Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Find XOR of two number without using XOR operator

  • Difficulty Level : Easy
  • Last Updated : 16 Nov, 2021

Given two integers, find XOR of them without using the XOR operator, i.e., without using ^ in C/C++.

Examples :  

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input:  x = 1, y = 2
Output: 3

Input:  x = 3, y = 5
Output: 6

A Simple Solution is to traverse all bits one by one. For every pair of bits, check if both are the same, set the corresponding bit like 0 in output, otherwise set as 1. 



C++




// C++ program to find XOR without using ^
#include <iostream>
using namespace std;
 
// Returns XOR of x and y
int myXOR(int x, int y)
{
    int res = 0; // Initialize result
     
    // Assuming 32-bit Integer
    for (int i = 31; i >= 0; i--)                    
    {
       // Find current bits in x and y
       bool b1 = x & (1 << i);
       bool b2 = y & (1 << i);
        
        // If both are 1 then 0 else xor is same as OR
        bool xoredBit = (b1 & b2) ? 0 : (b1 | b2);         
 
        // Update result
        res <<= 1;
        res |= xoredBit;
    }
    return res;
}
 
// Driver program to test above function
int main()
{
   int x = 3, y = 5;
   cout << "XOR is " << myXOR(x, y);
   return 0;
}

Java




// Java program to find XOR without using ^
import java.io.*;
 
class GFG{
   
// Returns XOR of x and y
static int myXOR(int x, int y)
{
     
    // Initialize result
    int res = 0;
 
    // Assuming 32-bit Integer
    for(int i = 31; i >= 0; i--)                    
    {
         
        // Find current bits in x and y
        int b1 = ((x & (1 << i)) == 0 ) ? 0 : 1
        int b2 = ((y & (1 << i)) == 0 ) ? 0 : 1
 
        // If both are 1 then 0 else xor is same as OR
        int xoredBit = ((b1 & b2) != 0) ? 0 : (b1 | b2);         
 
        // Update result
        res <<= 1;
        res |= xoredBit;
    }
    return res;
}
 
// Driver Code
public static void main (String[] args)
{
    int x = 3, y = 5;
     
    System.out.println("XOR is " + myXOR(x, y));
}
}
 
// This code is contributed by math_lover

Python3




# Python3 program to find XOR without using ^
 
# Returns XOR of x and y
def myXOR(x, y):
    res = 0 # Initialize result
 
    # Assuming 32-bit Integer
    for i in range(31, -1, -1):
         
        # Find current bits in x and y
        b1 = x & (1 << i)
        b2 = y & (1 << i)
        b1 = min(b1, 1)
        b2 = min(b2, 1)
 
        # If both are 1 then 0
        # else xor is same as OR
        xoredBit = 0
        if (b1 & b2):
            xoredBit = 0
        else:
            xoredBit = (b1 | b2)
 
        # Update result
        res <<= 1;
        res |= xoredBit
    return res
 
# Driver Code
x = 3
y = 5
print("XOR is", myXOR(x, y))
 
# This code is contributed by Mohit Kumar

C#




// C# program to find XOR
// without using ^
using System;
class GFG{
   
// Returns XOR of x and y
static int myXOR(int x,
                 int y)
{   
  // Initialize result
  int res = 0;
 
  // Assuming 32-bit int
  for(int i = 31; i >= 0; i--)                    
  {
    // Find current bits in x and y
    int b1 = ((x & (1 << i)) == 0 ) ?
               0 : 1; 
    int b2 = ((y & (1 << i)) == 0 ) ?
               0 : 1; 
 
    // If both are 1 then 0 else
    // xor is same as OR
    int xoredBit = ((b1 & b2) != 0) ?
                     0 : (b1 | b2);         
 
    // Update result
    res <<= 1;
    res |= xoredBit;
  }
  return res;
}
 
// Driver Code
public static void Main(String[] args)
{
  int x = 3, y = 5;
  Console.WriteLine("XOR is " +
                     myXOR(x, y));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program to find XOR without using ^
 
// Returns XOR of x and y
function myXOR(x, y)
{
    let res = 0; // Initialize result
     
    // Assuming 32-bit Integer
    for (let i = 31; i >= 0; i--)                    
    {
    // Find current bits in x and y
      let b1 = ((x & (1 << i)) == 0 ) ? 0 : 1; 
      let b2 = ((y & (1 << i)) == 0 ) ? 0 : 1; 
         
        // If both are 1 then 0 else xor is same as OR
        let xoredBit = (b1 & b2) ? 0 : (b1 | b2);        
 
        // Update result
        res <<= 1;
        res |= xoredBit;
    }
    return res;
}
 
// Driver program to test above function
 
let x = 3, y = 5;
document.write("XOR is " + myXOR(x, y));
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output
XOR is 6

Thanks to Utkarsh Trivedi for suggesting this solution.
 
A Better Solution can find XOR without using a loop. 
1) Find bitwise OR of x and y (Result has set bits where either x has set or y has set bit). OR of x = 3 (011) and y = 5 (101) is 7 (111)
2) To remove extra set bits find places where both x and y have set bits. The value of the expression “~x | ~y” has 0 bits wherever x and y both have set bits.
3) bitwise AND of “(x | y)” and “~x | ~y” produces the required result.

Below is the implementation. 

