Even or odd

Even or odd
You are encouraged to solve this task according to the task description, using any language you may know.

Test whether an integer is even or odd.

There is more than one way to solve this task:

• Use the even and odd predicates, if the language provides them.
• Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.
• Divide i by 2. The remainder equals 0 iff i is even. The remainder equals +1 or -1 iff i is odd.
• Use modular congruences:
• i ≡ 0 (mod 2) iff i is even.
• i ≡ 1 (mod 2) iff i is odd.

0815

` }:s:|=<:2:x~#:e:=/~%~<:20:~\$=<:73:x<:69:~\$~\$~<:20:~\$=^:o:<:65:x<:76:=\$=\$~\$<:6E:~\$<:a:~\$^:s:}:o:<:6F:x<:64:x~\$~\$\$<:a:~\$^:s: `

6502 Assembly

`         .lf  evenodd6502.lst        .cr  6502        .tf  evenodd6502.obj,ap1;------------------------------------------------------; Even or Odd for the 6502 by barrym95838 2014.12.10; Thanks to sbprojects.com for a very nice assembler!; The target for this assembly is an Apple II with;   mixed-case output capabilities.  Apple IIs like to;   work in '+128' ascii, and this version is tailored;   to that preference.; Tested and verified on AppleWin 1.20.0.0;------------------------------------------------------; Constant Section;CharIn   =   \$fd0c      ;Specific to the Apple IICharOut  =   \$fded      ;Specific to the Apple II;------------------------------------------------------; The main program;main    ldy  #sIntro-sbase        jsr  puts       ;Print Introloop    jsr  CharIn     ;Get a char from stdin        cmp  #\$83       ;Ctrl-C?        beq  done       ;  yes:  end program        jsr  CharOut    ;Echo char        ldy  #sOdd-sbase ;Pre-load odd string        lsr             ;LSB of char to carry flag        bcs  isodd        ldy  #sEven-sbaseisodd   jsr  puts       ;Print appropriate response        beq  loop       ;Always taken; Output NUL-terminated string @ offset Y;puts    lda  sbase,y    ;Get string char        beq  done       ;Done if NUL        jsr  CharOut    ;Output the char        iny             ;Point to next char        bne  puts       ;Loop up to 255 timesdone    rts             ;Return to caller;------------------------------------------------------; String Constants (in '+128' ascii, Apple II style);sbase:                  ;String base addresssIntro  .az     -"Hit any key (Ctrl-C to quit):",-#13sEven   .az     -" is even.",-#13sOdd    .az     -" is odd.",-#13;------------------------------------------------------        .en `

8th

The 'mod' method also works, but the bit method is fastest.

`: odd? \ n -- boolean    dup 1 n:band 1 n:= ;: even? \ n -- boolean     odd? not ;`

This could be shortened to:

` : even? \ n -- f  1 n:band not ;: odd? \ n -- f  even? not ; `

ABAP

` cl_demo_output=>display(  VALUE string_table(    FOR i = -5 WHILE i < 6 (      COND string(        LET r = i MOD 2 IN        WHEN r = 0 THEN |{ i } is even|        ELSE |{ i } is odd|      )    )  )). `
Output:
```Table
-5 is odd
-4 is even
-3 is odd
-2 is even
-1 is odd
0 is even
1 is odd
2 is even
3 is odd
4 is even
5 is odd
```

`-- Ada has bitwise operators in package Interfaces,-- but they work with Interfaces.Unsigned_*** types only.-- Use rem or mod for Integer types, and let the compiler-- optimize it.declare   N : Integer := 5;begin   if N rem 2 = 0 then      Put_Line ("Even number");   elseif N rem 2 /= 0 then      Put_Line ("Odd number");   else      Put_Line ("Something went really wrong!");   end if;end;`

Aime

`if (x & 1) {    # x is odd} else {    # x is even}`

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.win32
`# Algol 68 has a standard operator: ODD which returns TRUE if its integer  ## operand is odd and FALSE if it is even                                   ## E.g.:                                                                    # INT n;print( ( "Enter an integer: " ) );read( ( n ) );print( ( whole( n, 0 ), " is ", IF ODD n THEN "odd" ELSE "even" FI, newline ) ) `

ALGOL W

`begin    % the Algol W standard procedure odd returns true if its integer  %    % parameter is odd, false if it is even                           %    for i := 1, 1702, 23, -26    do begin        write( i, " is ", if odd( i ) then "odd" else "even" )    end for_iend.`
Output:
```             1   is odd
1702   is even
23   is odd
-26   is even
```

AntLang

`odd: {x mod 2}even: {1 - x mod 2}`

APL

The easiest way is probably to use modulo.

`      2|280      2|371`

AppleScript

`set nList to {3, 2, 1, 0, -1, -2, -3}repeat with n in nList    if (n / 2) = n / 2 as integer then        log "Value " & n & " is even."    else        log "Value " & n & " is odd."    end ifend repeat`
Output:
```(*Value 3 is odd.*)
(*Value 2 is even.*)
(*Value 1 is odd.*)
(*Value 0 is even.*)
(*Value -1 is odd.*)
(*Value -2 is even.*)
(*Value -3 is odd.*)```

Or, packaging reusable functions that can serve as arguments to filter etc (deriving even from mod, and odd from even):

`-- even :: Integral a => a -> Boolon even(n)    n mod 2 = 0end even -- odd :: Integral a => a -> Boolon odd(n)    not even(n)end odd  -- GENERIC FUNCTIONS FOR TEST ---------------------------------- -- filter :: (a -> Bool) -> [a] -> [a]on filter(f, xs)    tell mReturn(f)        set lst to {}        set lng to length of xs        repeat with i from 1 to lng            set v to item i of xs            if lambda(v, i, xs) then set end of lst to v        end repeat        return lst    end tellend filter -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Scripton mReturn(f)    if class of f is script then        f    else        script            property lambda : f        end script    end ifend mReturn  -- TEST ---------------------------------------------------------on run    set xs to [-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6]     {filter(even, xs), filter(odd, xs)}end run`
Output:
`{{-6, -4, -2, 0, 2, 4, 6}, {-5, -3, -1, 1, 3, 5}}`

Arendelle

```( input , "Please enter a number: " )

{ @input % 2 = 0 ,

"| @input | is even!"
,
"| @input | is odd!"
}```

AutoHotkey

Bitwise ops are probably most efficient:

`if ( int & 1 ){	; do odd stuff}else{	; do even stuff}`

AWK

`function isodd(x) {	return (x%2)!=0; } function iseven(x) {	return (x%2)==0; }`

BaCon

`' Even or oddOPTION MEMTYPE intSPLIT ARGUMENT\$ BY " " TO arg\$ SIZE dimn = IIF\$(dim < 2, 0, VAL(arg\$[1]))PRINT n, " is ", IIF\$(EVEN(n), "even", "odd")`
Output:
```prompt\$ ./even-or-odd 42
42 is even
prompt\$ ./even-or-odd 41
41 is odd```

BASIC

Applesoft BASIC

`10 INPUT "ENTER A NUMBER: ";N20 IF N/2 <> INT(N/2) THEN PRINT "THE NUMBER IS ODD":GOTO 4030 PRINT "THE NUMBER IS EVEN"40 END`
Works with: Commodore BASIC version 2.0

IS-BASIC

`100 DEF ODD(X)=MOD(X,2)110 INPUT PROMPT "Enter a number: ":X120 IF ODD(X) THEN130   PRINT X;"is odd."140 ELSE150   PRINT X;"is even."160 END IF`

Batch File

` @echo offset /p i=Insert number: ::bitwise andset /a "test1=%i%&1"::divide last character by 2set /a test2=%i:~-1%/2::moduloset /a test3=%i% %% 2 set testpause>nul `

BBC BASIC

Solutions using AND or MOD are restricted to 32-bit integers, so an alternative solution is given which works with a larger range of values.

`      IF FNisodd%(14) PRINT "14 is odd" ELSE PRINT "14 is even"      IF FNisodd%(15) PRINT "15 is odd" ELSE PRINT "15 is even"      IF FNisodd#(9876543210#) PRINT "9876543210 is odd" ELSE PRINT "9876543210 is even"      IF FNisodd#(9876543211#) PRINT "9876543211 is odd" ELSE PRINT "9876543211 is even"      END       REM Works for -2^31 <= n% < 2^31      DEF FNisodd%(n%) = (n% AND 1) <> 0       REM Works for -2^53 <= n# <= 2^53      DEF FNisodd#(n#) = n# <> 2 * INT(n# / 2)`
Output:
```14 is even
15 is odd
9876543210 is even
9876543211 is odd
```

bc

There are no bitwise operations, so this solution compares a remainder with zero. Calculation of i % 2 only works when scale = 0.

`i = -3 /* Assumes that i is an integer. */scale = 0if (i % 2 == 0) "i is even"if (i % 2) "i is odd"`

Befunge

`&2%52**"E"+,@`

Outputs E if even, O if odd.

Bracmat

Not the simplest solution, but the cheapest if the number that must be tested has thousands of digits.

`( ( even  =    . @( !arg       :   ?           [-2           ( 0           | 2           | 4           | 6           | 8           )       )  )& (odd=.~(even\$!arg))& ( eventest  =    .   out      \$ (!arg is (even\$!arg&|not) even)  )& ( oddtest  =    .   out      \$ (!arg is (odd\$!arg&|not) odd)  )& eventest\$5556& oddtest\$5556& eventest\$857234098750432987502398457089435& oddtest\$857234098750432987502398457089435)`
Output:
```5556 is even
5556 is not odd
857234098750432987502398457089435 is not even
857234098750432987502398457089435 is odd```

Brainf***

Assumes that input characters are an ASCII representation of a valid integer. Output is input mod 2.

