Even or odd

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

AutoHotkey

Bitwise ops are probably most efficient: <lang AHK>if ( int & 1 ){ ; do odd stuff }else{ ; do even stuff }</lang>

ABAP

<lang 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|
     )
   )
 )

). </lang>

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

<lang ada>-- 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;</lang>

Aime

<lang aime>if (x & 1) {

   # x is odd

} else {

   # x is even

}</lang>

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.win32

<lang algol68># 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 ) ) </lang>

ALGOL W

<lang algolw>begin

   % the Algol W standard procedure odd returns true if it's 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_i

end.</lang>

Output:

             1   is odd
          1702   is even
            23   is odd
           -26   is even

Arendelle

( input , "Please enter a number: " )

{ @input % 2 = 0 ,

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

AWK

<lang AWK>function isodd(x) { return (x%2)!=0; }

function iseven(x) { return (x%2)==0; }</lang>

BASIC

<lang basic>10 INPUT "ENTER A NUMBER: ";N 20 IF N/2 <> INT(N/2) THEN PRINT "THE NUMBER IS ODD":GOTO 40 30 PRINT "THE NUMBER IS EVEN" 40 END</lang>

Batch File

<lang dos> @echo off set /p i=Insert number:

bitwise and

set /a "test1=%i%&1"

divide last character by 2

set /a test2=%i:~-1%/2

modulo

set /a test3=%i% %% 2

set test pause>nul </lang>

BBC BASIC

Works with: BBC BASIC for Windows

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. <lang bbcbasic> 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)</lang>

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. <lang bc>i = -3

/* Assumes that i is an integer. */ scale = 0 if (i % 2 == 0) "i is even " if (i % 2) "i is odd "</lang>

Befunge

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. <lang bracmat>( ( even

& (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 )</lang>

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. <lang bf>,[>,----------] 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</lang>

If one need only determine rather than act on the parity of the input, the following is sufficient; it terminates either quickly or never. <lang bf>,[>,----------]<[--]</lang>

Burlesque

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): <lang c>if (x & 1) {

   /* x is odd */

} else {

   /* or not */

}</lang> If using long integer type from GMP (mpz_t), there are provided macros: <lang c>mpz_t x; ... if (mpz_even_p(x)) { /* x is even */ } if (mpz_odd_p(x)) { /* x is odd */ }</lang> The macros evaluate x more than once, so it should not be something with side effects.

C++

Test using the modulo operator, or use the c example from above. <lang cpp>bool isEven(int x) {

 return x % 2;

}</lang>

C#

<lang csharp>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;
       }
   }

}</lang>

Clojure

Standard predicates: <lang clojure>(if (even? some-var) (do-even-stuff)) (if (odd? some-var) (do-odd-stuff))</lang>

COBOL

<lang cobol> IF FUNCTION REM(Num, 2) = 0

          DISPLAY Num " is even."
      ELSE
          DISPLAY Num " is odd."
      END-IF</lang>

ColdFusion

 She's even

 He's odd

</cfif> </lang>

Common Lisp

Standard predicates: <lang lisp>(if (evenp some-var) (do-even-stuff)) (if (oddp some-other-var) (do-odd-stuff))</lang>

Component Pascal

BlackBox Component Builder <lang oberon2> 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) # 0 END Odd;

PROCEDURE CongruenceOdd(i: INTEGER): BOOLEAN; BEGIN RETURN ((i -1) MOD 2) = 0 END 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; </lang> 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

D

<lang d>void main() {

   import std.stdio, std.bigint;

   foreach (immutable i; -5 .. 6)
       writeln(i, " ", i & 1, " ", i % 2, " ", i.BigInt % 2);

}</lang>

Output:

Déjà Vu

<lang dejavu>even n:

   = 0 % n 2

odd:

   not even

!. odd 0 !. even 0 !. odd 7 !. even 7 </lang>

Output:

false
true
true
false

DWScript

Predicate: <lang delphi>var isOdd := Odd(i);</lang> Bitwise and: <lang delphi>var isOdd := (i and 1)<>0;</lang> Modulo: <lang delphi>var isOdd := (i mod 2)=1;</lang>

