Find XOR of two number without using XOR operator

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

Examples :  

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 same, set the corresponding bit as 0 in output, otherwise set as 1. 

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// 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;
}
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// 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
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# 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
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Output : 

XOR is 6

Thanks to Utkarsh Trivedi for suggesting this solution.
 
A Better Solution can find XOR without using 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 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. 

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// 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;
}
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// 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 contibuted by ajit
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# 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
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// 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 contibuted by m_kit
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<?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 contibuted by aj_36
?>
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Output : 

XOR is 6

Thanks to jitu_the_best for suggesting this solution. 

Alternate Solution : 

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// 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;
}
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// 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
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# 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
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// 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
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