`,[>,----------] Read until newline++<             Get a 2 and move into position[->-[>+>>]>     Do[+[-<+>]>+>>]   divmod<<<<<]          magic>[-]<++++++++   Clear and get an 8[>++++++<-]     to get a 48>[>+<-]>.       to get n % 2 to ASCII and print`

If one need only determine rather than act on the parity of the input, the following is sufficient; it terminates either quickly or never.

`,[>,----------]<[--]`

`2.%`

C

Test by bitwise and'ing 1, works for any builtin integer type as long as it's 2's compliment (it's always so nowadays):

`if (x & 1) {    /* x is odd */} else {    /* or not */}`

If using long integer type from GMP (`mpz_t`), there are provided macros:

`mpz_t x;...if (mpz_even_p(x)) { /* x is even */ }if (mpz_odd_p(x))  { /* x is odd */ }`

The macros evaluate `x` more than once, so it should not be something with side effects.

C#

`namespace RosettaCode{    using System;     public static class EvenOrOdd    {        public static bool IsEvenBitwise(this int number)        {            return (number & 1) == 0;        }         public static bool IsOddBitwise(this int number)        {            return (number & 1) != 0;        }         public static bool IsEvenRemainder(this int number)        {            int remainder;            Math.DivRem(number, 2, out remainder);            return remainder == 0;        }         public static bool IsOddRemainder(this int number)        {            int remainder;            Math.DivRem(number, 2, out remainder);            return remainder != 0;        }         public static bool IsEvenModulo(this int number)        {            return (number % 2) == 0;        }         public static bool IsOddModulo(this int number)        {            return (number % 2) != 0;        }    }}`

C++

Test using the modulo operator, or use the C example from above.

`bool isOdd(int x) {    return x % 2;} bool isEven(int x){    return !(x % 2);}`

A slightly more type-generic version, for C++11 and later. This should theoretically work for any type convertible to `int`:

` template < typename T > constexpr inline bool isEven( const T& v ){    return isEven( int( v ) );} template <> constexpr inline bool isEven< int >( const int& v ){    return (v & 1) == 0;} template < typename T > constexpr inline bool isOdd( const T& v ){    return !isEven(v);} `

Clojure

Standard predicates:

`(if (even? some-var) (do-even-stuff))(if (odd? some-var) (do-odd-stuff))`

COBOL

`       IF FUNCTION REM(Num, 2) = 0           DISPLAY Num " is even."       ELSE           DISPLAY Num " is odd."       END-IF`

CoffeeScript

`isEven = (x) -> !(x%2)`

ColdFusion

` function f(numeric n) {   return n mod 2?"odd":"even"} `

Common Lisp

Standard predicates:

`(if (evenp some-var) (do-even-stuff))(if (oddp some-other-var) (do-odd-stuff))`

Alternate solution

I use Allegro CL 10.1

` ;; Project : Even or odd (defun evenodd (nr)          (cond ((evenp nr) "even")          ((oddp nr) "odd")))(dotimes (n 10) (if (< n 1) (terpri))(if (< n 9) (format t "~a" " "))(write(+ n 1)) (format t "~a" ": ")(format t "~a" (evenodd (+ n 1))) (terpri)) `

Output:

```1: odd
2: even
3: odd
4: even
5: odd
6: even
7: odd
8: even
9: odd
10: even
```

Component Pascal

BlackBox Component Builder

` MODULE EvenOdd;IMPORT StdLog,Args,Strings; PROCEDURE BitwiseOdd(i: INTEGER): BOOLEAN;BEGIN	RETURN 0 IN BITS(i)END BitwiseOdd; PROCEDURE Odd(i: INTEGER): BOOLEAN;BEGIN	RETURN (i MOD 2) # 0END Odd; PROCEDURE CongruenceOdd(i: INTEGER): BOOLEAN;BEGIN	RETURN ((i -1) MOD 2) = 0END CongruenceOdd; PROCEDURE Do*;VAR	p: Args.Params;	i,done,x: INTEGER;BEGIN	Args.Get(p);	StdLog.String("Builtin function: ");StdLog.Ln;i := 0;	WHILE i < p.argc DO		Strings.StringToInt(p.args[i],x,done);		StdLog.String(p.args[i] + " is:> "); 		IF ODD(x) THEN StdLog.String("odd") ELSE StdLog.String("even") END;		StdLog.Ln;INC(i)	END;	StdLog.String("Bitwise: ");StdLog.Ln;i:= 0;	WHILE i < p.argc DO		Strings.StringToInt(p.args[i],x,done);		StdLog.String(p.args[i] + " is:> "); 		IF BitwiseOdd(x) THEN StdLog.String("odd") ELSE StdLog.String("even") END;		StdLog.Ln;INC(i)	END;	StdLog.String("Module: ");StdLog.Ln;i := 0;	WHILE i < p.argc DO		Strings.StringToInt(p.args[i],x,done);		StdLog.String(p.args[i] + " is:> "); 		IF Odd(x) THEN StdLog.String("odd") ELSE StdLog.String("even") END;		StdLog.Ln;INC(i)	END;	StdLog.String("Congruences: ");StdLog.Ln;i := 0;	WHILE i < p.argc DO		Strings.StringToInt(p.args[i],x,done);		StdLog.String(p.args[i] + " is:> "); 		IF CongruenceOdd(x) THEN StdLog.String("odd") ELSE StdLog.String("even") END;		StdLog.Ln;INC(i)	END;END Do; `

Execute: ^Q EvenOdd.Do 10 11 0 57 34 -23 -42~

Output:
```Builtin function:
10  is:> even
11  is:> odd
0  is:> even
57  is:> odd
34  is:> even
-23  is:> odd
-42 is:> even
Bitwise:
10  is:> even
11  is:> odd
0  is:> even
57  is:> odd
34  is:> even
-23  is:> odd
-42 is:> even
Module:
10  is:> even
11  is:> odd
0  is:> even
57  is:> odd
34  is:> even
-23  is:> odd
-42 is:> even
Congruences:
10  is:> even
11  is:> odd
0  is:> even
57  is:> odd
34  is:> even
-23  is:> odd
-42 is:> even
```

Crystal

`#Using bitwise shift  def isEven_bShift(n)    n == ((n >> 1) << 1)  end  def isOdd_bShift(n)    n != ((n >> 1) << 1)  end#Using modulo operator  def isEven_mod(n)    (n % 2) == 0  end  def isOdd_mod(n)    (n % 2) != 0  end# Using bitwise "and"  def isEven_bAnd(n)    (n & 1) ==  0  end  def isOdd_bAnd(n)    (n & 1) != 0  end puts isEven_bShift(7)puts isOdd_bShift(7) puts isEven_mod(12)puts isOdd_mod(12) puts isEven_bAnd(21)puts isOdd_bAnd(21) `
Output:
```false
true
true
false
false
true
```

D

`void main() {    import std.stdio, std.bigint;     foreach (immutable i; -5 .. 6)        writeln(i, " ", i & 1, " ", i % 2, " ", i.BigInt % 2);}`
Output:
```-5 1 -1 -1
-4 0 0 0
-3 1 -1 -1
-2 0 0 0
-1 1 -1 -1
0 0 0 0
1 1 1 1
2 0 0 0
3 1 1 1
4 0 0 0
5 1 1 1```

DCL

`\$! in DCL, for integers, the least significant bit determines the logical value, where 1 is true and 0 is false\$\$ i = -5\$ loop1:\$  if i then \$ write sys\$output i, " is odd"\$  if .not. i then \$ write sys\$output i, " is even"\$  i = i + 1\$  if i .le. 6 then \$ goto loop1`
Output:
```\$ @even_odd
-5 is odd
-4 is even
-3 is odd
-2 is even
-1 is odd
0 is even
1 is odd
2 is even
3 is odd
4 is even
5 is odd
6 is even```

DWScript

Predicate:

`var isOdd := Odd(i);`

Bitwise and:

`var isOdd := (i and 1)<>0;`

Modulo:

