'Goodbye';
 s2 := s1; // S2 points at the same string as S1
 s2 := s2 + ', World!'; // A new string is created for S2

 Writeln(s1);
 Writeln(s2);

end.</lang>

Output:

Goodbye
Goodbye, World!

DWScript

DWScript strings are value-type, from the language point of view, you can't have a reference to a String, no more than you can have a reference to an Integer or a Float (unless you wrap in an object of course).

Internally they're transparently implemented via either immutable reference or copy-on-write.

Dyalect

Strings in Dyalect are immutable:

<lang dyalect>var src = "foobar" var dst = src</lang>

Déjà Vu

In Déjà Vu, strings are immutable, so there really isn't a good reason to copy them. As such, no standard way of doing so is provided. However, one can still create a copy of a string by concatenating it with an empty string. <lang dejavu>local :orgininal "this is the original" local :scopy concat( original "" ) !. scopy</lang>

Output:

"this is the original"

E

E is a pass-references-by-value object-oriented language, and strings are immutable, so there is never a need for or benefit from copying a string. Various operations, such as taking the substring (run) from the beginning to the end (someString.run(0)) might create a copy, but this is not guaranteed.

EasyLang

<lang>a$ = "hello" b$ = a$</lang>

EchoLisp

Strings are immutable. A copy will return the same object. <lang scheme> (define-syntax-rule (string-copy s) (string-append s)) ;; copy = append nothing

   → #syntax:string-copy

(define s "abc") (define t (string-copy s))

   t → "abc"

(eq? s t) → #t ;; same reference, same object </lang>

EDSAC order code

Expects the final character of a string to be marked with a 1 in the least significant bit, as in Hello world/Line printer#EDSAC order code. The source string should be loaded at θ+34; it is copied into storage tank 6. The copy is then printed out. <lang edsac>[ Copy a string

 =============

 A program for the EDSAC

 Copies the source string into storage
 tank 6, which is assumed to be free,
 and then prints it from there

 Works with Initial Orders 2 ]

       T56K
       GK

[ 0 ] A34@ [ copy the string ] [ 1 ] T192F [ 2 ] H34@

[ 17 ] O192F [ print the copy ] [ 18 ] H192F

[ 30 ] ZF [ 31 ] PF [ 32 ] PD [ 33 ] P1F [ 34 ] *F

       RF
       OF
       SF
       EF
       TF
       TF
       AF
       !F
       CF
       OF
       DF
       ED

       EZPF</lang>

Output:

ROSETTA CODE

Elena

<lang elena> var src := "Hello"; var dst := src; // copying the reference var copy := src.clone(); // copying the content </lang>

Elixir

<lang elixir>src = "Hello" dst = src</lang>

Emacs Lisp

Erlang

<lang erlang>Src = "Hello". Dst = Src.</lang>

Euphoria

Works with: Euphoria version 4.0.3, 4.0.0 RC1 and later

Arrays in many languages are constrained to have a fixed number of elements, and those elements must all be of the same type. Euphoria eliminates both of those restrictions by defining all arrays (sequences) as a list of zero or more Euphoria objects whose element count can be changed at any time. When you retrieve a sequence element, it is not guaranteed to be of any type. You, as a programmer, need to check that the retrieved data is of the type you'd expect, Euphoria will not. The only thing it will check is whether an assignment is legal. For example, if you try to assign a sequence to an integer variable, Euphoria will complain at the time your code does the assignment.

<lang Euphoria>sequence first = "ABC" sequence newOne = first</lang>

F#

.NET strings are immutable, so it is usually not useful to make a deep copy. However if needed, it is possible using a static method of the System.String type: <lang fsharp>let str = "hello" let additionalReference = str let deepCopy = System.String.Copy( str )

printfn "%b" <| System.Object.ReferenceEquals( str, additionalReference ) // prints true printfn "%b" <| System.Object.ReferenceEquals( str, deepCopy ) // prints false</lang>

Factor

Factor strings are mutable but not growable. Strings will be immutable in a future release.

<lang factor>"This is a mutable string." dup ! reference "Let's make a deal!" dup clone ! copy "New" " string" append . ! new string

   "New string"</lang>

Factor string buffers (sbufs) are mutable and growable.

<lang factor>SBUF" Grow me!" dup " OK." append

   SBUF" Grow me!  OK."</lang>

Convert a string buffer to a string.

<lang factor>SBUF" I'll be a string someday." >string .

   "I'll be a string someday."</lang>

Forth

Forth strings are generally stored in memory as prefix counted string, where the first byte contains the string length. However, on the stack they are most often represented as <addr cnt> pairs. Thus the way you copy a string depends on where the source string comes from:

<lang forth>\ Allocate two string buffers create stringa 256 allot create stringb 256 allot

\ Copy a constant string into a string buffer s" Hello" stringa place

\ Copy the contents of one string buffer into another stringa count stringb place</lang>

Fortran

Because Fortran uses fixed length character strings if str1 is shorter than str2 then str2 is padded out with trailing spaces. If str1 is longer than str2 it is truncated to fit.

FreeBASIC

<lang freebasic>' FB 1.05.0 Win64

Dim s As String = "This is a string" Dim t As String = s ' a separate copy of the string contents has been made as can be seen from the addresses Print s, StrPtr(s) Print t, StrPtr(t) ' to refer to the same string a pointer needs to be used Dim u As String Ptr = @s Print Print *u, StrPtr(*u) Sleep</lang>

Output:

This is a string            10623504
This is a string            10623552

This is a string            10623504

Frink

Strings are immutable after construction, so "copying" a string just creates a new reference to a string. All string manipulation routines return a new string. <lang frink> a = "Monkey" b = a </lang>

FutureBasic

<lang> include "ConsoleWindow"

dim as Str15 s, c s = "Hello!" c = s print c </lang>

Output:

"Hello!"

