Even or odd
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
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.
Contents
- 1 0815
- 2 6502 Assembly
- 3 8th
- 4 ABAP
- 5 Ada
- 6 Aime
- 7 ALGOL 68
- 8 ALGOL W
- 9 AntLang
- 10 APL
- 11 AppleScript
- 12 Arendelle
- 13 AutoHotkey
- 14 AWK
- 15 BaCon
- 16 BASIC
- 17 Batch File
- 18 BBC BASIC
- 19 bc
- 20 Befunge
- 21 Bracmat
- 22 Brainf***
- 23 Burlesque
- 24 C
- 25 C#
- 26 C++
- 27 Clojure
- 28 COBOL
- 29 CoffeeScript
- 30 ColdFusion
- 31 Common Lisp
- 32 Component Pascal
- 33 Crystal
- 34 D
- 35 DCL
- 36 DWScript
- 37 Déjà Vu
- 38 EDSAC order code
- 39 Eiffel
- 40 Elixir
- 41 Emacs Lisp
- 42 Erlang
- 43 ERRE
- 44 Euphoria
- 45 Excel
- 46 F#
- 47 Factor
- 48 Fish
- 49 Forth
- 50 Fortran
- 51 FreeBASIC
- 52 Futhark
- 53 Gambas
- 54 GAP
- 55 Go
- 56 Groovy
- 57 Haskell
- 58 Icon and Unicon
- 59 J
- 60 Java
- 61 JavaScript
- 62 jq
- 63 Julia
- 64 K
- 65 Kotlin
- 66 L++
- 67 LabVIEW
- 68 Lang5
- 69 Lasso
- 70 LC3 Assembly
- 71 Liberty BASIC
- 72 Lingo
- 73 LiveCode
- 74 Logo
- 75 Logtalk
- 76 Lua
- 77 M2000 Interpreter
- 78 M4
- 79 Maple
- 80 Mathematica / Wolfram Language
- 81 MATLAB / Octave
- 82 Maxima
- 83 Mercury
- 84 MIPS Assembly
- 85 МК-61/52
- 86 ML
- 87 Modula-2
- 88 Neko
- 89 NESL
- 90 NetRexx
- 91 NewLISP
- 92 Nim
- 93 Oberon-2
- 94 Objeck
- 95 OCaml
- 96 Oforth
- 97 OOC
- 98 PARI/GP
- 99 Pascal
- 100 Perl
- 101 Perl 6
- 102 Phix
- 103 PHP
- 104 PicoLisp
- 105 Pike
- 106 PL/I
- 107 PowerShell
- 108 Processing
- 109 Prolog
- 110 PureBasic
- 111 Python
- 112 R
- 113 Racket
- 114 Rascal
- 115 REXX
- 116 Ring
- 117 Ruby
- 118 Run BASIC
- 119 Rust
- 120 Scala
- 121 Scheme
- 122 Seed7
- 123 SequenceL
- 124 SETL
- 125 Shen
- 126 Sidef
- 127 Smalltalk
- 128 SNOBOL4
- 129 SNUSP
- 130 SPL
- 131 SQL
- 132 SSEM
- 133 Standard ML
- 134 Stata
- 135 Swift
- 136 Symsyn
- 137 Tcl
- 138 TI-83 BASIC
- 139 TUSCRIPT
- 140 UNIX Shell
- 141 Ursa
- 142 உயிர்/Uyir
- 143 VBA
- 144 VBScript
- 145 Visual Basic .NET
- 146 WDTE
- 147 xEec
- 148 XLISP
- 149 Xojo
- 150 XPL0
- 151 Yabasic
- 152 zkl
- 153 zonnon
- 154 ZX Spectrum Basic
0815[edit]
}: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[edit]
.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 II
CharOut = $fded ;Specific to the Apple II
;------------------------------------------------------
; The main program
;
main ldy #sIntro-sbase
jsr puts ;Print Intro
loop 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-sbase
isodd 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 times
done rts ;Return to caller
;------------------------------------------------------
; String Constants (in '+128' ascii, Apple II style)
;
sbase: ;String base address
sIntro .az -"Hit any key (Ctrl-C to quit):",-#13
sEven .az -" is even.",-#13
sOdd .az -" is odd.",-#13
;------------------------------------------------------
.en
8th[edit]
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[edit]
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[edit]
-- 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[edit]
if (x & 1) {
# x is odd
} else {
# x is even
}
ALGOL 68[edit]
# 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[edit]
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_i
end.