`var isOdd := (i mod 2)=1;`

Déjà Vu

`even n:    = 0 % n 2 odd:    not even !. odd 0!. even 0!. odd 7!. even 7 `
Output:
```false
true
true
false```

EDSAC order code

This implementation uses the `C` (logical AND multiplier register with memory) order. It will cause the machine to print an E if the number stored at address θ+15 is even, or an O if it is odd. As an example, we shall test the number 37 (`P18D` in EDSAC encoding).

`[ Even or odd  ===========   A program for the EDSAC   Determines whether the number stored at  address [email protected] is even or odd, and prints  'E' or 'O' accordingly   Works with Initial Orders 2 ]        T56K   [ load point ]       GK     [ base address ]        [email protected]   [ print letter shift ]       [email protected]   [ clear accumulator ]       [email protected]   [ multiplier := n ]       [email protected]   [ acc +:= mult AND 1 ]       [email protected]   [ acc -:= 1 ]       [email protected]    [ branch on negative ]       [email protected]   [ print 'O' ]       ZF     [ halt ][ 8 ]  [email protected]   [ print 'E' ]       ZF     [ halt ] [ 10 ] P0F    [ used to clear acc ][ 11 ] *F     [ letter shift character ][ 12 ] P0D    [ const: 1 ][ 13 ] EF     [ character 'E' ][ 14 ] OF     [ character 'O' ][ 15 ] P18D   [ number to test: 37 ]        EZPF   [ branch to load point ]`
Output:
`O`

Eiffel

`--bit testingif i.bit_and (1) = 0 then	-- i is evenend --built-in bit testing (uses bit_and)if i.bit_test (0) then	-- i is oddend --integer remainder (modulo)if i \\ 2 = 0 then	-- i is evenend`

Elixir

`defmodule RC do  import Integer   def even_or_odd(n) when is_even(n), do: "#{n} is even"  def even_or_odd(n)                , do: "#{n} is odd"      # In second "def", the guard clauses of "is_odd(n)" is unnecessary.   # Another definition way  def even_or_odd2(n) do    if is_even(n), do: "#{n} is even", else: "#{n} is odd"  endend Enum.each(-2..3, fn n -> IO.puts RC.even_or_odd(n) end)`
Output:
```-2 is even
-1 is odd
0 is even
1 is odd
2 is even
3 is odd
```

Other ways to test even-ness:

`rem(n,2) == 0`

Emacs Lisp

With evenp and oddp

` (defun odd (n)  (if (oddp n) (format "%d is odd\n" n)    (format "%d is even\n" n))) (defun even (n)  (if (evenp n) (format "%d is even\n" n)    (format "%d is odd\n" n))) (progn  (insert (even 3) )  (insert (odd 2) ))) `

With mod

` (defun odd (n)  (if (= 1 (mod n 2) ) (format "%d is odd\n" n)    (format "%d is even\n" n))) (defun even (n)  (if (= 0 (mod n 2) ) (format "%d is even\n" n)    (format "%d is odd\n" n))) (progn  (insert (even 3) )  (insert (odd 2) )) `

Output:

```
3 is odd
2 is even
```

Erlang

Using Division by 2 Method

`%% Implemented by Arjun Sunel-module(even_odd).-export([main/0]). main()->	test(8). test(N) ->	if (N rem 2)==1 ->		io:format("odd\n");	true ->		io:format("even\n")	end.			 `

Using the least-significant bit method

` %% Implemented by Arjun Sunel-module(even_odd2).-export([main/0]). main()->	test(10). test(N) ->	if (N band 1)==1 ->		io:format("odd\n");	true ->		io:format("even\n")	end.				 `

ERRE

`PROGRAM ODD_EVEN ! works for -2^15 <= n% < 2^15 FUNCTION ISODD%(N%)      ISODD%=(N% AND 1)<>0END FUNCTION ! works for -2^38 <= n# <= 2^38FUNCTION ISODD#(N#)      ISODD#=N#<>2*INT(N#/2)END FUNCTION BEGIN  IF ISODD%(14) THEN PRINT("14 is odd") ELSE PRINT("14 is even") END IF  IF ISODD%(15) THEN PRINT("15 is odd") ELSE PRINT("15 is even") END IF  IF ISODD#(9876543210) THEN PRINT("9876543210 is odd") ELSE PRINT("9876543210 is even") END IF  IF ISODD#(9876543211) THEN PRINT("9876543211 is odd") ELSE PRINT("9876543211 is even") END IFEND PROGRAM `
Output:
```14 is even
15 is odd
9876543210 is even
9876543211 is odd
```

Euphoria

Using standard function

` include std/math.e for i = 1 to 10 do        ? {i, is_even(i)}end for `
Output:
```{1,0}
{2,1}
{3,0}
{4,1}
{5,0}
{6,1}
{7,0}
{8,1}
{9,0}
{10,1}
```

Excel

Use the MOD function

` =MOD(33;2)=MOD(18;2) `
Output:
```1
0
```

Use the ISEVEN function, returns TRUE or FALSE

` =ISEVEN(33)=ISEVEN(18) `
Output:
```FALSE
TRUE
```

Use the ISODD function, returns TRUE or FALSE

` =ISODD(33)=ISODD(18) `
Output:
```TRUE
FALSE
```

F#

Bitwise and:

`let isEven x =  x &&& 1 = 0`

Modulo:

`let isEven x =  x % 2 = 0`

Factor

The math vocabulary provides even? and odd? predicates. This example runs at the listener, which already uses the math vocabulary.

```( scratchpad ) 20 even? .
t
( scratchpad ) 35 even? .
f
( scratchpad ) 20 odd? .
f
( scratchpad ) 35 odd? .
t
```

Fish

This example assumes that the input command i returns an integer when one was inputted and that the user inputs a valid positive integer terminated by a newline.

`<v"Please enter a number:"a >l0)?!vo     v          <                        v    o<^      >i:a=?v>i:a=?v\$a*+^>"The number is even."ar>l0=?!^>             >      >2%0=?^"The number is odd."ar ^`

The actual computation is the 2%0= part. The rest is either user interface or parsing input.

Forth

`: odd? ( n -- ? ) 1 and ;`

Fortran

Please find the compilation and example run in the comments at the beginning of the FORTRAN 2008 source. Separating the bit 0 parity module from the main program enables reuse of the even and odd functions. Even and odd, with scalar and vector interfaces demonstrate the generic function capability of FORTRAN 90. Threading, stdin, and all-intrinsics are vestigial and have no influence here other than to confuse you.

` !-*- mode: compilation; default-directory: "/tmp/" -*-!Compilation started at Tue May 21 20:22:56!!a=./f && make \$a && OMP_NUM_THREADS=2 \$a < unixdict.txt!gfortran -std=f2008 -Wall -ffree-form -fall-intrinsics f.f08 -o f! n     odd    even!-6    F    T!-5    T    F!-4    F    T!-3    T    F!-2    F    T!-1    T    F! 0    F    T! 1    T    F! 2    F    T! 3    T    F! 4    F    T! 5    T    F! 6    F    T! -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6       n!  F  T  F  T  F  T  F  T  F  T  F  T  F     odd!  T  F  T  F  T  F  T  F  T  F  T  F  T    even!!Compilation finished at Tue May 21 20:22:56  module bit0parity   interface odd    module procedure odd_scalar, odd_list  end interface   interface even    module procedure even_scalar, even_list  end interface contains   logical function odd_scalar(a)    implicit none    integer, intent(in) :: a    odd_scalar = btest(a, 0)  end function odd_scalar   logical function even_scalar(a)    implicit none    integer, intent(in) :: a    even_scalar = .not. odd_scalar(a)  end function even_scalar   function odd_list(a) result(rv)    implicit none    integer, dimension(:), intent(in) :: a    logical, dimension(size(a)) :: rv    rv = btest(a, 0)  end function odd_list   function even_list(a) result(rv)    implicit none    integer, dimension(:), intent(in) :: a    logical, dimension(size(a)) :: rv    rv = .not. odd_list(a)  end function even_list end module bit0parity program oe  use bit0parity  implicit none  integer :: i  integer, dimension(13) :: j  write(6,'(a2,2a8)') 'n', 'odd', 'even'  write(6, '(i2,2l5)') (i, odd_scalar(i), even_scalar(i), i=-6,6)  do i=-6, 6    j(i+7) = i  end do  write(6, '((13i3),a8/(13l3),a8/(13l3),a8)') j, 'n', odd(j), 'odd', even(j), 'even'end program oe `

FreeBASIC

`' FB 1.05.0 Win64 Dim n As Integer Do  Print "Enter an integer or 0 to finish : ";  Input "", n  If n = 0 Then    Exit Do  ElseIf n Mod 2 = 0 Then    Print "Your number is even"    Print  Else    Print "Your number is odd"    Print  End ifLoop End`

Futhark

` fun main(x: int): bool = (x & 1) == 0 `

Gambas

`Public Sub Form_Open()Dim sAnswer, sMessage As String sAnswer = InputBox("Input an integer", "Odd or even") If IsInteger(sAnswer) Then  If Odd(Val(sAnswer)) Then sMessage = "' is an odd number"  If Even(Val(sAnswer)) Then sMessage = "' is an even number"Else  sMessage = "' does not compute!!"Endif Print "'" & sAnswer & sMessage End`

Output:

```'25' is an odd number
'100' is an even number
'Fred' does not compute!!
```

GAP

`IsEvenInt(n);IsOddInt(n);`

Go

`package main import (    "fmt"    "math/big") func main() {    test(-2)    test(-1)    test(0)    test(1)    test(2)    testBig("-222222222222222222222222222222222222")    testBig("-1")    testBig("0")    testBig("1")    testBig("222222222222222222222222222222222222")} func test(n int) {    fmt.Printf("Testing integer %3d:  ", n)    // & 1 is a good way to test    if n&1 == 0 {        fmt.Print("even ")    } else {        fmt.Print(" odd ")    }    // Careful when using %: negative n % 2 returns -1.  So, the code below    // works, but can be broken by someone thinking they can reverse the    // test by testing n % 2 == 1.  The valid reverse test is n % 2 != 0.    if n%2 == 0 {        fmt.Println("even")    } else {        fmt.Println(" odd")    }} func testBig(s string) {    b, _ := new(big.Int).SetString(s, 10)    fmt.Printf("Testing big integer %v:  ", b)    // the Bit function is the only sensible test for big ints.    if b.Bit(0) == 0 {        fmt.Println("even")    } else {        fmt.Println("odd")    }}`
Output:
```Testing integer  -2:  even even
Testing integer  -1:   odd  odd
Testing integer   0:  even even
Testing integer   1:   odd  odd
Testing integer   2:  even even
Testing big integer -222222222222222222222222222222222222:  even
Testing big integer -1:  odd
Testing big integer 0:  even
Testing big integer 1:  odd
Testing big integer 222222222222222222222222222222222222:  even
```

Groovy

Solution:

`def isOdd = { int i -> (i & 1) as boolean }def isEven = {int i -> ! isOdd(i) }`

Test:

`1.step(20, 2) { assert isOdd(it) } 50.step(-50, -2) { assert isEven(it) }`

`even` and `odd` functions are already included in the standard Prelude.

`Prelude> even 5FalsePrelude> even 42TruePrelude> odd 5TruePrelude> odd 42False`

Where even is derived from rem, and odd is derived from even:

`import Prelude hiding (even, odd) even, odd  :: (Integral a)  => a -> Booleven = (0 ==) . (`rem` 2) odd = not . even main :: IO ()main = print (even <\$> [0 .. 9])`
Output:
`[True,False,True,False,True,False,True,False,True,False]`

Icon and Unicon

One way is to check the remainder:

`procedure isEven(n)    return n%2 = 0end`

J

Modulo:

`   2 | 2 3 5 70 1 1 1   2|2 3 5 7 + (2^89x)-11 0 0 0`

Remainder:

`   (= <.&.-:) 2 3 5 71 0 0 0   (= <.&.-:) 2 3 5 7+(2^89x)-10 1 1 1`

Last bit in bit representation:

`   {:"1@#: 2 3 5 70 1 1 1   {:"1@#: 2 3 5 7+(2^89x)-11 0 0 0`

Bitwise and:

`   1 (17 b.) 2 3 5 70 1 1 1`

Note: as a general rule, the simplest expressions in J should be preferred over more complex approaches.

Java

Bitwise and:

`public static boolean isEven(int i){    return (i & 1) == 0;}`

Modulo:

`public static boolean isEven(int i){    return (i % 2) == 0;}`

Arbitrary precision bitwise:

`public static boolean isEven(BigInteger i){    return i.and(BigInteger.ONE).equals(BigInteger.ZERO);}`

Arbitrary precision bit test (even works for negative numbers because of the way `BigInteger` represents the bits of numbers):

`public static boolean isEven(BigInteger i){    return !i.testBit(0);}`

Arbitrary precision modulo:

`public static boolean isEven(BigInteger i){    return i.mod(BigInteger.valueOf(2)).equals(BigInteger.ZERO);}`

JavaScript

ES5

Bitwise:

`function isEven( i ) {  return (i & 1) === 0;} `

Modulo:

`function isEven( i ) {  return i % 2 === 0;} // Alternativefunction isEven( i ) {  return !(i % 2);}`

ES6

Lambda:

`// EMCAScript 6const isEven = x => !(x % 2)`

or, avoiding type coercion:

`(() => {    'use strict';     // even : Integral a => a -> Bool    const even = x => (x % 2) === 0;     // odd : Integral a => a -> Bool    const odd = x => !even(x);      // TEST ----------------------------------------    // range :: Int -> Int -> [Int]    const range = (m, n) =>        Array.from({            length: Math.floor(n - m) + 1        }, (_, i) => m + i);     // show :: a -> String    const show = JSON.stringify;     // xs :: [Int]    const xs = range(-6, 6);     return show([xs.filter(even), xs.filter(odd)]);})();`
Output:
`[[-6,-4,-2,0,2,4,6],[-5,-3,-1,1,3,5]]`

jq

In practice, to test whether an integer, i, is even or odd in jq, one would typically use: i % 2

For example, if it were necessary to have a strictly boolean function that would test if its input is an even integer, one could define:

`def is_even: type == "number" and floor == 0 and . % 2 == 0;`

The check that the floor is 0 is necessary as % is defined on floating point numbers.

"is_odd" could be similarly defined:

`def is_odd: type == "number" and floor == 0 and . % 2 == 1;`

Julia

Built-in functions:

`iseven(i), isodd(i)`

K

The following implementation uses the modulo of division by 2

` oddp: {:[x!2;1;0]} /Returns 1 if arg. is oddevenp: {~oddp[x]}  /Returns 1 if arg. is even Examples:   oddp 320   evenp 321 `

Kotlin

`// version 1.0.5-2 fun main(args: Array<String>) {    while (true) {        print("Enter an integer or 0 to finish : ")        val n = readLine()!!.toInt()        when {            n == 0     -> return            n % 2 == 0 -> println("Your number is even")            else       -> println("Your number is odd")        }    }}`

L++

`(defn bool isEven (int x) (return (% x 2)))`

LabVIEW

Using bitwise And
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

`: even?  2 % not ;: odd?  2 % ;1 even? .   # 01 odd? .    # 1`

Lasso

`define isoddoreven(i::integer) => {	#i % 2 ? return 'odd'	return 'even'}isoddoreven(12)`

LC3 Assembly

Prints EVEN if the number stored in NUM is even, otherwise ODD.

`      .ORIG      0x3000       LD         R0,NUM      AND        R1,R0,1      BRZ        EVEN       LEA        R0,ODD      BRNZP      DISP EVEN  LEA        R0,EVN DISP  PUTS       HALT NUM   .FILL      0x1C EVN   .STRINGZ   "EVEN\n"ODD   .STRINGZ   "ODD\n"       .END`

Liberty BASIC

`n=12 if n mod 2 = 0 then print "even" else print "odd"`

Lingo

`on even (n)  return n mod 2 = 0end on odd (n)  return n mode 2 <> 0end`

LiveCode

`function odd n    return (n bitand 1) = 1end odd function notEven n    return (n mod 2) = 1end notEven`

Logo

`to even? :num    output equal? 0 modulo :num 2end`

Logtalk

` :- object(even_odd).     :- public(test_mod/1).    test_mod(I) :-        (   I mod 2 =:= 0 ->            write(even), nl        ;   write(odd), nl        ).     :- public(test_bit/1).    test_bit(I) :-        (   I /\ 1 =:= 1 ->            write(odd), nl        ;   write(even), nl        ). :- end_object. `
Output:
` | ?- even_odd::test_mod(1).oddyes | ?- even_odd::test_mod(2).evenyes | ?- even_odd::test_bit(1).oddyes | ?- even_odd::test_bit(2).evenyes `

Lua

`-- test for even numberif n % 2 == 0 then  print "The number is even"end -- test for odd numberif not (n % 2 == 0) then  print "The number is odd"end`

M2000 Interpreter

Binary.Add take any numeric type, but value must be in range of 0 to 0xFFFFFFFF So Mod if a perfect choice, using it with Decimals (character @ indicate a Decimal type or literal). Variable a take the type of input. There is no reason here to write it as def Odd(a as decimal)= binary.and(Abs(a), 1)=1

Def used to define variables (an error occur if same variable exist), or to define one line local functions. If a function exist then replace code. This is the same for modules/functions, a newer definition alter an old definition with same name, in current module if they are local, or global if they defined as global, like this Function Global F(x) { code block here}.

A function F(x) {} is same as

```Function F {
code here
}
```

The same hold for Def Odd(a)=binary.and(Abs(a), 1)=1 Interpreter execute this:

```Function Odd {
=binary.and(Abs(a), 1)=1
}
```

So here is the task. Show an overflow from a decimal, then change function.

` Module CheckOdd {      Def Odd(a)= binary.and(Abs(a), 1)=1      Print Odd(-5), Odd(6), Odd(11)      Try {            Print Odd([email protected])      }      Print Error\$    ' overflow       def Odd(a)= Int(Abs(a)) mod 2 =1      Print Odd([email protected])      Print Odd(-5), Odd(6), Odd(11)}CheckOdd `

M4

`define(`even', `ifelse(eval(`\$1'%2),0,True,False)')define(`odd',  `ifelse(eval(`\$1'%2),0,False,True)') even(13)even(8) odd(5)odd(0)`

Maple

`EvenOrOdd := proc( x::integer )   if x mod 2 = 0 then      print("Even"):   else      print("Odd"):   end if:end proc:EvenOrOdd(9);`
`"Odd"`

Mathematica / Wolfram Language

`EvenQ[8]`

MATLAB / Octave

Bitwise And:

`   isOdd  =  logical(bitand(N,1));   isEven = ~logical(bitand(N,1)); `

Remainder of division by two

`   isOdd  =  logical(rem(N,2));   isEven = ~logical(rem(N,2)); `

Modulo: 2

`   isOdd  =  logical(mod(N,2));   isEven = ~logical(mod(N,2)); `

Maxima

`evenp(n);oddp(n);`

Mercury

Mercury's 'int' module provides tests for even/odd, along with all the operators that would be otherwise used to implement them.

`even(N)  % in a body, suceeeds iff N is even.odd(N).  % in a body, succeeds iff N is odd. % rolling our own::- pred even(int::in) is semidet. % It's an error to have all three in one module, mind; even/1 would fail to check as semidet.even(N) :- N mod 2 = 0.   % using division that truncates towards -infinityeven(N) :- N rem 2 = 0.   % using division that truncates towards zeroeven(N) :- N /\ 1 = 0.    % using bit-wise and.`

MIPS Assembly

This uses bitwise AND

` .data	even_str: .asciiz "Even"	odd_str: .asciiz "Odd" .text	#set syscall to get integer from user	li \$v0,5	syscall 	#perform bitwise AND and store in \$a0	and \$a0,\$v0,1 	#set syscall to print dytomh	li \$v0,4 	#jump to odd if the result of the AND operation	beq \$a0,1,oddeven:		#load even_str message, and print		la \$a0,even_str	syscall 	#exit program	li \$v0,10	syscall odd:	#load odd_str message, and print	la \$a0,odd_str	syscall 	#exit program	li \$v0,10	syscall `

МК-61/52

`/	2	{x}	ЗН`

Result: "0" - number is even; "1" - number is odd.

ML

mLite

`fun odd 		(x rem 2 = 1) = true 	| 	_ 	      = false; fun even 		(x rem 2 = 0) = true 	| 	_ 	      = false;	  `

Modula-2

`MODULE EvenOrOdd;FROM FormatString IMPORT FormatString;FROM Terminal IMPORT WriteString,ReadChar; VAR    buf : ARRAY[0..63] OF CHAR;    i : INTEGER;BEGIN    FOR i:=-5 TO 5 DO        FormatString("%i is even: %b\n", buf, i, i MOD 2 = 0);        WriteString(buf)    END;     ReadCharEND EvenOrOdd.`

Neko

`var number = 6; if(number % 2 == 0) {	\$print("Even");} else {	\$print("Odd");}`
Output:
`Even`

NESL

NESL provides evenp and oddp functions, but they wouldn't be hard to reimplement.

`function even(n) = mod(n, 2) == 0; % test the function by applying it to the first ten positive integers: %{even(n) : n in [1:11]};`
Output:
`it = [F, T, F, T, F, T, F, T, F, T] : [bool]`

NetRexx

`/* NetRexx */options replace format comments java crossref symbols nobinary say 'Val'.right(5)': mod  - ver  - pos  - bits'say '---'.right(5)': ---- + ---- + ---- + ----'loop nn = -15 to 15 by 3  say nn.right(5)':' eo(isEven(nn)) '-' eo(isEven(nn, 'v')) '-' eo(isEven(nn, 'p')) '-' eo(isEven(nn, 'b'))  end nnreturn -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-- Overloaded method.  Default is to use the remainder specialization belowmethod isEven(anInt, meth = 'R') public static returns boolean  select case meth.upper().left(1)    when 'R' then eo = isEvenRemainder(anInt)    when 'V' then eo = isEvenVerify(anInt)    when 'P' then eo = isEvenPos(anInt)    when 'B' then eo = isEvenBits(anInt)    otherwise     eo = isEvenRemainder(anInt) -- default    end  return eo -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~method isEvenRemainder(anInt) public static returns boolean  return anInt // 2 == 0 -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~method isEvenVerify(anInt) public static returns boolean  return anInt.right(1).verify('02468') == 0 -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~method isEvenPos(anInt) public static returns boolean  return '13579'.pos(anInt.right(1)) == 0 -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~method isEvenBits(anInt) public static returns boolean  return \(anInt.d2x(1).x2b().right(1)) -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~method eo(state = boolean) public static  if state then sv = 'Even'           else sv = 'Odd'  return sv.left(4) `
Output:
```  Val: mod  - ver  - pos  - bits
---: ---- + ---- + ---- + ----
-15: Odd  - Odd  - Odd  - Odd
-12: Even - Even - Even - Even
-9: Odd  - Odd  - Odd  - Odd
-6: Even - Even - Even - Even
-3: Odd  - Odd  - Odd  - Odd
0: Even - Even - Even - Even
3: Odd  - Odd  - Odd  - Odd
6: Even - Even - Even - Even
9: Odd  - Odd  - Odd  - Odd
12: Even - Even - Even - Even
15: Odd  - Odd  - Odd  - Odd
```

NewLISP

`(odd? 1)(even? 2)`

Nim

`# Least signficant bit:proc isOdd(i: int): bool = (i and 1) != 0proc isEven(i: int): bool = (i and 1) == 0 # Modulo:proc isOdd2(i: int): bool = (i mod 2) != 0proc isEven2(i: int): bool = (i mod 2) == 0 # Bit Shifting:proc isOdd3(n: int): bool = n != ((n shr 1) shl 1)proc isEven3(n: int): bool = n == ((n shr 1) shl 1) echo isEven(1)echo isOdd2(5)`

Oberon-2

Works with: oo2c
` MODULE EvenOrOdd;IMPORT   S := SYSTEM,  Out;VAR  x: INTEGER;  s: SET; BEGIN  x := 10;Out.Int(x,0);  IF ODD(x) THEN Out.String(" odd") ELSE Out.String(" even") END;   Out.Ln;   x := 11;s := S.VAL(SET,LONG(x));Out.Int(x,0);  IF 0 IN s THEN Out.String(" odd") ELSE Out.String(" even") END;  Out.Ln;     x := 12;Out.Int(x,0);  IF x MOD 2 # 0 THEN Out.String(" odd") ELSE Out.String(" even") END;  Out.LnEND EvenOrOdd. `
Output:
```10 even
11 odd
12 even
```

Objeck

`a := Console->ReadString()->ToInt();if(a % 2 = 0) {  "even"->PrintLine();}else {  "odd"->PrintLine();};`

OCaml

Modulo:

`let is_even d =  (d mod 2) = 0 let is_odd d =  (d mod 2) <> 0`

Bitwise and:

`let is_even d =  (d land 1) = 0 let is_odd d =  (d land 1) <> 0`

An instructive view on functional programming and recursion:

`(* hmm, only valid for N0 *)let rec myeven = function   | 0 -> true  | 1 -> false  | n -> myeven (n - 2) (* and here we have the not function in if form *)let myodd n = if myeven n then false else true`

Oforth

`12 isEven12 isOdd`

OOC

` // Using the modulo operatoreven: func (n: Int) -> Bool {  (n % 2) == 0} // Using bitwise andodd: func (n: Int) -> Bool {  (n & 1) == 1} `

PARI/GP

GP does not have a built-in predicate for testing parity, but it's easy to code:

`odd(n)=n%2;`

Alternately:

`odd(n)=bitand(n,1);`

PARI can use the same method as C for testing individual words. For multiprecision integers (t_INT), use `mpodd`. If the number is known to be nonzero, `mod2` is (insignificantly) faster.

Pascal

Built-in boolean function odd:

`isOdd := odd(someIntegerNumber);`

bitwise and:

`function isOdd(Number: integer): booleanbegin  isOdd := boolean(Number and 1)end;`

Dividing and multiplying by 2 and test on equality:

`function isEven(Number: integer): booleanbegin  isEven := (Number = ((Number div 2) * 2))end;`

Using built-in modulo

`function isOdd(Number: integer): booleanbegin  isOdd := boolean(Number mod 2)end;`

Perl

`for(0..10){    print "\$_ is ", qw(even odd)[\$_ % 2],"\n";}`

or

`print 6 % 2  ? 'odd' : 'even';   # prints even`

Perl 6

Perl 6 doesn't have a built-in for this, but with subsets it's easy to define a predicate for it.

`subset Even of Int where * %% 2;subset Odd of Int where * % 2; say 1 ~~ Even; # falsesay 1 ~~ Odd;  # truesay 1.5 ~~ Odd # false ( 1.5 is not an Int )`

Phix

and_bits(i,1) returns 1(true) for odd integers and 0(false) for even integers. remainder(i,2) could also validly be used, however "true" for odd numbers is actually 1 for positive odd integers and -1 for negative odd integers.

`for i = -5 to 5 do    ? {i, and_bits(i,1), remainder(i,2)}end for`
Output:
```{-5,1,-1}
{-4,0,0}
{-3,1,-1}
{-2,0,0}
{-1,1,-1}
{0,0,0}
{1,1,1}
{2,0,0}
{3,1,1}
{4,0,0}
{5,1,1}
```

PHP

` // using bitwise and to check least significant digitecho (2 & 1) ? 'odd' : 'even';echo (3 & 1) ? 'odd' : 'even'; // using moduloecho (3 % 2) ? 'odd' : 'even';echo (4 % 2) ? 'odd' : 'even'; `
Output:
```even
odd
odd
even```

PicoLisp

PicoLisp doesn't have a built-in predicate for that. Using 'bit?' is the easiest and most efficient. The bit test with 1 will return NIL if the number is even.

