How do I check if an integer is even or odd using bitwise operators

Consider what being "even" and "odd" means in "bit" terms. Since binary integer data is stored with bits indicating multiples of 2, the lowest-order bit will correspond to 2⁰, which is of course 1, while all of the other bits will correspond to multiples of 2 (2¹ = 2, 2² = 4, etc.). Gratuituous ASCII art:

NNNNNNNN |||||||| |||||||+−− bit 0, value = 1 (20) ||||||+−−− bit 1, value = 2 (21) |||||+−−−− bit 2, value = 4 (22) ||||+−−−−− bit 3, value = 8 (23) |||+−−−−−− bit 4, value = 16 (24) ||+−−−−−−− bit 5, value = 32 (25) |+−−−−−−−− bit 6, value = 64 (26) +−−−−−−−−− bit 7 (highest order bit), value = 128 (27) for unsigned numbers, value = -128 (-27) for signed numbers (2's complement) 

I've only shown 8 bits there, but you get the idea.

So you can tell whether an integer is even or odd by looking only at the lowest-order bit: If it's set, the number is odd. If not, it's even. You don't care about the other bits because they all denote multiples of 2, and so they can't make the value odd.

The way you look at that bit is by using the AND operator of your language. In C and many other languages syntactically derived from B (yes, B), that operator is &. In BASICs, it's usually And. You take your integer, AND it with 1 (which is a number with only the lowest-order bit set), and if the result is not equal to 0, the bit was set.

I'm intentionally not actually giving the code here, not only because I don't know what language you're using, but because you marked the question "homework." :-)

if (number & 1) number is odd else // (number & 1) == 0 number is even 

For example, let's take integer 25, which is odd. In binary 25 is 00011001. Notice that the least significant bit b0 is 1.

00011001 00000001 (00000001 is 1 in binary) & -------- 00000001 

Just a footnote to Jim's answer.

In C#, unlike C, bitwise AND returns the resulting number, so you'd want to write:

if ((number & 1) == 1) { // It's odd } 

if(x & 1) // '&' is a bit-wise AND operator printf("%d is ODD\n", x); else printf("%d is EVEN\n", x); 

Examples:

 For 9: 9 -> 1 0 0 1 1 -> & 0 0 0 1 ------------------- result-> 0 0 0 1 

So 9 AND 1 gives us 1, as the right most bit of every odd number is 1.

 For 14: 14 -> 1 1 1 0 1 -> & 0 0 0 1 ------------------ result-> 0 0 0 0 

So 14 AND 1 gives us 0, as the right most bit of every even number is 0.

Also in Java you will have to use if((number&1)==1){//then odd}, because in Java and C# like languages the int is not casted to boolean. You'll have to use the relational operators to return a boolean value i.e true and false unlike C and C++ like languages which treats non-zero value as true.

You can do it simply using bitwise AND & operator.

if(num & 1) { //I am odd number. } 

Read more over here - Checking even odd using bitwise operator in C

Check Number is Even or Odd using XOR Operator Number = 11 1011 - 11 in Binary Format ^ 0001 - 1 in Binary Format ---- 1010 - 10 in Binary Format Number = 14 1110 - 14 in Binary Format ^ 0001 - 1 in Binary Format ---- 1111 - 15 in Binary Format AS It can observe XOR Of a number with 1, increments it by 1 if it is even, decrements it by 1 if it is odd. 

Code:

if((n^1) == (n+1)) cout<<"even\n"; else cout<<"odd\n"; 

#include <iostream> #include <algorithm> #include <vector> void BitConvert(int num, std::vector<int> &array){ while (num > 0){ array.push_back(num % 2); num = num / 2; } } void CheckEven(int num){ std::vector<int> array; BitConvert(num, array); if (array[0] == 0) std::cout << "Number is even"; else std::cout << "Number is odd"; } int main(){ int num; std::cout << "Enter a number:"; std::cin >> num; CheckEven(num); std::cout << std::endl; return 0; } 

In Java,

if((num & 1)==0){ //its an even num } //otherwise its an odd num 

This is an old question, however the other answers have left this out.

In addition to using num & 1, you can also use num | 1 > num.

This works because if a number is odd, the resulting value will be the same since the original value num will have started with the ones bit set, however if the original value num was even, the ones bit won't have been set, so changing it to a 1 will make the new value greater by one.

Approach 1: Short and no need for explicit comparison with 1

if (number & 1) { // number is odd } else { // number is even } 

Approach 2: Needs an extra bracket and explicit comparison with 0

if((num & 1) == 0){ // Note: Bracket is MUST around num & 1 // number is even } else { // number is odd } 

What would happen if I miss the bracket in the above code

if(num & 1 == 0) { } // wrong way of checking even or not!! 

becomes

if(num & (1 == 0)) { } // == is higher precedence than & 

https://en.cppreference.com/w/cpp/language/operator_precedence