Fast isEven() function

Just a note for myself...

This is a way I used to know if int is even or odd:

(i % 2) == 0;

And here's a faster way

((i * 0.5) - (i >> 1)) == 0;

I was afraid that first part

i * 0.5;

can give 1.0000001, for example, due to round error. But it sems that integer numbers divided by 2 without presision lost. If someone have another information, please let me know.

Also the best performance gain experienced if above code used inline. Calling a method like this

private function isEven(i:int):Boolean{
 return ((i * 0.5) - (i >> 1)) == 0;
}

is 10 times slower!

Upd.: method that you can find in comment below or here - great blog - is even faster. I definetely will use this approach.