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This is a really simple one, but when I spoke to a developer recently he had completely overlooked the simplest way to discover if an integer is odd or even, and was performing quite a convoluted formula to get the answer.
He had completely overlooked using the built in modulus operator (%) to find the result. This provides the remainder after division, so for an even number after dividing by 2 the remainder will be 0, otherwise it will be 1.
The simplest way to do this is
int myvar;
if(myVar % 2==0)
{ // its even
}
else
{ // its odd
}
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24 November 2009 IB
C#
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