Check whether number is even or odd

Question

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.

if ((x % 2) == 0) {
   // even
} else {
   // odd
}

If the remainder when you divide by 2 is 0, it's even. % is the operator to get a remainder.

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.

Every even number is divisible by two, regardless of if it's a decimal (but the decimal, if present, must also be even). So you can use the % (modulo) operator, which divides the number on the left by the number on the right and returns the remainder...

boolean isEven(double num) { return ((num % 2) == 0); }

I would recommend

Java Puzzlers: Traps, Pitfalls, and Corner Cases Book by Joshua Bloch and Neal Gafter

There is a briefly explanation how to check if number is odd. First try is something similar what @AseemYadav tried:

public static boolean isOdd(int i) {
     return i % 2 == 1;
}

but as was mentioned in book:

when the remainder operation returns a nonzero result, it has the same sign as its left operand

so generally when we have negative odd number then instead of 1 we'll get -1 as result of i%2. So we can use @Camilo solution or just do:

public static boolean isOdd(int i) {
     return i % 2 != 0;
}

but generally the fastest solution is using AND operator like @lucasmo write above:

public static boolean isOdd(int i) {
     return (i & 1) != 0;
}

@Edit It also worth to point Math.floorMod(int x, int y); which deals good with negative the dividend but also can return -1 if the divisor is negative

Least significant bit (rightmost) can be used to check if the number is even or odd. For all Odd numbers, rightmost bit is always 1 in binary representation.

public static boolean checkOdd(long number){
   return ((number & 0x1) == 1);
}