Gambas

Note that the DIM statement is required in Gambas.

Click this link to run this code <lang gambas>Public Sub main() Dim src As String Dim dst As String

src = "Hello" dst = src

Print src Print dst End</lang>

GAP

<lang gap>#In GAP strings are lists of characters. An affectation simply copy references a := "more"; b := a; b{[1..4]} := "less"; a;

"less"

Here is a true copy

a := "more"; b := ShallowCopy(a); b{[1..4]} := "less"; a;

"more"</lang>

GML

<lang GML>src = "string"; dest = src;</lang>

Go

Just use assignment: <lang go>src := "Hello" dst := src</lang> Strings in Go are immutable. Because of this, there is no need to distinguish between copying the contents and making an additional reference. Technically, Go strings are immutable byte slices. A slice is an object that contains a reference to an underlying array. In the assignment shown above, a new slice object is created for dst. Its internal reference is likely to point to the same underlying array as src, but the language does not specify this behavior or make any guarantees about it.

<lang go>package main

import "fmt"

func main() { // creature string var creature string = "shark" // point to creature var pointer *string = &creature // creature string fmt.Println("creature =", creature) // creature = shark // creature location in memory fmt.Println("pointer =", pointer) // pointer = 0xc000010210 // creature through the pointer fmt.Println("*pointer =", *pointer) // *pointer = shark // set creature through the pointer *pointer = "jellyfish" // creature through the pointer fmt.Println("*pointer =", *pointer) // *pointer = jellyfish // creature string fmt.Println("creature =", creature) // creature = jellyfish }</lang>

Groovy

The dynamics of references and object creation are very much the same as in Java. However, the meaning of the equality (==) operator is different in Groovy, so we show those differences here, even though they are not relevant to the actual copying.

Example and counter-example: <lang groovy>def string = 'Scooby-doo-bee-doo' // assigns string object to a variable reference def stringRef = string // assigns another variable reference to the same object def stringCopy = new String(string) // copies string value into a new object, and assigns to a third variable reference</lang>

Test Program: <lang groovy>assert string == stringRef // they have equal values (like Java equals(), not like Java ==) assert string.is(stringRef) // they are references to the same objext (like Java ==) assert string == stringCopy // they have equal values assert ! string.is(stringCopy) // they are references to different objects (like Java !=)</lang>

Caveat Lector: Strings are immutable objects in Groovy, so it is wasteful and utterly unnecessary to ever make copies of them within a Groovy program.

GUISS

<lang guiss>Start.Programs,Accessories,Notepad, Type:Hello world[pling],Highlight:Hello world[pling], Menu,Edit,Copy,Menu,Edit,Paste</lang>

Haskell

In Haskell, every value is immutable, including Strings. So one never needs to copy them; references are shared.

HicEst

<lang hicest>src = "Hello World" dst = src</lang>

i

<lang i>//Strings are immutable in 'i'. software { a = "Hello World" b = a //This copies the string.

a += "s"

print(a) print(b) } </lang>

Icon and Unicon

Strings in Icon are immutable. <lang icon>procedure main()

   a := "qwerty"
   b := a
   b[2+:4] := "uarterl"
   write(a," -> ",b)

end</lang>

Under the covers 'b' is created as a reference to the same string as 'a'; the sub-string assignment creates a new copy of the string. However, there is no way to tell this in the language. While most of the time this is transparent, programs that create very long strings through repeated concatenation need to avoid generating intermediate strings. Instead using a list and concatenating at the last minute can perform much better.

Note that strings are indicated using double quotes. However, single quotes are another type called character sets or csets.

J

<lang j>src =: 'hello' dest =: src</lang>

J has copy-on-write semantics. So both src and dest are references to the same memory, until src changes, at which time dest retains a copy of the original value of src.

Java

In Java, Strings are immutable, so it doesn't make that much difference to copy it. <lang java>String src = "Hello"; String newAlias = src; String strCopy = new String(src);

//"newAlias == src" is true //"strCopy == src" is false //"strCopy.equals(src)" is true</lang>

Instead, maybe you want to create a StringBuffer (mutable string) from an existing String or StringBuffer: <lang java>StringBuffer srcCopy = new StringBuffer("Hello");</lang>

JavaScript

Objects can be copied in JavaScript via simple reassignment. Changes to the properties of one will be reflected in the other: <lang javascript>var container = {myString: "Hello"}; var containerCopy = container; // Now both identifiers refer to the same object

containerCopy.myString = "Goodbye"; // container.myString will also return "Goodbye"</lang>

If you copy property values with reassignment, such as properties of the global object (window in browsers), only the value will be copied and not the reference <lang javascript>var a = "Hello"; var b = a; // Same as saying window.b = window.a

b = "Goodbye" // b contains a copy of a's value and a will still return "Hello"</lang>

Joy

<lang joy>"hello" dup</lang>

Strings are immutable.

jq

jq is a functional language and all data types, including strings, are immutable. If a string were to be copied (e.g. by exploding and imploding it), the resultant string would be equal in all respects to the original, and from the jq programmer's perspective, the two would be identical.

jq does however have a type of variable, though their values actually don't change -- they are just context-dependent. For example, consider the sequence of steps in the following function:<lang jq>def demo:

 "abc" as $s    # assignment of a string to a variable
 | $s as $t     # $t points to the same string as $s
 | "def" as $s  # This $s shadows the previous $s
 | $t           # $t still points to "abc"

demo </lang>

Output:

"abc"

Julia

Strings are immutable in Julia. Assignment of one string valued variable to another is effectively a copy, as subsequent changes to either variable have no effect on the other. <lang Julia> s = "Rosetta Code" t = s

println("s = \"", s, "\" and, after \"t = s\", t = \"", t, "\"")

s = "Julia at "*s

println("s = \"", s, "\" and, after this change, t = \"", t, "\"") </lang>

Output:

s = "Rosetta Code" and, after "t = s", t = "Rosetta Code"
s = "Julia at Rosetta Code" and, after this change, t = "Rosetta Code"

KonsolScript

<lang KonsolScript>Var:String str1 = "Hello"; Var:String str2 = str1;</lang>

Kotlin

<lang scala>val s = "Hello" val alias = s // alias === s val copy = "" + s // copy !== s</lang>

LabVIEW

In LabVIEW, one can simply wire an input to more than one output.
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.