`: (bit? 1 3)-> 1  # Odd : (bit? 1 4)-> NIL  # Even`

Pike

`> int i = 73;> (i&1);Result: 1> i%2;Result: 1`

PL/I

`i = iand(i,1)`

The result is 1 when i is odd, and 0 when i is even.

PowerShell

Works with: PowerShell version 2

Predicate

A predicate can be used with BigInteger objects. Even/odd predicates to not exist for basic value types. Type accelerator [bigint] can be used in place of [System.Numerics.BigInteger].

` \$IsOdd  = -not ( [bigint]\$N ).IsEven\$IsEven =      ( [bigint]\$N ).IsEven `

Least significant digit

` \$IsOdd  = [boolean]( \$N -band 1 )\$IsEven = [boolean]( \$N -band 0 ) `

Remainder

Despite being known as a modulus operator, the % operator in PowerShell actually returns a remainder. As such, when testing negative numbers it returns the true modulus result minus M. In this specific case, it returns -1 for odd negative numbers. Thus we test for not zero for odd numbers.

` \$IsOdd  = \$N % 2 -ne 0\$IsEven = \$N % 2 -eq 0 `

Processing

` boolean isEven(int i){  return i%2 == 0;} boolean isOdd(int i){  return i%2 == 1;} `

Prolog

Prolog does not provide special even or odd predicates as one can simply write "0 is N mod 2" to test whether the integer N is even. To illustrate, here is a predicate that can be used both to test whether an integer is even and to generate the non-negative even numbers:

`   even(N) :-     (between(0, inf, N); integer(N) ),     0 is N mod 2. `

Least Significant Bit

If N is a positive integer, then lsb(N) is the offset of its least significant bit, so we could write:

`   odd(N) :- N = 0 -> false; 0 is lsb(abs(N)). `

PureBasic

`;use last bit methodisOdd = i & 1         ;isOdd is non-zero if i is oddisEven = i & 1 ! 1    ;isEven is non-zero if i is even ;use modular methodisOdd = i % 2         ;isOdd is non-zero if i is oddisEven = i % 2 ! 1    ;isEven is non-zero if i is even`

Python

Python: Using the least-significant bit method

`>>> def is_odd(i): return bool(i & 1) >>> def is_even(i): return not is_odd(i) >>> [(j, is_odd(j)) for j in range(10)][(0, False), (1, True), (2, False), (3, True), (4, False), (5, True), (6, False), (7, True), (8, False), (9, True)]>>> [(j, is_even(j)) for j in range(10)][(0, True), (1, False), (2, True), (3, False), (4, True), (5, False), (6, True), (7, False), (8, True), (9, False)]>>> `

Python: Using modular congruences

`>> def is_even(i):        return (i % 2) == 0 >>> is_even(1)False>>> is_even(2)True>>>`

R

`is.even <- function(x) !is.odd(x) is.odd <- function(x) intToBits(x)[1] == 1#oris.odd <- function(x) x %% 2 == 1`

Racket

With built in predicates:

`(even? 6) ; -> true(even? 5) ; -> false(odd? 6) ; -> false(odd? 5) ; -> true `

With modular arithmetic:

`(define (my-even? x)  (= (modulo x 2) 0))  (define (my-odd? x)  (= (modulo x 2) 1))`

Rascal

`public bool isEven(int n) = (n % 2) == 0;public bool isOdd(int n) = (n % 2) == 1;`

Or with block quotes:

`public bool isEven(int n){return (n % 2) == 0;}public bool isOdd(int n){return (n % 2) == 1;}`

REXX

Programming note:   division by   1   (one)   in REXX is a way to normalize a number:

• by removing a superfluous leading   +   sign
• by removing superfluous leading zeroes
• by removing superfluous trailing zeroes
• by removing a trailing decimal point
• possible converting an exponentiated number
• possible rounding the number to the current digits

Programming note:   the last method is the fastest method in REXX to determine oddness/evenness.
It requires a sparse stemmed array     !.     be defined in the program's prologue (or elsewhere).
This method gets its speed from   not   using any BIF and   not   performing any (remainder) division.

Some notes on programming styles:   If (execution) speed isn't an issue, then the 1st test method
shown would be the simplest   (in terms of coding the concisest/tightest/smallest code).   The other test
methods differ mostly in programming techniques, mostly depending on the REXX programmer's style.
The last method shown is the fastest algorithm, albeit it might be a bit obtuse (without comments) to a
novice reader of the REXX language   (and it requires additional REXX statement baggage).

`/*REXX program tests and displays if an integer is  even or odd  using different styles.*/!.=0;   do j=0  by 2  to 8;   !.j=1;   end       /*assign  0,2,4,6,8  to a "true" value.*/                                                 /* [↑]  assigns even digits to  "true".*/numeric digits 1000                              /*handle most huge numbers from the CL.*/parse arg x _ .                                  /*get an argument from the command line*/if x==''               then call terr "no integer input (argument)."if _\=='' | arg()\==1  then call terr "too many arguments: "          _  arg(2)if \datatype(x, 'N')   then call terr "argument isn't numeric: "      xif \datatype(x, 'W')   then call terr "argument isn't an integer: "   xy=abs(x)/1                                       /*in case  X  is negative or malformed,*/                                                 /* [↑]  remainder of neg # might be -1.*/                                                 /*malformed #s: 007  9.0  4.8e1  .21e2 */call tell 'remainder method (oddness)'if y//2  then say  x  'is odd'         else say  x  'is even'                                                 /* [↑]  uses division to get remainder.*/ call tell 'rightmost digit using BIF (not evenness)'_=right(y, 1)if pos(_, 86420)==0  then say x 'is odd'                     else say x 'is even'                                                 /* [↑]  uses 2 BIF (built─in functions)*/ call tell 'rightmost digit using BIF (evenness)'_=right(y, 1)if pos(_, 86420)\==0  then say x 'is even'                      else say x 'is odd'                                                 /* [↑]  uses 2 BIF (built─in functions)*/ call tell 'even rightmost digit using array (evenness)'_=right(y, 1)if !._  then say x 'is even'        else say x 'is odd'                                                 /* [↑]  uses a BIF (built─in function).*/ call tell 'remainder of division via function invoke (evenness)'if even(y)  then say x 'is even'            else say x 'is odd'                                                 /* [↑]  uses (even) function invocation*/ call tell 'remainder of division via function invoke (oddness)'if odd(y)  then say x 'is odd'           else say x 'is even'                                                 /* [↑]  uses (odd)  function invocation*/ call tell 'rightmost digit using BIF (not oddness)'_=right(y, 1)if pos(_, 13579)==0  then say x 'is even'                     else say x 'is odd'                                                 /* [↑]  uses 2 BIF (built─in functions)*/ call tell 'rightmost (binary) bit (oddness)'if right(x2b(d2x(y)), 1)  then say x 'is odd'                          else say x 'is even'                                                 /* [↑]  requires extra numeric digits. */ call tell 'parse statement using BIF (not oddness)'parse var  y   ''  -1  _                         /*obtain last decimal digit of the Y #.*/if pos(_, 02468)==0  then say x 'is odd'                     else say x 'is even'                                                 /* [↑]  uses a BIF (built─in function).*/ call tell 'parse statement using array (evenness)'parse var  y   ''  -1  _                         /*obtain last decimal digit of the Y #.*/if !._  then say  x  'is even'        else say  x  'is odd'                                                 /* [↑]  this is the fastest algorithm. */exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/even:                     return \( arg(1)//2 )  /*returns "evenness" of arg, version 1.*/even:                     return    arg(1)//2==0 /*   "         "      "  "      "    2.*/even: parse arg '' -1 _;  return !._             /*   "         "      "  "      "    3.*/                                                 /*last version shown is the fastest.   */odd:                      return   arg(1)//2     /*returns  "oddness" of the argument.  */tell: say;   say center('using the' arg(1), 79, "═");                    returnterr: say;   say '***error***';     say;    say arg(1);    say;          exit 13`

output   when using the input of:   0

```═════════════════════using the remainder method (oddness)══════════════════════
0 is even

══════════════using the rightmost digit using BIF (not evenness)═══════════════
0 is even

════════════════using the rightmost digit using BIF (evenness)═════════════════
0 is even

═════════════using the even rightmost digit using array (evenness)═════════════
0 is even

════════using the remainder of division via function invoke (evenness)═════════
0 is even

═════════using the remainder of division via function invoke (oddness)═════════
0 is even

═══════════════using the rightmost digit using BIF (not oddness)═══════════════
0 is even

══════════════════using the rightmost (binary) bit (oddness)═══════════════════
0 is even

═══════════════using the parse statement using BIF (not oddness)═══════════════
0 is even

═══════════════using the parse statement using array (evenness)════════════════
0 is even
```

output   when using the input of:   9876543210987654321098765432109876543210987654321

```═════════════════════using the remainder method (oddness)══════════════════════
9876543210987654321098765432109876543210987654321 is odd

(rest of the output was elided.)
```

output   when using the input of:   .6821e4

```═════════════════════using the remainder method (oddness)══════════════════════
.8621e4 is odd

(rest of the output was elided.)
```

output   when using the input of:   -9411

```═════════════════════using the remainder method (oddness)══════════════════════
-9411 is odd

(rest of the output was elided.)
```

