Check if a Number is Odd or Even using Bitwise Operators

Given a number N, the task is to check whether the number is even or odd using Bitwise Operators.
Examples: 
 

Input: N = 11 
Output: Odd
Input: N = 10 
Output: Even 
 

 

Following Bitwise Operators can be used to check if a number is odd or even:
 

1. Using Bitwise XOR operator: 
The idea is to check whether the last bit of the number is set or not. If the last bit is set then the number is odd, otherwise even. 
As we know bitwise XOR Operation of the Number by 1 increment the value of the number by 1 if the number is even otherwise it decrements the value of the number by 1 if the value is odd.



Below is the implementation of the above approach:
 

C++

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// C++ program to check for even or odd
// using Bitwise XOR operator
 
#include <iostream>
using namespace std;
 
// Returns true if n is even, else odd
bool isEven(int n)
{
 
    // n^1 is n+1, then even, else odd
    if (n ^ 1 == n + 1)
        return true;
    else
        return false;
}
 
// Driver code
int main()
{
    int n = 100;
    isEven(n)
? cout << "Even"
: cout << "Odd";
 
    return 0;
}

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Java

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// Java program to check for even or odd
// using Bitwise XOR operator
class GFG
{
 
    // Returns true if n is even, else odd
    static boolean isEven(int n)
    {
 
        // n^1 is n+1, then even, else odd
        if ((n ^ 1) == n + 1)
            return true;
        else
            return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 100;
        System.out.print(isEven(n) == true ? "Even" : "Odd");
    }
}
 
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program to check for even or odd
# using Bitwise XOR operator
 
# Returns true if n is even, else odd
def isEven( n) :
 
    # n^1 is n+1, then even, else odd
    if (n ^ 1 == n + 1) :
        return True;
    else :
        return False;
 
# Driver code
if __name__ == "__main__" :
    n = 100;
    print("Even") if isEven(n) else print( "Odd");
 
# This code is contributed by AnkitRai01

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C#

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// C# program to check for even or odd
// using Bitwise XOR operator
using System;
 
class GFG
{
 
    // Returns true if n is even, else odd
    static bool isEven(int n)
    {
 
        // n^1 is n+1, then even, else odd
        if ((n ^ 1) == n + 1)
            return true;
        else
            return false;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 100;
        Console.Write(isEven(n) == true ? "Even" : "Odd");
    }
}
 
// This code is contributed by Rajput-Ji

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Output

Even

2. Using Bitwise AND operator: 
The idea is to check whether the last bit of the number is set or not. If last bit is set then the number is odd, otherwise even.
As we know bitwise AND Operation of the Number by 1 will be 1, If it is odd because the last bit will be already set. Otherwise it will give 0 as output. 

Below is the implementation of the above approach: 
 

C++

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// C++ program to check for even or odd
// using Bitwise AND operator
#include <iostream>
using namespace std;
   
// Returns true if n is even, else odd
bool isEven(int n)
{
    // n&1 is 1, then odd, else even
    return (!(n & 1));
}
   
// Driver code
int main()
{
    int n = 101;
    isEven(n)
    ? cout << "Even"
    : cout << "Odd";
    return 0;
}

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Java

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// Java program to check for even or odd
// using Bitwise AND operator
class GFG
{
     
// Returns true if n is even, else odd
static boolean isEven(int n)
{
    // n&1 is 1, then odd, else even
    return ((n & 1)!=1);
}
     
// Driver code
public static void main(String[] args)
{
    int n = 101;
    System.out.print(isEven(n) == true ? "Even" : "Odd");
}
}
 
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 program to check for even or odd
# using Bitwise AND operator
 
# Returns true if n is even, else odd
def isEven(n) :
 
    # n&1 is 1, then odd, else even
    return (not(n & 1));
 
# Driver code
if __name__ == "__main__" :
 
    n = 101;
    if isEven(n) :
        print("Even")
    else:
        print("Odd");
 
# This code is contributed by AnkitRai01

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C#

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// C# program to check for even or odd
// using Bitwise AND operator
using System;
 
class GFG
{
     
    // Returns true if n is even, else odd
    static bool isEven(int n)
    {
        // n&1 is 1, then odd, else even
        return ((n & 1) != 1);
    }
         
    // Driver code
    public static void Main()
    {
        int n = 101;
        Console.Write(isEven(n) == true ? "Even" : "Odd");
    }
}
 
// This code is contributed by AnkitRai01

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Output

Odd

3. Using Bitwise OR operator: The idea is to check whether the last bit of the number is set or not. If the last bit is set then the number is odd, otherwise even. As we know bitwise OR Operation of the Number by 1 increment the value of the number by 1 if the number is even otherwise it will remain unchanged. So, if after OR operation of number with 1 gives a result which is greater than the number then it is even and we will return true otherwise it is odd and we will return false.

Below is the implementation of the above approach: 

C++

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// C++ program to check for even or odd
// using Bitwise OR operator
 
#include <iostream>
using namespace std;
 
// Returns true if n is even, else odd
bool isEven(int n)
{
 
    // n|1 is greater than n, then even, else odd
    if ((n | 1) > n)
        return true;
    else
        return false;
}
 
// Driver code
int main()
{
    int n = 100;
    isEven(n)
    ? cout << "Even"
    : cout << "Odd";
 
    return 0;
}

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Java

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// Java program to check for even or odd
// using Bitwise OR operator
class GFG
{
    // Returns true if n is even, else odd
    static boolean isEven(int n)
    {
 
        // n|1 is greater than n, then even, else odd
        if ((n | 1) > n)
            return true;
        else
            return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 100;
        System.out.print(isEven(n) == true ? "Even" : "Odd");
    }
}

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Python3

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# Python3 program to check for even or odd
# using Bitwise OR operator
 
# Returns true if n is even, else odd
def isEven( n) :
 
    # n|1 is greater then n, then even, else odd
    if (n | 1 > n) :
        return True;
    else :
        return False;
 
# Driver code
if __name__ == "__main__" :
    n = 100;
    print("Even") if isEven(n) else print( "Odd");

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C#

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// C# program to check for even or odd
// using Bitwise XOR operator
using System;
 
class GFG
{
 
    // Returns true if n is even, else odd
    static bool isEven(int n)
    {
 
        // n|1 is greater then n, then even, else odd
        if ((n | 1) > n)
            return true;
        else
            return false;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 100;
        Console.Write(isEven(n) == true ? "Even" : "Odd");
    }
}
 
// This code is contributed by Rajput-Ji

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Output

Even

Time Complexity: O(1)

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