Lang5

<lang lang5>'hello dup</lang>

Lasso

While other datatypes like arrays require ->asCopy & ->asCopyDeep methods, assigning strings creates a copy, not a reference, as is seen below. <lang Lasso>local(x = 'I saw a rhino!') local(y = #x)

x //I saw a rhino!

'\r'

y //I saw a rhino!

'\r\r'

x = 'I saw one too'
x //I saw one too

'\r'

y //I saw a rhino!

'\r\r'

y = 'it was grey.'
x //I saw one too

'\r'

y //it was grey.</lang>

Latitude

Strings are immutable in Latitude, so it is seldom necessary to explicitly copy one. However, a copy can be distinguished from the original using === <lang latitude>a := "Hello". b := a. c := a clone. println: a == b. ; True println: a == c. ; True println: a === b. ; True println: a === c. ; False</lang>

LC3 Assembly

Copying a string is the same as copying any other zero-terminated array. This program copies the string at SRC to COPY, then prints the copy to show it has worked. <lang lc3asm> .ORIG 0x3000

       LEA        R1,SRC
       LEA        R2,COPY

LOOP LDR R3,R1,0

       STR        R3,R2,0
       BRZ        DONE
       ADD        R1,R1,1
       ADD        R2,R2,1
       BRNZP      LOOP

DONE LEA R0,COPY

       PUTS

       HALT

SRC .STRINGZ "What, has this thing appeared again tonight?"

COPY .BLKW 128

       .END</lang>

Output:

What, has this thing appeared again tonight?

LFE

<lang lisp>(let* ((a '"data assigned to a")

      (b a)) 
 (: io format '"Contents of 'b': ~s~n" (list b)))</lang>

Output:

Contents of 'b': data assigned to a

One can also use set to copy a sting when one is in the LFE REPL:

Liberty BASIC

<lang lb>src$ = "Hello" dest$ = src$ print src$ print dest$ </lang>

Lingo

<lang lingo>str = "Hello world!" str2 = str</lang>

Syntax-wise strings are not immuatable in Lingo. You can alter an existing string without new assignment:

<lang lingo>put "X" before str put "X" after str put "X" into char 6 of str put str -- "XHellX world!X"</lang>

But memory-wise they are immutable: Lingo internally stores references to strings, and as soon as a string is altered, a new copy is created on the fly, so other references to the original string are not affected by the change.

Lisaac

<lang Lisaac>+ scon : STRING_CONSTANT; + svar : STRING;

scon := "sample"; svar := STRING.create 20; svar.copy scon; svar.append "!\n";

svar.print;</lang> STRING_CONSTANT is immutable, STRING is not.

Little

<lang C>string a = "A string"; string b = a; a =~ s/$/\./; puts(a); puts(b);</lang>

LiveCode

<lang LiveCode>put "foo" into bar put bar into baz answer bar && baz</lang>

Copies are nearly always made, on function calls parameters may be passed by reference (pointer) by prepending @ to a parameter in the function definition, however this is the only case where it is usually performed.

Logo

As a functional language, words are normally treated as symbols and cannot be modified. The EQUAL? predicate compares contents instead of identity. In UCB Logo the .EQ predicate tests for "thing" identity. <lang logo>make "a "foo make "b "foo print .eq :a :b ; true, identical symbols are reused

make "c :a print .eq :a :c ; true, copy a reference

make "c word :b "|| ; force a copy of the contents of a word by appending the empty word print equal? :b :c ; true print .eq :b :c ; false</lang>

Lua

Lua strings are immutable, so only one reference to each string exists. <lang lua> a = "string" b = a print(a == b) -->true print(b) -->string</lang>

Maple

In Maple, you cannot really copy a string in the sense that there can be two copies of the string in memory. As soon as you create a second copy of a string that already exists, it get turned into a reference to the first copy. However, you can copy a reference to a string by a simple assignment statement. <lang Maple> > s := "some string";

                          s := "some string"

> t := "some string";

                          t := "some string"

> evalb( s = t ); # they are equal

                                 true

> addressof( s ) = addressof( t ); # not just equal data, but the same address in memory

                       3078334210 = 3078334210

> u := t: # copy reference </lang>

Mathematica / Wolfram Language

<lang Mathematica>a="Hello World" b=a</lang>

MATLAB

<lang MATLAB>string1 = 'Hello'; string2 = string1;</lang>

Maxima

<lang maxima>/* It's possible in Maxima to access individual characters by subscripts, but it's not the usual way. Also, the result is "Lisp character", which cannot be used by other Maxima functions except cunlisp. The usual way to access characters is charat, returning a "Maxima character" (actually a one characte string). With the latter, it's impossible to modify a string in place, thus scopy is of little use. */

a: "loners"$ b: scopy(a)$ c: a$

c[2]: c[5]$

a; "losers"

b; "loners"

c; "losers"</lang>

MAXScript

<lang maxscript>str1 = "Hello" str2 = copy str1</lang>

Metafont

Metafont will always copy a string (does not make references).

