Check whether number is even or odd
Solution 1:
You can use the modulus operator, but that can be slow. If it's an integer, you can do:
if ( (x & 1) == 0 ) { even... } else { odd... }
This is because the low bit will always be set on an odd number.
Solution 2:
if ((x % 2) == 0) {
// even
} else {
// odd
}
Solution 3:
If the remainder when you divide by 2 is 0, it's even. % is the operator to get a remainder.
Solution 4:
The remainder operator, %, will give you the remainder after dividing by a number.
So n % 2 == 0 will be true if n is even and false if n is odd.