Comment on I wish
DarkMessiah@lemmy.world 1 year ago
Just in case anyone was looking for a decent way to do it…
if (((number/2) - round(number/2)) == 0) return true; else return false;
Or whatever the rounding function is in your language of choice.
Every bit aside for the ones bit is even, all you have to do is yet the ones bit(the far left) for it being a 1 or 0. Which is the fastest and least amount of code needed.
use bitwise &
bool isEven(int n)
{// n&1 is true, then odd, or !n&1 is true for even return (!(n & 1));}
Huh?
return number % 2 == 0
That’s the only sane solution.
Do note how I said “a decent” way, not “the best” way. Get that huh outta here.
Or modulo %
private bool IsEven(int number){ return number % 2 ? false : true; }
number % 2 == 0 and // If integer (number & 0b1) == 0
Are the only sane ways to do this. No need to floor.
If you are using floats, you really do not want to have an isEven function …
Whats the alternative a macro? An inline function is perfectly fine for checking if a nunber is even. Compiler will probably optimize it to a single and instruction.
No. The alternative is to not use a float. Testing if a float is even simply does not make sense.
Even testing two floats for equality rarely makes sense.
Take out the else and I’m in
Valid point.
AnxiousOtter@lemmy.world 1 year ago
Modulo operator my dude.