<lang metafont>string s, a; s := "hello"; a := s; s := s & " world"; message s; % writes "hello world" message a; % writes "hello" end</lang>

MiniScript

<lang MiniScript>phrase = "hi" copy = phrase print phrase print copy</lang>

MIPS Assembly

This does a full copy of the string, not just copying the pointer to the string's contents. <lang mips>.data

.text

strcpy:

 addi $sp, $sp, -4
 sw $s0, 0($sp)
 add $s0, $zero, $zero

L1:

 add $t1, $s0, $a1
 lb $t2, 0($t1)
 add $t3, $s0, $a0
 sb $t2, 0($t3)
 beq $t2, $zero, L2
 addi $s0, $s0, 1
 j L1

L2:

 lw $s0, 0($sp)
 addi $sp, $sp, 4
 jr $ra

</lang>

Mirah

<lang mirah>src = "Hello" new_alias = src

puts 'interned strings are equal' if src == new_alias

str_copy = String.new(src) puts 'non-interned strings are not equal' if str_copy != src puts 'compare strings with equals()' if str_copy.equals(src) </lang>

Modula-3

Strings in Modula-3 have the type TEXT. <lang modula3>VAR src: TEXT := "Foo"; VAR dst: TEXT := src;</lang>

MUMPS

<lang>SET S1="Greetings, Planet" SET S2=S1</lang>

Nanoquery

<lang nanoquery>a = "Hello" b = a</lang>

Neko

<lang Neko>var src = "Hello" var dst = src</lang>

Nemerle

Nemerle gives you the option of declaring a variable - even a string - as mutable, so the caveats of languages with only immutable strings don't necessarily apply. However, Nemerle binds the value of the string to the new name when copying; to sort of emulate copying a reference you can use lazy evaluation. <lang Nemerle>using System; using System.Console; using Nemerle;

module StrCopy {

   Main() : void
   {
       mutable str1 = "I am not changed";      // str1 is bound to literal
       def str2 = lazy(str1);                  // str2 will be bound when evaluated
       def str3 = str1;                        // str3 is bound to value of str1
       str1 = "I am changed";                  // str1 is bound to new literal
       Write($"$(str1)\n$(str2)\n$(str3)\n");  // str2 is bound to value of str1
       // Output: I am changed
       //         I am changed
       //         I am not changed
   }

}</lang>

NetRexx

In addition to the string capabilities provided by the Java String libraries (see Java for some examples) NetRexx provides comprehensive string capabilities through the built-in Rexx type. Rexx strings can be copied by simple assignment; as follows: <lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary

s1 = 'This is a Rexx string' s2 = s1

s2 = s2.changestr(' ', '_')

say s1 say s2</lang> In this example a string is created, the string is copied then the copy is modified with the changestr built-in function. Finally both strings are displayed to confirm that the original string wasn't modified by the call to changestr.

Output:

This is a Rexx string
This_is_a_Rexx_string

NewLISP

<lang NewLISP>(define (assert f msg) (if (not f) (println msg)))

(setq s "Greetings!" c (copy s)) (reverse c) ; Modifies c in place.

(assert (= s c) "Strings not equal.")

another way
Nehal-Singhal 2018-05-25

> (setq a "abcd") "abcd" > (setq b a) "abcd" > b "abcd" > (= a b) true

</lang>

Nim

<lang nim>var

 c = "This is a string"
 d = c # Copy c into a new string</lang>

NS-HUBASIC

<lang NS-HUBASIC>10 A$ = "HELLO" 20 B$ = A$ 30 A$ = "HI" 40 PRINT A$, B$</lang>

Oberon-2

<lang oberon2>MODULE CopyString; TYPE String = ARRAY 128 OF CHAR; VAR a,b: String;

BEGIN a := "plain string"; COPY(a,b); END CopyString.</lang>

Objeck

<lang objeck>a := "GoodBye!"; b := a;</lang>

Objective-C

Immutable strings - since they are immutable, you may get the same instance with its references count increased. Or, you can get a copy which is mutable if you use mutableCopy. Remember that both copy and mutableCopy return a retained instance. You can also get a copy by doing [NSString stringWithString:] or [[NSString alloc] initWithString:].

Note that both copy and initWithString:/stringWithString: are optimized to return the original string object (possibly retained) if it is immutable.

<lang objc>NSString *original = @"Literal String"; NSString *new = [original copy]; NSString *anotherNew = [NSString stringWithString:original]; NSString *newMutable = [original mutableCopy];</lang>

Mutable strings - you can get either new mutable (if you use mutableCopy) or immutable (if you use copy) string:

<lang objc>NSMutableString *original = [NSMutableString stringWithString:@"Literal String"]; NSString *immutable = [original copy]; NSString *anotherImmutable = [NSString stringWithString:original]; NSMutableString *mutable = [original mutableCopy];</lang>

Copying a CString into an NSString:

<lang objc>const char *cstring = "I'm a plain C string"; NSString *string = [NSString stringWithUTF8String:cstring];</lang>

Copying from data, possibly not null terminated:

<lang objc>char bytes[] = "some data"; NSString *string = [[NSString alloc] initWithBytes:bytes length:9 encoding:NSASCIIStringEncoding];</lang>

And of course, if a C string is needed, you can use standard functions like strcpy.

