How can I write a bash function that checks if an integer is even or not?

Question

How can I write a shell predicate function event that takes an integer argument and tests if it's even or not?

Answer 1

Use the modulo operator

Example:

#!/bin/bash

iseven() {
    return $(($1 % 2))
}

if iseven 10; then echo 10 is even; else echo 10 is odd; fi
if iseven 11; then echo 11 is even; else echo 11 is odd; fi

Answer 2

Check the remainder:

var1=7
[ $((var1%2)) -ne 0 ] && echo "odd"
var2=4
[ $((var2%2)) -eq 0 ] && echo "even"

...is the way to go. Although I am confused with your later question edit asking your "predicate function event".

Answer 3

Actually bash has a bitwise AND operator, so a bitwise AND with 1 is the same conceptually with useing % 2 to get a remainder mod 2

[ $((var & 1)) -ne 0 ] && echo "odd"
[ $((var & 1)) -eq 0 ] && echo "even"

(and save microseconds by skipping a division)

EDIT: removed expression with ! that somehow was history and a logical not