Ring

` size = 10for i = 1 to size    if i % 2 = 1 see "" + i + " is odd" + nl    else see "" + i + " is even" + nl oknext `

Ruby

`print "evens: "p -5.upto(5).select(&:even?)print "odds: "p -5.upto(5).select(&:odd?)`
Output:
```evens: [-4, -2, 0, 2, 4]
odds: [-5, -3, -1, 1, 3, 5]```

Other ways to test even-ness:

`n & 1 == 0quotient, remainder = n.divmod(2); remainder == 0 # The next way only works when n.to_f/2 is exact.# If Float is IEEE double, then -2**53 .. 2**53 must include n.n.to_f/2 == n/2 # You can use the bracket operator to access the i'th bit# of a Fixnum or Bignum (i = 0 means least significant bit)n[0].zero?`

Run BASIC

`for i = 1 to 10  if i and 1 then print i;" is odd" else print i;" is even"next i`
```1 is odd
2 is even
3 is odd
4 is even
5 is odd
6 is even
7 is odd
8 is even
9 is odd
10 is even
```

Rust

Checking the last significant digit:

`let is_odd = |x: i32| x & 1 == 1;let is_even = |x: i32| x & 1 == 0;`

Using modular congruences:

`let is_odd = |x: i32| x % 2 != 0;let is_even = |x: i32| x % 2 == 0;`

Scala

`def isEven( v:Int ) : Boolean = v % 2 == 0def isOdd( v:Int ) : Boolean = v % 2 != 0`

Accept any numeric type as an argument:

`def isEven( v:Number ) : Boolean = v.longValue % 2 == 0def isOdd( v:Number ) : Boolean = v.longValue % 2 != 0`
Output:
```isOdd( 81 )                     // Results in true
isEven( BigInt(378) )           // Results in true
isEven( 234.05003513013145 )    // Results in true```

Scheme

`even?` and `odd?` functions are built-in (R4RS, R5RS, and R6RS):

`> (even? 5)#f> (even? 42)#t> (odd? 5)#t> (odd? 42)#f`

Seed7

Test whether an integer or bigInteger is odd:

`odd(aNumber)`

Test whether an integer or bigInteger is even:

`not odd(aNumber)`

SequenceL

`even(x) := x mod 2 = 0;odd(x) := x mod 2 = 1;`
Output:
```cmd:>even(1 ... 10)
[false,true,false,true,false,true,false,true,false,true]
cmd:>odd(1 ... 10)
[true,false,true,false,true,false,true,false,true,false]
```

SETL

SETL provides built-in even and odd functions. This short program illustrates their use.

`xs := {1..10};evens := {x in xs | even( x )};odds := {x in xs | odd( x )};print( evens );print( odds );`
Output:
```{2 4 6 8 10}
{1 3 5 7 9}```

Shen

Mutual Recursion:

`(define even?    0 -> true    X -> (odd? (- X 1))) (define odd?    0 -> false    X -> (even? (- X 1)))`

Modulo:

`(define even? X -> (= 0 (shen.mod X 2))) (define odd? X -> (not (= 0 (shen.mod X 2))))`

Sidef

Built-in methods:

`var n = 42;say n.is_odd;       # falsesay n.is_even;      # true`

Checking the last significant digit:

`func is_odd(n)  { n&1 == 1 };func is_even(n) { n&1 == 0 };`

Using modular congruences:

`func is_odd(n)  { n%2 == 1 };func is_even(n) { n%2 == 0 };`

Smalltalk

Using the built in methods on Number class:

`5 even5 odd`

even is implemented as follows:

`Number>>even	^((self digitAt: 1) bitAnd: 1) = 0 `

SNOBOL4

Works with: Macro SNOBOL4 in C
Works with: Spitbol
Works with: SNOBOL4+
`      DEFINE('even(n)')                         :(even_end)even  even = (EQ(REMDR(n, 2), 0) 'even', 'odd') :(RETURN)even_end       OUTPUT = "-2 is " even(-2)      OUTPUT = "-1 is " even(-1)      OUTPUT = "0 is " even(0)      OUTPUT = "1 is " even(1)      OUTPUT = "2 is " even(2)END`
Output:
```-2 is even
-1 is odd
0 is even
1 is odd
2 is even

```

SNUSP

` \$====!/?\==even#      - -#odd==\?/ `

SPL

`> n, 0..9  ? #.even(n), #.output(n," even")  ? #.odd(n), #.output(n," odd")<`
Output:
```0 even
1 odd
2 even
3 odd
4 even
5 odd
6 even
7 odd
8 even
9 odd
```

SQL

Database vendors can't agree on how to get a remainder. This should work for many, including Oracle. For others, including MS SQL Server, try "int % 2" instead of "mod(int, 2)".

`-- Setup a table with some integersCREATE TABLE ints(INT INTEGER);INSERT INTO ints VALUES (-1);INSERT INTO ints VALUES (0);INSERT INTO ints VALUES (1);INSERT INTO ints VALUES (2); -- Are they even or odd?SELECT  INT,  CASE MOD(INT, 2) WHEN 0 THEN 'Even' ELSE 'Odd' ENDFROM  ints;`
Output:
```       INT CASE
---------- ----
-1 Odd
0 Even
1 Odd
2 Even```

SSEM

The SSEM doesn't provide AND, but for once the instruction set does allow the problem to be solved quite elegantly (albeit extravagantly slowly). Load the value of ${\displaystyle n}$ into storage address 15. The first three instructions test whether ${\displaystyle n}$ is positive, and replace it with its negation if it isn't. We then loop, subtracting 2 each time and testing whether we have got down either to 0 or to 1. When we have, the computer will halt with the accumulator storing 0 if ${\displaystyle n}$ was even or 1 if it was odd.

Note that the constant 2, stored at address 14, does double service: it is the operand for the Sub. instruction at address 6 and also the jump target returning to the top of the main loop (which is at address 2 + 1 = 3).

For larger positive or smaller negative values of ${\displaystyle n}$, you should be ready with something else to do while the machine is working: a test run took several minutes to confirm that 32,769 was odd.

`11110000000000100000000000000000   0. -15 to c00000000000000110000000000000000   1. Test11110000000001100000000000000000   2. c to 1511110000000000100000000000000000   3. -15 to c00001000000001100000000000000000   4. c to 1600001000000000100000000000000000   5. -16 to c01110000000000010000000000000000   6. Sub. 1411110000000001100000000000000000   7. c to 1510110000000000010000000000000000   8. Sub. 1300000000000000110000000000000000   9. Test01110000000000000000000000000000  10. 14 to CI11110000000000100000000000000000  11. -15 to c00000000000001110000000000000000  12. Stop10000000000000000000000000000000  13. 101000000000000000000000000000000  14. 2`

Standard ML

`fun even n =   n mod 2 = 0; fun odd n =   n mod 2 <> 0; (* bitwise and *) type werd = Word.word; fun evenbitw(w: werd) =   Word.andb(w, 0w2) = 0w0; fun oddbitw(w: werd) =   Word.andb(w, 0w2) <> 0w0;`

Stata

`matafunction iseven(n) {	return(mod(n,2)==0)} function isodd(n) {	return(mod(n,2)==1)}end`

Swift

`func isEven(n:Int) -> Bool {     // Bitwise check    if (n & 1 != 0) {        return false    }     // Mod check    if (n % 2 != 0) {        return false    }    return true}`

Symsyn

` n : 23  if n bit 0    'n is odd' [] else    'n is even' [] `

Tcl

`package require Tcl 8.5 # Bitwise test is the most efficientproc tcl::mathfunc::isOdd x  { expr {\$x & 1} }proc tcl::mathfunc::isEven x { expr {!(\$x & 1)} } puts " # O E"puts 24:[expr isOdd(24)],[expr isEven(24)]puts 49:[expr isOdd(49)],[expr isEven(49)]`
Output:
``` # O E
24:0,1
49:1,0
```

TI-83 BASIC

TI-83 BASIC does not have a modulus operator.