OCaml

<lang ocaml>let src = "foobar"</lang>
Before OCaml 4.02 (2014), strings were mutable and explicit deep copying was needed: <lang ocaml>let dst = String.copy src</lang>
Between 4.02 and 4.06 (2017), immutable strings were optionally enabled via a flag: -safe-string. A Bytes module was added to provide safe and unsafe mutable views on strings. The two modules were synonymous unless the aforementioned flag was added. <lang ocaml>(* Transition-period synonymy between types, explicit type annotations are just for emphasis *) let dst1 : string = Bytes.copy (src : bytes) let dst2 : bytes = Bytes.copy (src : string) (* fails to compile with -safe-string *)</lang>
After 4.06, immutable strings became the default, Bytes still exists, but its type is now distinct. The only way to get mutable strings and type synonymy back is at configure-time on the compiler itself.
String.copy issues a deprecation warning, and a (shallow) copy would simply be an assignment by default: <lang ocaml>let dst = src</lang> To get a mutable deep-copy still, just convert the string to bytes via Bytes.of_string, which copies for safety, or String.sub/map/init/.. for an immutable copy.

depending on your compiler version, choose the example accordingly.

Octave

Oforth

To make a copy of the reference, just dup the string <lang Oforth>"abcde" dup</lang>

There is no need to copy a string content as strings are immutable. If really needed : <lang Oforth>StringBuffer new "abcde" << </lang>

ooRexx

<lang ooRexx>/* Rexx ***************************************************************

16.05.2013 Walter Pachl
- - - - /

s1 = 'This is a Rexx string' s2 = s1 /* does not copy the string */

Say 's1='s1 Say 's2='s2 i1=s1~identityhash; Say 's1~identityhash='i1 i2=s2~identityhash; Say 's2~identityhash='i2

s2 = s2~changestr('*', '*') /* creates a modified copy */

Say 's1='s1 Say 's2='s2 i1=s1~identityhash; Say 's1~identityhash='i1 i2=s2~identityhash; Say 's2~identityhash='i2</lang>

Output:

s1=This is a Rexx string
s2=This is a Rexx string
s1~identityhash=17587366586244
s2~identityhash=17587366586244
s1=This is a Rexx string
s2=This is a Rexx string
s1~identityhash=17587366586244
s2~identityhash=17587366588032

OxygenBasic

<lang oxygenbasic> string s, t="hello" s=t </lang>

PARI/GP

Assignment in GP always copies. <lang parigp>s1=s</lang>

In PARI, strings can be copied and references can be made. <lang C>GEN string_copy = gcopy(string); GEN string_ref = string;</lang>

Pascal

<lang pascal>program in,out;

type

  pString = ^string;

var

  s1,s2 : string ;
  pStr  : pString ;

begin

  /* direct copy */
  s1 := 'Now is the time for all good men to come to the aid of their party.'
  s2 := s1 ;

  writeln(s1);
  writeln(s2);

  /* By Reference */
  pStr := @s1 ;
  writeln(pStr^);

  pStr := @s2 ;
  writeln(pStr^);

end;</lang>

Perl

To copy a string, just use ordinary assignment:

<lang perl>my $original = 'Hello.'; my $new = $original; $new = 'Goodbye.'; print "$original\n"; # prints "Hello."</lang>

To create a reference to an existing string, so that modifying the referent changes the original string, use a backslash:

<lang perl>my $original = 'Hello.'; my $ref = \$original; $$ref = 'Goodbye.'; print "$original\n"; # prints "Goodbye."</lang>

If you want a new name for the same string, so that you can modify it without dereferencing a reference, assign a reference to a typeglob:

<lang perl>my $original = 'Hello.'; our $alias; local *alias = \$original; $alias = 'Good evening.'; print "$original\n"; # prints "Good evening."</lang>

Note that our $alias, though in most cases a no-op, is necessary under stricture. Beware that local binds dynamically, so any subroutines called in this scope will see (and possibly modify!) the value of $alias assigned here.

To make a lexical variable that is an alias of some other variable, the Lexical::Alias module can be used: <lang perl>use Lexical::Alias; my $original = 'Hello.'; my $alias; alias $alias, $original; $alias = 'Good evening.'; print "$original\n"; # prints "Good evening."</lang>

Phix

Library: Phix/basics

Use of strings is utterly intuitive with no unexpected side effects. For example

string one = "feed"
string two = one     -- (two becomes "feed", one remains "feed")
two[2..3] = "oo"     -- (two becomes "food", one remains "feed")
one[1] = 'n'         -- (two remains "food", one becomes "need")
?{one,two}

Output:

{"need","food"}

Phix variables are reference counted (except for integers). When a simple copy is made, it increases the reference count and shares the data, making it very fast on large sequences and long strings. Attempts to modify any data with a reference count greater than one cause a copy to be made, and all other variables are left unchanged. Strings can be modified "in situ", no problem.

PHP

<lang php>$src = "Hello"; $dst = $src;</lang>

PicoLisp

<lang PicoLisp>(setq Str1 "abcdef") (setq Str2 Str1) # Create a reference to that symbol (setq Str3 (name Str1)) # Create new symbol with name "abcdef"</lang>

Pike

<lang pike>int main(){

  string hi = "Hello World.";
  string ih = hi;

}</lang>

PL/I

<lang pli> declare (s1, s2) character (20) varying;

  s1 = 'now is the time';
  s2 = s1;</lang>

Pop11

In Pop11 normal data are represented by references, so plain assignment will copy references. To copy data one has to use copy procedure:

<lang pop11>vars src, dst; 'Hello' -> src; copy(src) -> dst;</lang>

One can also combine assignment (initialization) with variable declarations:

<lang pop11>vars src='Hello'; vars dst=copy(src);</lang>

PostScript

In PostScript, <lang postscript>(hello) dup length string copy</lang>

PowerShell

Since PowerShell uses .NET behind the scenes and .NET strings are immutable you can simply assign the same string to another variable without breaking anything: <lang powershell>$str = "foo" $dup = $str</lang> To actually create a copy the Clone() method can be used: <lang powershell>$dup = $str.Clone()</lang>

ProDOS

<lang ProDOS>editvar /newvar /value=a /userinput=1 /title=Enter a string to be copied: editvar /newvar /value=b /userinput=1 /title=Enter current directory of the string: editvar /newvar /value=c /userinput=1 /title=Enter file to copy to: copy -a- from -b- to -c- </lang>