- Output:
1 is odd 1702 is even 23 is odd -26 is even
AntLang[edit]
odd: {x mod 2}
even: {1 - x mod 2}
APL[edit]
The easiest way is probably to use modulo.
2|28
0
2|37
1
AppleScript[edit]
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 if
end 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 -> Bool
on even(n)
n mod 2 = 0
end even
-- odd :: Integral a => a -> Bool
on 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 tell
end filter
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property lambda : f
end script
end if
end 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[edit]
( input , "Please enter a number: " ) { @input % 2 = 0 , "| @input | is even!" , "| @input | is odd!" }
AutoHotkey[edit]
Bitwise ops are probably most efficient:
if ( int & 1 ){
; do odd stuff
}else{
; do even stuff
}
AWK[edit]
function isodd(x) {
return (x%2)!=0;
}
function iseven(x) {
return (x%2)==0;
}
BaCon[edit]
' Even or odd
OPTION MEMTYPE int
SPLIT ARGUMENT$ BY " " TO arg$ SIZE dim
n = 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[edit]
Applesoft BASIC[edit]
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
IS-BASIC[edit]
100 DEF ODD(X)=MOD(X,2)
110 INPUT PROMPT "Enter a number: ":X
120 IF ODD(X) THEN
130 PRINT X;"is odd."
140 ELSE
150 PRINT X;"is even."
160 END IF
Batch File[edit]
@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
BBC BASIC[edit]
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[edit]
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 = 0
if (i % 2 == 0) "i is even
"
if (i % 2) "i is odd
"
Befunge[edit]
&2%52**"E"+,@
Outputs E if even, O if odd.
Bracmat[edit]
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***[edit]
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.
,[>,----------]<[--]
Burlesque[edit]
2.%
C[edit]
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#[edit]
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++[edit]
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[edit]
Standard predicates:
(if (even? some-var) (do-even-stuff))
(if (odd? some-var) (do-odd-stuff))
COBOL[edit]
IF FUNCTION REM(Num, 2) = 0
DISPLAY Num " is even."
ELSE
DISPLAY Num " is odd."
END-IF
CoffeeScript[edit]
isEven = (x) -> !(x%2)
ColdFusion[edit]
function f(numeric n) {
return n mod 2?"odd":"even"
}
Common Lisp[edit]
Standard predicates:
(if (evenp some-var) (do-even-stuff))
(if (oddp some-other-var) (do-odd-stuff))
Alternate solution[edit]
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[edit]
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) # 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;
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[edit]
#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[edit]
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[edit]
$! 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[edit]
Predicate:
var isOdd := Odd(i);
Bitwise and:
var isOdd := (i and 1)<>0;
Modulo:
var isOdd := (i mod 2)=1;
Déjà Vu[edit]
even n:
= 0 % n 2
odd:
not even
!. odd 0
!. even 0
!. odd 7
!. even 7
- Output:
false true true false
EDSAC order code[edit]
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[edit]
--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
Elixir[edit]
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"
end
end
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[edit]
With evenp and oddp[edit]
(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[edit]
(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[edit]
Using Division by 2 Method[edit]
%% 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[edit]
%% 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[edit]
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
- Output:
14 is even 15 is odd 9876543210 is even 9876543211 is odd
Euphoria[edit]
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[edit]
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#[edit]
Bitwise and:
let isEven x =
x &&& 1 = 0
Modulo:
let isEven x =
x % 2 = 0
Factor[edit]
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[edit]
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[edit]
: odd? ( n -- ? ) 1 and ;
Fortran[edit]
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[edit]
' 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"
Else
Print "Your number is odd"
End if
Loop
End
Futhark[edit]
fun main(x: int): bool = (x & 1) == 0
Gambas[edit]
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[edit]
IsEvenInt(n);
IsOddInt(n);
Go[edit]
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[edit]
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) }
Haskell[edit]
even
and odd
functions are already included in the standard Prelude.