Eiffel

<lang Eiffel>--bit testing if i.bit_and (1) = 0 then -- i is even end

--built-in bit testing (uses bit_and) if i.bit_test (0) then -- i is odd end

--integer remainder (modulo) if i \\ 2 = 0 then -- i is even end</lang>

Erlang

Using Division by 2 Method

<lang erlang>%% 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. </lang>

Using the least-significant bit method

<lang erlang> %% 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. </lang>

ERRE

<lang ERRE>PROGRAM ODD_EVEN

! works for -2^15 <= n% < 2^15

FUNCTION ISODD%(N%)

     ISODD%=(N% AND 1)<>0

END FUNCTION

! works for -2^38 <= n# <= 2^38 FUNCTION 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 IF

END PROGRAM </lang>

Output:

14 is even
15 is odd
9876543210 is even
9876543211 is odd

Euphoria

Using standard function <lang Euphoria> include std/math.e

for i = 1 to 10 do

       ? {i, is_even(i)}

end for </lang>

Output:

{1,0}
{2,1}
{3,0}
{4,1}
{5,0}
{6,1}
{7,0}
{8,1}
{9,0}
{10,1}

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. <lang Fish><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 ^</lang>

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

Forth

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. <lang FORTRAN> !-*- 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 </lang>

F#

Bitwise and: <lang fsharp>let isEven x =

 x &&& 1 = 0</lang>

Modulo: <lang fsharp>let isEven x =

 x % 2 = 0</lang>

GAP

<lang gap>IsEvenInt(n); IsOddInt(n);</lang>

Go

<lang 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")
   }

}</lang>

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: <lang groovy>def isOdd = { int i -> (i & 1) as boolean } def isEven = {int i -> ! isOdd(i) }</lang> Test: <lang groovy>1.step(20, 2) { assert isOdd(it) }

50.step(-50, -2) { assert isEven(it) }</lang>

Haskell

even and odd functions are already included in the standard Prelude. <lang haskell>Prelude> even 5 False Prelude> even 42 True Prelude> odd 5 True Prelude> odd 42 False</lang>

Icon and Unicon

One way is to check the remainder: <lang unicon>procedure isEven(n)

   return n%2 = 0

end</lang>

J

Modulo: <lang j> 2 | 2 3 5 7 0 1 1 1

  2|2 3 5 7 + (2^89x)-1

1 0 0 0</lang> Remainder: <lang j> (= <.&.-:) 2 3 5 7 1 0 0 0

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

0 1 1 1</lang> Last bit in bit representation: <lang j> {:"1@#: 2 3 5 7 0 1 1 1

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

1 0 0 0</lang> Bitwise and: <lang j> 1 (17 b.) 2 3 5 7 0 1 1 1</lang> Note: as a general rule, the simplest expressions in J should be preferred over more complex approaches.

Java

Bitwise and: <lang java>public static boolean isEven(int i){

   return (i & 1) == 0;

}</lang> Modulo: <lang java>public static boolean isEven(int i){

   return (i % 2) == 0;

}</lang> Arbitrary precision bitwise: <lang java>public static boolean isEven(BigInteger i){

   return i.and(BigInteger.ONE).equals(BigInteger.ZERO);

}</lang> Arbitrary precision bit test (even works for negative numbers because of the way BigInteger represents the bits of numbers): <lang java>public static boolean isEven(BigInteger i){

   return !i.testBit(0);

}</lang> Arbitrary precision modulo: <lang java>public static boolean isEven(BigInteger i){

   return i.mod(BigInteger.valueOf(2)).equals(BigInteger.ZERO);

}</lang>

JavaScript

Bitwise: <lang javascript>function isEven( i ) {

 return (i & 1) === 0;

} </lang> Modulo: <lang javascript>function isEven( i ) {

 return i % 2 === 0;

} </lang>

Julia

Built-in functions: <lang julia>iseven(i), isodd(i)</lang>

L++

<lang lisp>(defn bool isEven (int x) (return (% x 2)))</lang>

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

<lang lang5>: even? 2 % not ;

odd? 2 % ;

1 even? . # 0 1 odd? . # 1</lang>

Liberty BASIC

<lang lb>n=12

if n mod 2 = 0 then print "even" else print "odd"</lang>

Lasso

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

Logo

<lang logo>to even? :num

   output equal? 0 modulo :num 2

end</lang>

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.

