this post was submitted on 01 Nov 2023
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Just in case anyone was looking for a decent way to do it...
Or whatever the rounding function is in your language of choice.
EDIT: removed unnecessary else.
Modulo operator my dude.
Every bit aside for the ones bit is even. All you have to do is get the ones bit(the far right) for it being a 1 or 0. Which is the fastest and least amount of code needed.
use bitwise &
Or modulo %
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.
Are the only sane ways to do this. No need to floor. Although If its C and you can't modulo floats then (number/2 == floor(number/2))
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.
What is the correct output of isEven((.2 + .4) ×10)
Hint: (.2 + .4) x 10 != 6
It does if it dosen't have a decimal. If it has decimal then it automatically isn't and the function will return false. Are you talking about cases like 0.1 + 0.2 equaling 0.3000000004 because that is just due to the nature of floats and there is nothing a function can do other than use larger floats for more accuracy.
Take out the
elseand I'm inValid point.