Prelude> even 5
False
Prelude> even 42
True
Prelude> odd 5
True
Prelude> odd 42
False
Where even is derived from rem, and odd is derived from even:
import Prelude hiding (even, odd)
even, odd
:: (Integral a)
=> a -> Bool
even = (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[edit]
One way is to check the remainder:
procedure isEven(n)
return n%2 = 0
end
J[edit]
Modulo:
2 | 2 3 5 7
0 1 1 1
2|2 3 5 7 + (2^89x)-1
1 0 0 0
Remainder:
(= <.&.-:) 2 3 5 7
1 0 0 0
(= <.&.-:) 2 3 5 7+(2^89x)-1
0 1 1 1
Last bit in bit representation:
{:"1@#: 2 3 5 7
0 1 1 1
{:"1@#: 2 3 5 7+(2^89x)-1
1 0 0 0
Bitwise and:
1 (17 b.) 2 3 5 7
0 1 1 1
Note: as a general rule, the simplest expressions in J should be preferred over more complex approaches.
Java[edit]
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[edit]
ES5[edit]
Bitwise:
function isEven( i ) {
return (i & 1) === 0;
}
Modulo:
function isEven( i ) {
return i % 2 === 0;
}
// Alternative
function isEven( i ) {
return !(i % 2);
}
ES6[edit]
Lambda:
// EMCAScript 6
const 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[edit]
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[edit]
Built-in functions:
iseven(i), isodd(i)
K[edit]
The following implementation uses the modulo of division by 2
oddp: {:[x!2;1;0]} /Returns 1 if arg. is odd
evenp: {~oddp[x]} /Returns 1 if arg. is even
Examples:
oddp 32
0
evenp 32
1
Kotlin[edit]
// 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++[edit]
(defn bool isEven (int x) (return (% x 2)))
LabVIEW[edit]
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[edit]
: even? 2 % not ;
: odd? 2 % ;
1 even? . # 0
1 odd? . # 1
Lasso[edit]
define isoddoreven(i::integer) => {
#i % 2 ? return 'odd'
return 'even'
}
isoddoreven(12)
LC3 Assembly[edit]
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[edit]
n=12
if n mod 2 = 0 then print "even" else print "odd"
Lingo[edit]
on even (n)
return n mod 2 = 0
end
on odd (n)
return n mode 2 <> 0
end
LiveCode[edit]
function odd n
return (n bitand 1) = 1
end odd
function notEven n
return (n mod 2) = 1
end notEven
Logo[edit]
to even? :num
output equal? 0 modulo :num 2
end
Logtalk[edit]
:- 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).
odd
yes
| ?- even_odd::test_mod(2).
even
yes
| ?- even_odd::test_bit(1).
odd
yes
| ?- even_odd::test_bit(2).
even
yes
Lua[edit]
-- 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
M2000 Interpreter[edit]
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 { Read x code here }
The same hold for Def Odd(a)=binary.and(Abs(a), 1)=1 Interpreter execute this:
Function Odd { Read a =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(212121[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[edit]
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[edit]
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[edit]
EvenQ[8]
MATLAB / Octave[edit]
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[edit]
evenp(n);
oddp(n);
Mercury[edit]
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 -infinity
even(N) :- N rem 2 = 0. % using division that truncates towards zero
even(N) :- N /\ 1 = 0. % using bit-wise and.
MIPS Assembly[edit]
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,odd
even:
#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[edit]
/ 2 {x} ЗН
Result: "0" - number is even; "1" - number is odd.
ML[edit]
mLite[edit]
fun odd
(x rem 2 = 1) = true
| _ = false
;
fun even
(x rem 2 = 0) = true
| _ = false
;
Modula-2[edit]
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;
ReadChar
END EvenOrOdd.