</lang>

Output:

<lang text> | ?- even_odd::test_mod(1). odd yes

| ?- even_odd::test_mod(2). even yes

| ?- even_odd::test_bit(1). odd yes

| ?- even_odd::test_bit(2). even yes </lang>

Lua

<lang lua>-- test for even number if n % 2 == 0 then

 print "The number is even"

end

-- test for odd number if not (n % 2 == 0) then

 print "The number is odd"

end</lang>

M4

<lang 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)</lang>

Maple

<lang Maple>EvenOrOdd := proc( x::integer )

  if x mod 2 = 0 then
     print("Even"):
  else
     print("Odd"):
  end if:

end proc: EvenOrOdd(9);</lang>

"Odd"

Mathematica

<lang Mathematica>EvenQ[8]</lang>

MATLAB / Octave

Bitwise And: <lang Matlab> isOdd = logical(bitand(N,1));

  isEven = ~logical(bitand(N,1)); </lang>

Remainder of division by two <lang Matlab> isOdd = logical(rem(N,2));

  isEven = ~logical(rem(N,2)); </lang>

Modulo: 2 <lang Matlab> isOdd = logical(mod(N,2));

  isEven = ~logical(mod(N,2)); </lang>

Maxima

<lang maxima>evenp(n); oddp(n);</lang>

Mercury

Mercury's 'int' module provides tests for even/odd, along with all the operators that would be otherwise used to implement them. <lang Mercury>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 -infinity even(N) :- N rem 2 = 0. % using division that truncates towards zero even(N) :- N /\ 1 = 0. % using bit-wise and.</lang>

МК-61/52

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

ML

mLite

<lang ocaml>fun odd (x rem 2 = 1) = true | _ = false

fun even (x rem 2 = 0) = true | _ = false

</lang>

NetRexx

<lang 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 nn

return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -- Overloaded method. Default is to use the remainder specialization below method 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)

</lang>

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

Nim

<lang nim># Least signficant bit: proc isOdd(i: int): bool = (i and 1) != 0 proc isEven(i: int): bool = (i and 1) == 0

Modulo:

proc isOdd2(i: int): bool = (i mod 2) != 0 proc isEven2(i: int): bool = (i mod 2) == 0

echo isEven(1) echo isOdd2(5)</lang>

Objeck

<lang objeck>a := Console->ReadString()->ToInt(); if(a % 2 = 0) {

 "even"->PrintLine();

} else {

 "odd"->PrintLine();

};</lang>

OCaml

Modulo: <lang ocaml>let is_even d =

 (d mod 2) = 0

let is_odd d =

 (d mod 2) <> 0</lang>

Bitwise and: <lang ocaml>let is_even d =

 (d land 1) = 0

let is_odd d =

 (d land 1) <> 0</lang>

Oforth

<lang Oforth>12 isEven 12 isOdd</lang>

OOC

<lang ooc> // Using the modulo operator even: func (n: Int) -> Bool {

 (n % 2) == 0

}

// Using bitwise and odd: func (n: Int) -> Bool {

 (n & 1) == 1

} </lang>

PARI/GP

GP does not have a built-in predicate for testing parity, but it's easy to code: <lang parigp>odd(n)=n%2;</lang> Alternately: <lang parigp>odd(n)=bitand(n,1);</lang> 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: <lang pascal>isOdd := odd(someIntegerNumber);</lang> bitwise and: <lang pascal>function isOdd(Number: integer): boolean begin

 isOdd := boolean(Number and 1)

end;</lang> Dividing and multiplying by 2 and test on equality: <lang pascal>function isEven(Number: integer): boolean begin

 isEven := (Number = ((Number div 2) * 2))

end;</lang> Using built-in modulo <lang pascal>function isOdd(Number: integer): boolean begin

 isOdd := boolean(Number mod 2)

end;</lang>

Perl

<lang perl>for(0..10){

   print "$_ is ", qw(even odd)[$_ % 2],"\n";

}</lang> or <lang perl>print 6 % 2 ? 'odd' : 'even'; # prints even</lang>

Perl 6

Perl 6 doesn't have a built-in for this, but with subsets it's easy to define a predicate for it. <lang perl6>subset Even of Int where * %% 2; subset Odd of Int where * % 2;

say 1 ~~ Even; # false say 1 ~~ Odd; # true say 1.5 ~~ Odd # false ( 1.5 is not an Int )</lang>

PHP

<lang php> // using bitwise and to check least significant digit echo (2 & 1) ? 'odd' : 'even'; echo (3 & 1) ? 'odd' : 'even';

// using modulo echo (3 % 2) ? 'odd' : 'even'; echo (4 % 2) ? 'odd' : 'even'; </lang>

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. <lang PicoLisp>: (bit? 1 3) -> 1 # Odd

(bit? 1 4)

-> NIL # Even</lang>

Pike

<lang Pike>> int i = 73; > (i&1); Result: 1 > i%2; Result: 1</lang>

PL/I

<lang PL/I>i = iand(i,1)</lang> The result is 1 when i is odd, and 0 when i is even.

