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views:

138

answers:

5

I feel stupid asking such a simple question, but is there an easy way to determine whether an Integer is even or odd?

+7  A: 
if ((n % 2) == 0) {
    // number is even
}

else {
    // number is odd
}
Richard Fearn
+3  A: 

You can use modular division (technically in Java it acts as a strict remainder operator; the link has more discussion):

if ( ( n % 2 ) == 0 ) {
    //Is even
} else {
    //Is odd
}
eldarerathis
+3  A: 

If you do a bitwise-and with 1, you can detect whether the least significant bit is 1. If it is, the number is odd, otherwise even.

In C-ish languages, bool odd = mynum & 1;

This is faster (performance-wise) than mod, if that's a concern.

mtrw
wouldn't even be true when mynum is odd?
Bill James
I think this is flawed. You need to rename your variable to odd.
Anton
@Bill, @Anton - sorry, I had posted with the wrong sense for the result. I thought I had edited before anyone caught me...
mtrw
+4  A: 

It's not exactly android specific, but a pretty standard function would be:

boolean isOdd( int val ) { return (val & 0x01) != 0; }
Bill James
A: 

When somehow % as an operator doesn't exist, you can use the AND operator:

oddness = (n & 1) ? 'odd' : 'even'
littlegreen
yeah, that was a typo, sorry!
littlegreen
shouldn't it be double quotes?
st0le