Prolog

Values in Prolog are immutable so unifying with a variable that already has the value of a string will effectively copy that string. You cannot reassign a value once it has been unified, it is not logical to have a value equal more than one thing. <lang prolog>?- A = "A test string", A = B. A = B, B = "A test string".</lang>

PureBasic

<lang PureBasic>src$ = "Hello" dst$ = src$</lang>

Python

Works with: Python version 2.3, 2.4, and 2.5

Since strings are immutable, all copy operations return the same string, with the reference count increased as appropriate

<lang python>>>> src = "hello" >>> a = src >>> b = src[:] >>> import copy >>> c = copy.copy(src) >>> d = copy.deepcopy(src) >>> src is a is b is c is d True</lang>

To actually copy a string:

<lang python>>>> a = 'hello' >>> b = .join(a) >>> a == b True >>> b is a ### Might be True ... depends on "interning" implementation details! False</lang>

As a result of object "interning" some strings such as the empty string and single character strings like 'a' may be references to the same object regardless of copying. This can potentially happen with any Python immutable object and should be of no consequence to any proper code.

Be careful with is - use it only when you want to compare the identity of the object. To compare string values, use the == operator. For numbers and strings any given Python interpreter's implementation of "interning" may cause the object identities to coincide. Thus any number of names to identical numbers or strings might become references to the same objects regardless of how those objects were derived (even if the contents were properly "copied" around). The fact that these are immutable objects makes this a reasonable behavior.

Quackery

Translation of: Joy

Strings are immutable. <lang Quackery>$ "hello" dup</lang>

R

Copy a string by value: <lang R>str1 <- "abc" str2 <- str1</lang>

Racket

lang racket

(let* ([s1 "Hey"]

      [s2 s1]
      [s3 (string-copy s1)]
      [s4 s3])
 (printf "s1 and s2 refer to ~a strings\n"
         (if (eq? s1 s2) "the same" "different")) ; same
 (printf "s1 and s3 refer to ~a strings\n"
         (if (eq? s1 s3) "the same" "different")) ; different
 (printf "s3 and s4 refer to ~a strings\n"
         (if (eq? s3 s4) "the same" "different")) ; same
 (string-fill! s3 #\!)
 (printf "~a~a~a~a\n" s1 s2 s3 s4)) ; outputs "HeyHey!!!!!!"

</lang>

Raku

(formerly Perl 6)

There is no special handling needed to copy a string; just assign it to a new variable: <lang perl6>my $original = 'Hello.'; my $copy = $original; say $copy; # prints "Hello." $copy = 'Goodbye.'; say $copy; # prints "Goodbye." say $original; # prints "Hello."</lang>

You can also bind a new variable to an existing one so that each refers to, and can modify the same string. <lang perl6>my $original = 'Hello.'; my $bound := $original; say $bound; # prints "Hello." $bound = 'Goodbye.'; say $bound; # prints "Goodbye." say $original; # prints "Goodbye."</lang>

Raven

Copy a string by reference:

Copy a string by value:

REBOL

<lang REBOL>REBOL [

   Title: "String Copy"
   URL: http://rosettacode.org/wiki/Copy_a_string

]

x: y: "Testing." y/2: #"X" print ["Both variables reference same string:" mold x "," mold y]

x: "Slackeriffic!" print ["Now reference different strings:" mold x "," mold y]

y: copy x ; String copy here! y/3: #"X" ; Modify string. print ["x copied to y, then modified:" mold x "," mold y]

y: copy/part x 7 ; Copy only the first part of y to x. print ["Partial copy:" mold x "," mold y]

y: copy/part skip x 2 3 print ["Partial copy from offset:" mold x "," mold y]</lang>

Output:

Script: "String Copy" (16-Dec-2009)
Both variables reference same string: "TXsting." , "TXsting."
Now reference different strings: "Slackeriffic!" , "TXsting."
x copied to y, then modified: "Slackeriffic!" , "SlXckeriffic!"
Partial copy: "Slackeriffic!" , "Slacker"
Partial copy from offset: "Slackeriffic!" , "ack"

Red

<lang Red> Red[] originalString: "hello wordl" copiedString: originalString

OR

copiedString2: copy originalString </lang>

Retro

<lang Retro>'this_is_a_string dup s:temp</lang>

REXX

The example shows how to copy the contents of one string into another string.

Note that delimiters for literal strings, REXX accepts either of:

' (an apostrophe)
" (a double quote)

Also note that all REXX values (variables) are stored as (varying length) character strings. <lang rexx>src = "this is a string" dst = src</lang>

Ring

<lang ring> cStr1 = "Hello!" # create original string cStr2 = cStr1 # make new string from original </lang>

RLaB

<lang RLaB>>> s1 = "A string" A string >> s2 = s1 A string</lang>

Robotic

<lang robotic> set "$string1" to "This is a string" set "$string2" to "$string1"

"&$string2&"

</lang>

Ruby

In Ruby, String are mutable. <lang ruby>original = "hello" reference = original # copies reference copy1 = original.dup # instance of original.class copy2 = String.new(original) # instance of String

original << " world!" # append p reference #=> "hello world!" p copy1 #=> "hello" p copy2 #=> "hello"</lang>

There is a method of Object#clone, too, in the copy of the object. <lang ruby>original = "hello".freeze # prevents further modifications copy1 = original.dup # copies contents (without status) copy2 = original.clone # copies contents (with status) p copy1.frozen? #=> false p copy1 << " world!" #=> "hello world!" p copy2.frozen? #=> true p copy2 << " world!" #=> can't modify frozen String (RuntimeError)</lang>