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: <lang prolog>

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

</lang>

Least Significant Bit

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

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

</lang>

PureBasic

<lang PureBasic>;use last bit method isOdd = i & 1 ;isOdd is non-zero if i is odd isEven = i & 1 ! 1 ;isEven is non-zero if i is even

use modular method

isOdd = i % 2 ;isOdd is non-zero if i is odd isEven = i % 2 ! 1 ;isEven is non-zero if i is even</lang>

Python

Python: Using the least-significant bit method

<lang python>>>> 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)] >>> </lang>

Python: Using modular congruences

<lang python>>> def is_even(i):

       return (i % 2) == 0

>>> is_even(1) False >>> is_even(2) True >>></lang>

R

<lang R>is.even <- function(x) !is.odd(x)

is.odd <- function(x) intToBits(x)[1] == 1

or

is.odd <- function(x) x %% 2 == 1</lang>

Racket

With built in predicates: <lang Racket>(even? 6) ; -> true (even? 5) ; -> false (odd? 6) ; -> false (odd? 5) ; -> true </lang>

With modular arithmetic: <lang Racket>(define (my-even? x)

 (= (modulo x 2) 0))

(define (my-odd? x)

 (= (modulo x 2) 1))</lang>

Rascal

<lang rascal>public bool isEven(int n) = (n % 2) == 0; public bool isOdd(int n) = (n % 2) == 1;</lang> Or with block quotes: <lang rascal>public bool isEven(int n){return (n % 2) == 0;} public bool isOdd(int n){return (n % 2) == 1;}</lang>

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 a 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.
This method gets its speed from not using any BIF and not doing any division. <lang rexx>/*REXX program displays if an integer is even or odd. */ /*REXX program displays if an integer is even or odd. */ !.=0; !.0=1; !.2=1; !.4=1; !.6=1; !.8=1 /*assign even digits to "true".*/ numeric digits 1000 /*handle most big 'uns from the CL*/ parse arg x _ . /*get arg(s) from the command line*/ if x== then call terr 'no input' if _\== | arg()\==1 then call terr 'too many arguments: ' arg(1) if \datatype(x,'N') then call terr x " isn't numeric" if \datatype(x,'W') then call terr x " isn't an integer" y=abs(x)/1 /*just in case X is negative, */

                                     /*(remainder of neg # might be -1)*/
                   /*══════════════════════════════════════════════════*/
                   say center('test using remainder method',40,'─')

if y//2 then say x 'is odd'

        else say  x  'is even'
                   /*══════════════════════════════════════════════════*/

say; say center('test rightmost digit for evenness',40,'─') _=right(y,1) if pos(_,02468)==0 then say x 'is odd'

                   else say  x  'is even'
                   /*══════════════════════════════════════════════════*/

say; say center('test rightmost digit for oddness',40,'─') if pos(right(y,1),13579)==0 then say x 'is even'

                            else say  x  'is odd'
                   /*══════════════════════════════════════════════════*/

say; say center('test rightmost (binary) bit',40,'─') if right(x2b(d2x(y)),1) then say x 'is odd'

                        else say  x  'is even'
                   /*══════════════════════════════════════════════════*/

say; say center('test using parse statement',40,'─') parse var y -1 _ /*obtain last decimal digit of Y.*/ if !._ then say x 'is even'

       else say  x  'is odd'

exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────TERR subroutine─────────────────────*/ terr: say; say '***error!***'; say; say arg(1); say; exit 13</lang>

Output:

when using the input of

0

─────────────modulo method──────────────
0 is even

───test rightmost digit for evenness────
0 is even

────test rightmost digit for oddness────
0 is even

──────test rightmost (binary) bit───────
0 is even

───────test using parse statement───────
0 is even

out0put when using the input of: 9876543210987654321098765432109876543210987654321

─────────────modulo method──────────────
9876543210987654321098765432109876543210987654321 is odd

