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# Can anyone explain a formula for determining odd and even numbers in loops.

Specifically I want to omit a user input in the loop if the number is odd.

2 Antworten

+ 1

A much faster test than using modulo is to check the number's lowest bit. If the bit is 1, the number is odd. If it is 0, the number is even.
You can do this by using the bitwise & (and) operator.
if (num&1==1) //odd
if (num&1==0) //even

+ 8

You need to use modulus (%) operator, if num%2==0 number is even, if not it's an odd number.
So you could use while loop, or for in if you are working with arrays, for of if you are working with objects