Run BASIC

<lang runbasic>origString$ = "Hello!" ' create original string newString$ = origString$ ' make new strig from original</lang>

Rust

<lang rust>fn main() {

   let s1 = "A String";
   let mut s2 = s1;

   s2 = "Another String";

   println!("s1 = {}, s2 = {}", s1, s2);

}</lang>

Output: <lang>s1 = A String, s2 = Another String</lang>

Sather

<lang sather>class MAIN is

 main is
   s  ::= "a string";
   s1 ::= s;
   -- s1 is a copy
 end;

end;</lang>

Scala

<lang scala> val src = "Hello"

 // Its actually not a copy but a reference
 // That is not a problem because String is immutable
 // In fact its a feature
 val des = src
 assert(src eq des) // Proves the same reference is used. 
 // To make a real copy makes no sense.
 // Actually its hard to make a copy, the compiler is too smart.
 // mkString, toString makes also not a real copy
 val cop = src.mkString.toString
 assert((src eq cop))                 // Still no copyed image
 val copy = src.reverse.reverse       // Finally double reverse makes a copy
 assert(src == copy && !(src eq copy))// Prove, but it really makes no sense.</lang>

Scheme

<lang scheme>(define dst (string-copy src))</lang>

Seed7

<lang seed7>var string: dest is "";

dest := "Hello";</lang>

SenseTalk

<lang sensetalk>(* In SenseTalk, assignments normally always make copies of values. *)

put "glorious" into myWord put myWord into yourWord

(* Assignments can also be made by reference if desired. *)

put a reference to myWord into myRef set another to refer to myRef

put "ly" after myWord put "in" before another

put "myWord: " & myWord put "yourWord: " & yourWord put "myRef: " & myRef put "another: " & another </lang>

Output:

myWord:   ingloriously
yourWord: glorious
myRef:    ingloriously
another:  ingloriously

Shiny

<lang shiny>src: 'hello' cpy: src</lang>

Sidef

<lang ruby>var original = "hello"; # new String object var reference = original; # points at the original object var copy1 = String.new(original); # creates a new String object var copy2 = original+; # ==//==</lang>

Simula

<lang simula>BEGIN

   TEXT ORIGINAL, REFERENCE, COPY1;

   ORIGINAL :- "THIS IS CONSTANT TEXT";
   ORIGINAL.SETPOS(1);
   REFERENCE :- ORIGINAL;

! RUN TIME ERROR: ! ORIGINAL.PUTCHAR('X'); ! "copy-a-string.sim", line 9: ./copy-a-string: Putchar: Constant text object

   OUTTEXT(ORIGINAL);
   OUTIMAGE;

   ! CONTENT EQUAL? => T ;
   OUTTEXT(IF ORIGINAL = REFERENCE THEN "T" ELSE "F");
   OUTIMAGE;

   ! SAME TEXT OBJECT? => T ;
   OUTTEXT(IF ORIGINAL == REFERENCE THEN "T" ELSE "F");
   OUTIMAGE;

   COPY1 :- COPY(ORIGINAL);
   COPY1.SETPOS(1);
   COPY1.PUTCHAR('X');
   OUTTEXT(COPY1);
   OUTIMAGE;

END;</lang>

Output:

THIS IS CONSTANT TEXT
T
T
XHIS IS CONSTANT TEXT

Slate

<lang slate>[ | :s | s == s copy] applyTo: {'hello'}. "returns False"</lang>

Smalltalk

<lang smalltalk>|s1 s2| "bind the var s1 to the object string on the right" s1 := 'i am a string'. "bind the var s2 to the same object..." s2 := s1. "bind s2 to a copy of the object bound to s1" s2 := (s1 copy).</lang>

SNOBOL4

copy a to b

         b = a = "test"
         output = a
         output = b

change the copy

         b "t" = "T"
         output = b

end</lang>

Output:

  test
  test
  Test

Standard ML

In Standard ML, strings are immutable, so you don't copy it.

Instead, maybe you want to create a CharArray.array (mutable string) from an existing string: <lang sml>val src = "Hello"; val srcCopy = CharArray.array (size src, #"x"); (* 'x' is just dummy character *) CharArray.copyVec {src = src, dst = srcCopy, di = 0}; src = CharArray.vector srcCopy; (* evaluates to true *)</lang>

or from another CharArray.array: <lang sml>val srcCopy2 = CharArray.array (CharArray.length srcCopy, #"x"); (* 'x' is just dummy character *) CharArray.copy {src = srcCopy, dst = srcCopy2, di = 0};</lang>

Swift

Just use assignment: <lang swift>var src = "Hello" var dst = src</lang> Strings in Swift are value types, so assigning copies the string.

Tcl

<lang tcl>set src "Rosetta Code" set dst $src</lang> Tcl copies strings internally when needed. To be exact, it uses a basic value model based on simple objects that are immutable when shared (i.e., when they have more than one effective reference to them); when unshared, they can be changed because the only holder of a reference has to be the code requesting the change. At the script level, this looks like Tcl is making a copy when the variable is assigned as above, but is more efficient in the common case where a value is not actually modified.

TI-83 BASIC

<lang ti83b>:"Rosetta Code"→Str1

Str1→Str2</lang>

TI-89 BASIC

<lang ti89b>:"Rosetta Code"→str1

str1→str2</lang>

Toka

<lang toka>" hello" is-data a a string.clone is-data b</lang>

Trith

Strings are immutable character sequences, so copying a string just means duplicating the reference at the top of the stack: <lang trith>"Hello" dup</lang>

TUSCRIPT

<lang tuscript>$$ MODE TUSCRIPT str="Hello" dst=str</lang>

UNIX Shell

<lang bash>foo="Hello" bar=$foo # This is a copy of the string</lang>

Ursa

<lang ursa>decl string a b set a "hello" set b a</lang>

V

dup really makes a reference, but the language is functional, so the string is immutable.