───test rightmost digit for evenness────
9876543210987654321098765432109876543210987654321 is odd

────test rightmost digit for oddness────
9876543210987654321098765432109876543210987654321 is odd

──────test rightmost (binary) bit───────
9876543210987654321098765432109876543210987654321 is odd

───────test using parse statement───────
9876543210987654321098765432109876543210987654321 is odd

Output:

when using the input of

00067.00

────────test using modulo method────────
00067.00 is odd

───test rightmost digit for evenness────
00067.00 is odd

────test rightmost digit for oddness────
00067.00 is odd

──────test rightmost (binary) bit───────
00067.00 is odd

───────test using parse statement───────
00067.00 is odd

Output:

when using the input of

-9411

─────────────modulo method──────────────
-9411 is odd

───test rightmost digit for evenness────
-9411 is odd

────test rightmost digit for oddness────
-9411 is odd

──────test rightmost (binary) bit───────
-9411 is odd

───────test using parse statement───────
-9411 is odd

Ruby

Ruby 1.8.7 added Integer#even? and Integer#odd? as new methods.

Works with: Ruby version 1.8.7

<lang ruby>print "evens: " p -5.upto(5).select {|n| n.even?} print "odds: " p -5.upto(5).select {|n| n.odd?}</lang>

Output:

evens: [-4, -2, 0, 2, 4]
odds: [-5, -3, -1, 1, 3, 5]

Other ways to test even-ness: <lang ruby>n & 1 == 0 quotient, 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?</lang>

Run BASIC

<lang runbasic>for i = 1 to 10

 if i and 1 then print i;" is odd" else print i;" is even"

next i</lang>

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: <lang rust>let is_odd = |x: i32| x & 1 == 1; let is_even = |x: i32| x & 1 == 0;</lang>

Using modular congruences: <lang rust>let is_odd = |x: i32| x % 2 != 0; let is_even = |x: i32| x % 2 == 0;</lang>

Scala

<lang scala>def isEven( v:Int ) : Boolean = v % 2 == 0 def isOdd( v:Int ) : Boolean = v % 2 != 0</lang> Accept any numeric type as an argument: <lang scala>def isEven( v:Number ) : Boolean = v.longValue % 2 == 0 def isOdd( v:Number ) : Boolean = v.longValue % 2 != 0</lang>

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): <lang scheme>> (even? 5)

f

> (even? 42)

t

> (odd? 5)

t

> (odd? 42)

f</lang>

Seed7

Test whether an integer or bigInteger is odd: <lang seed7>odd(aNumber)</lang> Test whether an integer or bigInteger is even: <lang seed7>not odd(aNumber)</lang>

Sidef

Built-in methods: <lang ruby>var n = 42; say n.is_odd; # false say n.is_even; # true</lang>

Checking the last significant digit: <lang ruby>func is_odd(n) { n&1 == 1 }; func is_even(n) { n&1 == 0 };</lang>

Using modular congruences: <lang ruby>func is_odd(n) { n%2 == 1 }; func is_even(n) { n%2 == 0 };</lang>

Smalltalk

Using the built in methods on Number class:

even is implemented as follows: <lang smalltalk>Number>>even ^((self digitAt: 1) bitAnd: 1) = 0 </lang>

SNUSP

<lang SNUSP> $====!/?\==even#

- -

odd==\?/

</lang>

Swift

<lang Swift>func isEven(n:Int) -> Bool {

   // Bitwise check
   if (n & 1 != 0) {
       return false
   }
   
   // Mod check
   if (n % 2 != 0) {
       return false
   }
   return true

}</lang>

Tcl

<lang tcl>package require Tcl 8.5

Bitwise test is the most efficient

proc 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)]</lang>

Output:

 # O E
24:0,1
49:1,0

TUSCRIPT

<lang tuscript>$$ MODE TUSCRIPT LOOP n=-5,5 x=MOD(n,2) SELECT x CASE 0 PRINT n," is even" DEFAULT PRINT n," is odd" ENDSELECT ENDLOOP</lang>

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

XPL0

<lang 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);
       ];

]</lang>

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

zkl

<lang zkl>[-3..4].pump(fcn(n){ println(n," is ",n.isEven and "even" or "odd") })</lang> 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

<lang zkl>[-3..4].apply("isEven").println();</lang>

Output:

L(False,True,False,True,False,True,False,True)