What is the quickest way to check whether a number is divisible by 2
Someone started and deleted a thread. For those looking for the answer:
check and see if the lowest bit is set
if (number and &b1) > 0 if it is its an odd number else if it isnt its even & therefor divisible by 2 end if
dunno if this is quicker than mod but it should be
I tested, they are about the same.
if ( number mod 2 ) = 0 then // it's even else // it's odd end if