<lang v>"hello" dup</lang>

VBA

This program copies string in variable a to variable b. Mutating variable a subsequently doesn't alter variable b. Variable b is not a reference. <lang vb>Sub copystring()

   a = "Hello World!"
   b = a
   a = "I'm gone"
   Debug.Print b
   Debug.Print a

End Sub</lang>

Output:

Hello World!
I'm gone

Vim Script

<lang vim>let str1 = "original string" let str2 = str1 let str1 = "new string"

echo "String 1:" str1 echo "String 2:" str2</lang>

Output:

String 1: new string                                                            
String 2: original string

Visual Basic .NET

Platform: .NET

Works with: Visual Basic .NET version 9.0+

<lang vbnet>'Immutable Strings Dim a = "Test string" Dim b = a 'reference to same string Dim c = New String(a.ToCharArray) 'new string, normally not used

'Mutable Strings Dim x As New Text.StringBuilder("Test string") Dim y = x 'reference Dim z = New Text.StringBuilder(x.ToString) 'new string</lang>

Alternatively, you can use, with all versions of the .NET framework: <lang vbnet>Dim a As String = "Test String" Dim b As String = String.Copy(a) ' New string</lang>

Wren

A string in Wren is an immutable array of bytes.

Although technically a reference type, this means there is no need to distinguish between copying the contents of a string and making an additional reference. We can therefore just use assignment to copy a string. <lang ecmascript>var s = "wren" var t = s System.print("Are 's' and 't' equal? %(s == t)")</lang>

Output:

Are 's' and 't' equal? true

X86 Assembly

Works with: nasm

creating a second 0 terminated string with the same content: <lang asm> section .data

   string db "Hello World", 0

section .bss

   string2 resb 12

section .text global _main _main:

   mov ecx, 0
   looping:
       mov al, [string + ecx]
       mov [string2 + ecx], al
       inc ecx
       cmp al, 0 ;copy until we find the terminating 0
       je end
       jmp looping
   end:
       xor eax, eax
       ret

</lang>

creating a second string; first byte signals length of string <lang asm> section .data

   string db 11,"Hello World"

section .bss

   string2 resb 12

section .text global _main _main:

   xor ecx, ecx ;clear ecx
   mov cl, [string]
   mov [string2], cl ;copy byte signaling length
   mov edx, 1
   looping: ;copy each single byte
       mov al, [string + edx]
       mov [string2 + edx], al
       inc edx
       dec ecx
       cmp ecx, 0
       jg looping
   xor eax, eax
   ret

</lang>

X86-64 Assembly

UASM 2.52

<lang asm> option casemap:none option literals:on

printf proto :dword, :VARARG exit proto :dword

.data

 s   db "Goodbye, World!",0

.data?

 d   db 20 dup (?)
 dp  dq ?
 tb  dd ?

.code main proc

 lea rsi, s                    ;; Put the address of var S into the source index(RSI)
 xor rcx, rcx                  ;; Zero out RCX
 _getsize:
   inc rcx                     ;; Advanced the index by 1
   cmp byte ptr [rsi+rcx],0    ;; check the current byte for terminating 0
   jne _getsize                ;; nope, jump back and check again

 mov tb, ecx                   ;; tb = Total bytes, Keep a copy of the size of the string
 lea rsi, s                    ;; Copy the address of s into the source index(rsi)
 lea rdi, d                    ;; Copy the address of d into the destination index(rdi)
 rep movsb                     ;; Copy bytes from ESI to EDI until RCX is 0

 lea rax, s                    ;; Get the address of S
 mov dp, rax                   ;; Copy it from RAX to dp

 mov rbx,rdi                   ;; Make a copy of RDI, cause over writes due to ABI call args T_T
 invoke printf, CSTR("-> s (0x%x) = %s",10), rsi, addr s
 invoke printf, CSTR("-> d (0x%x) = %s",10), rbx, addr d
 invoke printf, CSTR("-> dp (0x%x) = %s",10), addr dp, dp
 invoke printf, CSTR("-> bytes copied: %i",10), tb
 xor rsi, rsi
 call exit
 ret

main endp

end </lang>

Output:

-> s (0x40303f) = Goodbye, World!
-> d (0x40309f) = Goodbye, World!
-> dp (0x4030a4) = Goodbye, World!
-> bytes copied: 15

XPL0

The default method of terminating strings is to set the most significant bit of the last character. An alternative is to use the 'string 0' command to specify zero-terminated strings. The string copy routine from the standard library is shown.

<lang XPL0>proc StrCopy(A, B); \Copy string: A --> B char A, B; \Strings: B must already have enough space "Reserved" int I; \Beware if strings overlap for I:= 0 to -1>>1-1 do

   [B(I):= A(I);
   if A(I) >= $80 then return
   ];

char S1, S2, S3(13); [S1:= "Hello, world!"; \S1 now points to the string S2:= S1; \S2 now also points to the string StrCopy(S1, S3); \S3 points to a separate copy of the string ]</lang>

Zig

<lang zig>const std = @import("std");

const debug = std.debug; const mem = std.mem;

test "copy a string" {

   const source = "A string.";

   // Variable `dest1` will have the same type as `source`, which is
   // `*const [9:0]u8`.
   const dest1 = source;

   // Variable `dest2`'s type is [9]u8.
   //
   // The difference between the two is that `dest1` string is null-terminated,
   // while `dest2` is not.
   var dest2: [source.len]u8 = undefined;
   mem.copy(u8, dest2[0..], source[0..]);

   debug.assert(mem.eql(u8, dest1[0..], "A string."));
   debug.assert(mem.eql(u8, dest2[0..], "A string."));

}</lang>