I feel stupid asking such a simple question, but is there an easy way to determine whether an Integer is even or odd?
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27
It's not android specific, but a standard function would be:
boolean isOdd( int val ) { return (val & 0x01) != 0; }
Many compilers convert modulo (%
) operations to their bitwise counterpart automatically, but this method is also compatible with older compilers.

Abandoned Cart
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billjamesdev
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11
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
}

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eldarerathis
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4
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
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wouldn't even be true when mynum is odd? – billjamesdev Sep 29 '10 at 21:06
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I think this is flawed. You need to rename your variable to odd. – Anton Sep 29 '10 at 21:16
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@Bill, @Anton - sorry, I had posted with the wrong sense for the result. I thought I had edited before anyone caught me... – mtrw Sep 29 '10 at 21:20
1
When somehow %
as an operator doesn't exist, you can use the AND operator:
oddness = (n & 1) ? 'odd' : 'even'

littlegreen
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1
Similar to others, but here's how I did it.
boolean isEven = i %2 ==0;

seekingStillness
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