C++




// C++ program to find XOR without using ^
#include <iostream>
using namespace std;
 
// Returns XOR of x and y
int myXOR(int x, int y)
{
   return (x | y) & (~x | ~y);
}
 
// Driver program to test above function
int main()
{
   int x = 3, y = 5;
   cout << "XOR is " << myXOR(x, y);
   return 0;
}

Java




// Java program to find
// XOR without using ^
import java.io.*;
 
class GFG
{
 
// Returns XOR of x and y
static int myXOR(int x, int y)
{
    return (x | y) &
           (~x | ~y);
}
 
// Driver Code
public static void main (String[] args)
{
    int x = 3, y = 5;
    System.out.println("XOR is "+
                      (myXOR(x, y)));
}
}
 
// This code is contributed by ajit

Python3




# Python 3 program to
# find XOR without using ^
 
# Returns XOR of x and y
def myXOR(x, y):
    return ((x | y) &
            (~x | ~y))
 
# Driver Code
x = 3
y = 5
print("XOR is" ,
       myXOR(x, y))
 
# This code is contributed
# by Smitha

C#




// C# program to find
// XOR without using ^
using System;
 
class GFG
{
     
// Returns XOR of x and y
static int myXOR(int x, int y)
{
    return (x | y) &
           (~x | ~y);
}
 
// Driver Code
static public void Main ()
{
    int x = 3, y = 5;
    Console.WriteLine("XOR is "+
                     (myXOR(x, y)));
}
}
 
// This code is contributed by m_kit

PHP




<?php
// PHP program to find
// XOR without using ^
 
// Returns XOR of x and y
function myXOR($x, $y)
{
    return ($x | $y) & (~$x | ~$y);
}
 
// Driver Code
$x = 3;
$y = 5;
 
echo "XOR is " , myXOR($x, $y);
 
// This code is contributed by aj_36
?>

Javascript




<script>
// Javascript program to find XOR without using ^
 
// Returns XOR of x and y
function myXOR(x, y)
{
   return (x | y) & (~x | ~y);
}
 
// Driver program to test above function
   let x = 3, y = 5;
   document.write("XOR is " + myXOR(x, y));
 
// This code is contributed by subham348.
</script>
Output
XOR is 6

Thanks to jitu_the_best for suggesting this solution. 

Alternate Solution : 

C++




// C++ program to find XOR without using ^
#include <iostream>
using namespace std;
 
// Returns XOR of x and y
int myXOR(int x, int y)
{
   return (x & (~y)) | ((~x )& y);
}
 
// Driver program to test above function
int main()
{
   int x = 3, y = 5;
   cout << "XOR is " << myXOR(x, y);
   return 0;
}

Java




// Java program to find XOR without using ^
import java.io.*;
  
class GFG
{
 
// Returns XOR of x and y
static int myXOR(int x, int y)
{
return (x & (~y)) | ((~x )& y);
}
 
// Driver Code
public static void main (String[] args)
{
  
int x = 3, y = 5;
System.out.println("XOR is "+
                      (myXOR(x, y)));
}
}
 
// This code is contributed by shivanisinghss2110

Python3




# Python3 program to
# Returns XOR of x and y
def myXOR(x, y):
    return (x & (~y)) | ((~x )& y)
 
# Driver Code
x = 3
y = 5
print("XOR is" ,
    myXOR(x, y))
 
# This code is contributed by shivanisinghss2110

C#




// C# program to find XOR without using ^
using System;
 
class GFG{
 
// Returns XOR of x and y
static int myXOR(int x, int y)
{
    return (x & (~y)) | ((~x )& y);
}
 
// Driver program to test above function
public static void Main()
{
    int x = 3, y = 5;
    Console.WriteLine("XOR is " +myXOR(x, y));
}
}
 
// This code is contributed by shivansinghss2110

Javascript




<script>
 
// Javascript program to find XOR without using ^
 
// Returns XOR of x and y
function myXOR(x, y)
{
   return (x & (~y)) | ((~x ) & y);
}
 
// Driver code
let x = 3, y = 5;
 
document.write("XOR is " + myXOR(x, y));
 
// This code is contributed by subhammahato348
 
</script>
Output
XOR is 6

Another Solution: we can simply use one of the properties of the XOR bitwise operator i.e. a+b = a^b + 2*(a&b), with the help of this we can do the same for an operator variant also.

C++14




// C++ program to find XOR without using ^
#include <iostream>
using namespace std;
 
int XOR(int x, int y) { return (x + y - (2 * (x & y))); }
 
int main()
{
    int x = 3, y = 5;
    cout << XOR(x, y) << endl;
    return 0;
}
// this code is contributed by vishu05

Java




// Java program to find XOR without using ^
 
class GFG {
 
    static int XOR(int x, int y) {
        return (x + y - (2 * (x & y)));
    }
 
    public static void main(String[] args) {
        int x = 3, y = 5;
        System.out.print(XOR(x, y) + "\n");
    }
}
 
// This code is contributed by umadevi9616

Python3




# Python3 program to return XOR of x and y without ^ operator
def XOR(x, y):
    return (x+y - (2*(x & y)))
 
 
# Driver Code
x = 3
y = 5
print("XOR of",x,'&',y,'is:',
      XOR(x, y))
 
# This code is contributed by vishu05

C#




// C# program to find XOR without using ^
using System;
 
class GFG{
 
static int XOR(int x, int y)
{
    return(x + y - (2 * (x & y)));
}
 
// Driver code
public static void Main(String[] args)
{
    int x = 3, y = 5;
   
    Console.Write(XOR(x, y) + "\n");
}
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
// javascript program to find XOR without using ^
   function XOR(x , y) {
        return (x + y - (2 * (x & y)));
    }
 
     
        var x = 3, y = 5;
        document.write(XOR(x, y) + "\n");
 
// This code contributed by umadevi9616
</script>
Output
6

Time Complexity: O(1) i.e. simple calculation of arithmetic and bitwise operator.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!