` If fPart(.5AnsThenDisp "ODDElseDisp "EVENEnd `

TUSCRIPT

`\$\$ MODE TUSCRIPTLOOP n=-5,5x=MOD(n,2)SELECT xCASE 0PRINT n," is even"DEFAULTPRINT n," is odd"ENDSELECTENDLOOP`
Output:
```-5 is odd
-4 is even
-3 is odd
-2 is even
-1 is odd
0 is even
1 is odd
2 is even
3 is odd
4 is even
5 is odd
```

UNIX Shell

`iseven() {    [[ \$((\$1%2)) -eq 0 ]] && return 0    return 1}`

Ursa

`decl int inputset input (in int console)if (= (mod input 2) 1)        out "odd" endl consoleelse        out "even" endl consoleend if`

Output:

```123
odd```

உயிர்/Uyir

`முதன்மை என்பதின் வகை எண் பணி {{        எ இன் வகை எண்{\$5} = 0;        படை வகை சரம்;         "எண்ணைக் கொடுங்கள்? ") ஐ திரை.இடு;         எ = எண்{\$5} ஐ விசை.எடு;         ஒருக்கால் (எ.இருமம்(0) == 1) ஆகில் {                படை = "ஒற்றை";        } இல்லையேல் {                படை = "இரட்டை ";        }         {எ, " ஒரு ", படை, "ப்படை எண் ஆகும்"} என்பதை திரை.இடு;         முதன்மை  = 0;}};`

VBA

```4 ways = 4 Functions :
IsEven ==> Use the even and odd predicates
IsEven2 ==> Check the least significant digit. With binary integers, i bitwise-and 1 equals 0 iff i is even
IsEven3 ==> Divide i by 2. The remainder equals 0 if i is even.
IsEven4 ==> Use modular congruences```
` Option Explicit Sub Main_Even_Odd()Dim i As Long     For i = -50 To 48 Step 7        Debug.Print i & " : IsEven ==> " & IIf(IsEven(i), "is even", "is odd") _         & " " & Chr(124) & " IsEven2 ==> " & IIf(IsEven2(i), "is even", "is odd") _         & " " & Chr(124) & " IsEven3 ==> " & IIf(IsEven3(i), "is even", "is odd") _         & " " & Chr(124) & " IsEven4 ==> " & IIf(IsEven4(i), "is even", "is odd")    NextEnd Sub Function IsEven(Number As Long) As Boolean'Use the even and odd predicates    IsEven = (WorksheetFunction.Even(Number) = Number)End Function Function IsEven2(Number As Long) As Boolean'Check the least significant digit.'With binary integers, i bitwise-and 1 equals 0 iff i is even, or equals 1 iff i is odd.Dim lngTemp As Long    lngTemp = CLng(Right(CStr(Number), 1))    If (lngTemp And 1) = 0 Then IsEven2 = TrueEnd Function Function IsEven3(Number As Long) As Boolean'Divide i by 2.'The remainder equals 0 if i is even.Dim sngTemp As Single    sngTemp = Number / 2    IsEven3 = ((Int(sngTemp) - sngTemp) = 0)End Function Function IsEven4(Number As Long) As Boolean'Use modular congruences    IsEven4 = (Number Mod 2 = 0)End Function `
Output:
```-50 : IsEven ==> is even | IsEven2 ==> is even | IsEven3 ==> is even | IsEven4 ==> is even
-43 : IsEven ==> is odd | IsEven2 ==> is odd | IsEven3 ==> is odd | IsEven4 ==> is odd
-36 : IsEven ==> is even | IsEven2 ==> is even | IsEven3 ==> is even | IsEven4 ==> is even
-29 : IsEven ==> is odd | IsEven2 ==> is odd | IsEven3 ==> is odd | IsEven4 ==> is odd
-22 : IsEven ==> is even | IsEven2 ==> is even | IsEven3 ==> is even | IsEven4 ==> is even
-15 : IsEven ==> is odd | IsEven2 ==> is odd | IsEven3 ==> is odd | IsEven4 ==> is odd
-8 : IsEven ==> is even | IsEven2 ==> is even | IsEven3 ==> is even | IsEven4 ==> is even
-1 : IsEven ==> is odd | IsEven2 ==> is odd | IsEven3 ==> is odd | IsEven4 ==> is odd
6 : IsEven ==> is even | IsEven2 ==> is even | IsEven3 ==> is even | IsEven4 ==> is even
13 : IsEven ==> is odd | IsEven2 ==> is odd | IsEven3 ==> is odd | IsEven4 ==> is odd
20 : IsEven ==> is even | IsEven2 ==> is even | IsEven3 ==> is even | IsEven4 ==> is even
27 : IsEven ==> is odd | IsEven2 ==> is odd | IsEven3 ==> is odd | IsEven4 ==> is odd
34 : IsEven ==> is even | IsEven2 ==> is even | IsEven3 ==> is even | IsEven4 ==> is even
41 : IsEven ==> is odd | IsEven2 ==> is odd | IsEven3 ==> is odd | IsEven4 ==> is odd
48 : IsEven ==> is even | IsEven2 ==> is even | IsEven3 ==> is even | IsEven4 ==> is even```

VBScript

` Function odd_or_even(n)	If n Mod 2 = 0 Then		odd_or_even = "Even"	Else		odd_or_even = "Odd"	End IfEnd Function WScript.StdOut.Write "Please enter a number: "n = WScript.StdIn.ReadLineWScript.StdOut.Write n & " is " & odd_or_even(CInt(n))WScript.StdOut.WriteLine `
Output:
```C:\>cscript /nologo odd_or_even.vbs
6 is Even

C:\>cscript /nologo odd_or_even.vbs
9 is Odd

C:\>cscript /nologo odd_or_even.vbs
-1 is Odd
```

WDTE

`let s => import 'stream';let str => import 'strings'; let evenOrOdd n => (	let even n => == (% n 2) 0;	switch n {		even => 'even';		default => 'odd';	};); s.range 10-> s.map (@ s n => str.format '{} is {}.' n (evenOrOdd n))-> s.map (io.writeln io.stdout)-> s.drain;`

xEec

` >100 p i# jz-1 o# t h#1 ms jz2003 p >0110 h#2 r ms t h#1 ms p jz1002 h? jz2003 p jn0110 h#10 o\$ p jn100 >2003 p p h#0 h#10 h\$d h\$d h\$o h#32 h\$s h\$i h#32 jn0000 >1002 p p h#0 h#10 h\$n h\$e h\$v h\$e h#32 h\$s h\$i h#32 >0000 o\$ p jn0000 jz100 `

XLISP

XLISP provides EVENP and ODDP, or, if you prefer, EVEN? and ODD?; if one wanted to reimplement them, it could be done like this (or in other ways).

`(defun my-evenp (x)    (= (logand x 1) 0) ) (defun my-oddp (x)    (/= (logand x 1) 0) )`

Xojo

` For num As Integer = 1 To 5  If num Mod 2 = 0 Then    MsgBox(Str(num) + " is even.")  Else    MsgBox(Str(num) + " is odd.")  End IfNext `
Output:
```1 is odd.
2 is even.
3 is odd.
4 is even.
5 is odd.
```

XPL0

`include c:\cxpl\codes;int I;[for I:= -4 to +3 do        [IntOut(0, I);        Text(0, if I&1 then " is odd   " else " is even  ");         Text(0, if rem(I/2)#0 then "odd" else "even");        CrLf(0);        ];]`
Output:
```-4 is even  even
-3 is odd   odd
-2 is even  even
-1 is odd   odd
0 is even  even
1 is odd   odd
2 is even  even
3 is odd   odd
```

Yabasic

Translation of: Phix
`for i = -5 to 5    print i, and(i,1), mod(i,2)next `

zkl

`[-3..4].pump(fcn(n){ println(n," is ",n.isEven and "even" or "odd") })`

Ints have isEven and isOdd properties. pump, in this case, is the same as apply/map without aggregating a result.

Output:
```-3 is odd
-2 is even
-1 is odd
0 is even
1 is odd
2 is even
3 is odd
4 is even
```
`[-3..4].apply("isEven").println();`
Output:
`L(False,True,False,True,False,True,False,True)`

zonnon

` module Main;var	x: integer;	s: set;begin	x := 10;writeln(x:3," is odd?",odd(x));	s := set(s);writeln(x:3," is odd?",0 in s); (* check right bit *)	x := 11;writeln(x:3," is odd?",odd(x));	s := set(x);writeln(x:3," is odd?",0 in s); (* check right bit *)end Main. `
Output:
``` 10 is odd? false
10 is odd? false
11 is odd?  true
11 is odd?  true
```

ZX Spectrum Basic

`10 FOR n=-3 TO 4: GO SUB 30: NEXT n20 STOP 30 LET odd=FN m(n,2)40 PRINT n;" is ";("Even" AND odd=0)+("Odd" AND odd=1)50 RETURN 60 DEF FN m(a,b)=a-INT (a/b)*b`