Neko[edit]
var number = 6;
if(number % 2 == 0) {
$print("Even");
} else {
$print("Odd");
}
- Output:
Even
NESL[edit]
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[edit]
/* 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)
- 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[edit]
(odd? 1)
(even? 2)
Nim[edit]
# 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
# 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[edit]
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.Ln
END EvenOrOdd.
- Output:
10 even 11 odd 12 even
Objeck[edit]
a := Console->ReadString()->ToInt();
if(a % 2 = 0) {
"even"->PrintLine();
}
else {
"odd"->PrintLine();
};
OCaml[edit]
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[edit]
12 isEven
12 isOdd
OOC[edit]
// Using the modulo operator
even: func (n: Int) -> Bool {
(n % 2) == 0
}
// Using bitwise and
odd: func (n: Int) -> Bool {
(n & 1) == 1
}
PARI/GP[edit]
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[edit]
Built-in boolean function odd:
isOdd := odd(someIntegerNumber);
bitwise and:
function isOdd(Number: integer): boolean
begin
isOdd := boolean(Number and 1)
end;
Dividing and multiplying by 2 and test on equality:
function isEven(Number: integer): boolean
begin
isEven := (Number = ((Number div 2) * 2))
end;
Using built-in modulo
function isOdd(Number: integer): boolean
begin
isOdd := boolean(Number mod 2)
end;
Perl[edit]
for(0..10){
print "$_ is ", qw(even odd)[$_ % 2],"\n";
}
or
print 6 % 2 ? 'odd' : 'even'; # prints even
Perl 6[edit]
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; # false
say 1 ~~ Odd; # true
say 1.5 ~~ Odd # false ( 1.5 is not an Int )
Phix[edit]
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[edit]
// 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';
- Output:
even odd odd even
PicoLisp[edit]
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[edit]
> int i = 73;
> (i&1);
Result: 1
> i%2;
Result: 1
PL/I[edit]
i = iand(i,1)
The result is 1 when i is odd, and 0 when i is even.
PowerShell[edit]
Predicate[edit]
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[edit]
$IsOdd = [boolean]( $N -band 1 )
$IsEven = [boolean]( $N -band 0 )
Remainder[edit]
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[edit]
boolean isEven(int i){
return i%2 == 0;
}
boolean isOdd(int i){
return i%2 == 1;
}
Prolog[edit]
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[edit]
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[edit]
;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
Python[edit]
Python: Using the least-significant bit method[edit]
>>> 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[edit]
>> def is_even(i):
return (i % 2) == 0
>>> is_even(1)
False
>>> is_even(2)
True
>>>
R[edit]
is.even <- function(x) !is.odd(x)
is.odd <- function(x) intToBits(x)[1] == 1
#or
is.odd <- function(x) x %% 2 == 1
Racket[edit]
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[edit]
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[edit]
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: " x
if \datatype(x, 'W') then call terr "argument isn't an integer: " x
y=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, "═"); return
terr: 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[edit]
size = 10
for i = 1 to size
if i % 2 = 1 see "" + i + " is odd" + nl
else see "" + i + " is even" + nl ok
next
Ruby[edit]
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 == 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?
Run BASIC[edit]
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[edit]
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[edit]
def isEven( v:Int ) : Boolean = v % 2 == 0
def isOdd( v:Int ) : Boolean = v % 2 != 0
Accept any numeric type as an argument:
def isEven( v:Number ) : Boolean = v.longValue % 2 == 0
def 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[edit]
even?
and odd?
functions are built-in (R4RS, R5RS, and R6RS):
> (even? 5)
#f
> (even? 42)
#t
> (odd? 5)
#t
> (odd? 42)
#f
Seed7[edit]
Test whether an integer or bigInteger is odd:
odd(aNumber)
Test whether an integer or bigInteger is even:
not odd(aNumber)
SequenceL[edit]
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[edit]
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[edit]
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[edit]
Built-in methods:
var n = 42;
say n.is_odd; # false
say 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[edit]
Using the built in methods on Number class:
5 even
5 odd
even is implemented as follows:
Number>>even
^((self digitAt: 1) bitAnd: 1) = 0
SNOBOL4[edit]
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[edit]
$====!/?\==even#
- -
#odd==\?/
SPL[edit]
> 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[edit]
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 integers
CREATE 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' END
FROM
ints;
- Output:
INT CASE ---------- ---- -1 Odd 0 Even 1 Odd 2 Even
SSEM[edit]
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 into storage address 15. The first three instructions test whether 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 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 , 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 c
00000000000000110000000000000000 1. Test
11110000000001100000000000000000 2. c to 15
11110000000000100000000000000000 3. -15 to c
00001000000001100000000000000000 4. c to 16
00001000000000100000000000000000 5. -16 to c
01110000000000010000000000000000 6. Sub. 14
11110000000001100000000000000000 7. c to 15
10110000000000010000000000000000 8. Sub. 13
00000000000000110000000000000000 9. Test
01110000000000000000000000000000 10. 14 to CI
11110000000000100000000000000000 11. -15 to c
00000000000001110000000000000000 12. Stop
10000000000000000000000000000000 13. 1
01000000000000000000000000000000 14. 2
Standard ML[edit]
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[edit]
mata
function iseven(n) {
return(mod(n,2)==0)
}
function isodd(n) {
return(mod(n,2)==1)
}
end
Swift[edit]
func isEven(n:Int) -> Bool {
// Bitwise check
if (n & 1 != 0) {
return false
}
// Mod check
if (n % 2 != 0) {
return false
}
return true
}
Symsyn[edit]
n : 23
if n bit 0
'n is odd' []
else
'n is even' []
Tcl[edit]
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)]
- Output:
# O E 24:0,1 49:1,0
TI-83 BASIC[edit]
TI-83 BASIC does not have a modulus operator.
If fPart(.5Ans
Then
Disp "ODD
Else
Disp "EVEN
End
TUSCRIPT[edit]
$$ 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
- 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[edit]
iseven() {
[[ $(($1%2)) -eq 0 ]] && return 0
return 1
}
Ursa[edit]
decl int input
set input (in int console)
if (= (mod input 2) 1)
out "odd" endl console
else
out "even" endl console
end if
Output:
123 odd
உயிர்/Uyir[edit]
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[edit]
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")
Next
End 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 = True
End 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[edit]
Function odd_or_even(n)
If n Mod 2 = 0 Then
odd_or_even = "Even"
Else
odd_or_even = "Odd"
End If
End Function
WScript.StdOut.Write "Please enter a number: "
n = WScript.StdIn.ReadLine
WScript.StdOut.Write n & " is " & odd_or_even(CInt(n))
WScript.StdOut.WriteLine
- Output:
C:\>cscript /nologo odd_or_even.vbs Please enter a number: 6 6 is Even C:\>cscript /nologo odd_or_even.vbs Please enter a number: 9 9 is Odd C:\>cscript /nologo odd_or_even.vbs Please enter a number: -1 -1 is Odd
Visual Basic .NET[edit]
Module Module1
Sub Main()
Dim str As String
Dim num As Integer
While True
Console.Write("Enter and integer or 0 to finish: ")
str = Console.ReadLine()
If Integer.TryParse(str, num) Then
If num = 0 Then
Exit While
End If
If num Mod 2 = 0 Then
Console.WriteLine("Even")
Else
Console.WriteLine("Odd")
End If
Else
Console.WriteLine("Bad input.")
End If
End While
End Sub
End Module
WDTE[edit]
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[edit]
>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[edit]
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[edit]
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 If
Next
- Output:
1 is odd. 2 is even. 3 is odd. 4 is even. 5 is odd.
XPL0[edit]
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[edit]
for i = -5 to 5
print i, and(i,1), mod(i,2)
next
zkl[edit]
[-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[edit]
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[edit]
10 FOR n=-3 TO 4: GO SUB 30: NEXT n
20 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
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