Dado um número, a tarefa é alternar todos os bits pares de um número.
Exemplos: 
 

Input : 10
Output : 0
binary representation 1 0 1 0
after toggle          0 0 0 0 


Input : 20
Output : 30
binary representation 1 0 1 0 0
after toggle          1 1 1 1 0

1. Primeiro gere um número que contenha bits de posição pares. 
2. Pegue XOR com o número original. Observe que 1 ^ 1 = 0 e 1 ^ 0 = 1.
Vamos entender essa abordagem com o código a seguir. 
 

// CPP code to Toggle all even
// bit of a number
#include <iostream>
using namespace std;
 
// Returns a number which has all even
// bits of n toggled.
int evenbittogglenumber(int n)
{
    // Generate number form of 101010
    // ..till of same order as n
    int res = 0, count = 0;
    for (int temp = n; temp > 0; temp >>= 1) {
 
        // if bit is even then generate
        // number and or with res
        if (count % 2 == 1)
            res |= (1 << count);     
 
        count++;
    }
 
    // return toggled number
    return n ^ res;
}
 
// Driver code
int main()
{
    int n = 11;
    cout << evenbittogglenumber(n);
    return 0;
}

// Java code to Toggle all
// even bit of a number
import java.io.*;
 
class GFG {
     
    // Returns a number which has
    // all even bits of n toggled.
    static int evenbittogglenumber(int n)
    {
        // Generate number form of 101010
        // ..till of same order as n
        int res = 0, count = 0;
        for (int temp = n; temp > 0;
                               temp >>= 1)
        {
            // if bit is even then generate
            // number and or with res
            if (count % 2 == 1)
                res |= (1 << count);     
      
            count++;
        }
      
        // return toggled number
        return n ^ res;
    }
      
    // Driver code
    public static void main(String args[])
    {
        int n = 11;
        System.out.println(evenbittogglenumber(n));
    }
}
 
// This code is contributed by Nikita Tiwari.

# Python code to Toggle all
# even bit of a number
 
# Returns a number which has all even
# bits of n toggled.
def evenbittogglenumber(n) :
 
    # Generate number form of 101010
    # ..till of same order as n
    res = 0
    count = 0
    temp = n
 
    while (temp > 0) :
         
        # if bit is even then generate
        # number and or with res
        if (count % 2 == 1) :
            res = res | (1 << count)    
  
        count = count + 1
        temp >>= 1
         
  
    # return toggled number
    return n ^ res
  
# Driver code
n = 11
print(evenbittogglenumber(n))
 
#This code is contributed by Nikita Tiwari.

// C# code to Toggle all
// even bit of a number
using System;
  
class GFG {
      
    // Returns a number which has
    // all even bits of n toggled.
    static int evenbittogglenumber(int n)
    {
        // Generate number form of 101010
        // ..till of same order as n
        int res = 0, count = 0;
         
        for (int temp = n; temp > 0;
                               temp >>= 1)
        {
            // if bit is even then generate
            // number and or with res
            if (count % 2 == 1)
                res |= (1 << count);     
       
            count++;
        }
       
        // return toggled number
        return n ^ res;
    }
       
    // Driver code
    public static void Main()
    {
         
        int n = 11;
         
        Console.WriteLine(evenbittogglenumber(n));
    }
}
  
// This code is contributed by Anant Agarwal.

<?php
// php code to Toggle all
// even bit of a number
 
// Returns a number which has
// all even bits of n toggled.
function evenbittogglenumber($n)
{
     
    // Generate number form of 101010
    // ..till of same order as n
    $res = 0;
    $count = 0;
    for ($temp = $n; $temp > 0; $temp >>= 1)
    {
 
        // if bit is even then generate
        // number and or with res
        if ($count % 2 == 1)
            $res |= (1 << $count);    
 
        $count++;
    }
 
    // return toggled number
    return $n ^ $res;
}
 
    // Driver code
    $n = 11;
    echo evenbittogglenumber($n);
 
// This code is contributed by mits
?>

<script>
 
// JavaScript program to Toggle all
// even bit of a number
 
    // Returns a number which has
    // all even bits of n toggled.
    function evenbittogglenumber(n)
    {
        // Generate number form of 101010
        // ..till of same order as n
        let res = 0, count = 0;
           
        for (let temp = n; temp > 0;
                               temp >>= 1)
        {
            // if bit is even then generate
            // number and or with res
            if (count % 2 == 1)
                res |= (1 << count);     
         
            count++;
        }
         
        // return toggled number
        return n ^ res;
    }
 
 
// Driver code
 
    let n = 11;
           
        document.write(evenbittogglenumber(n));
 
</script>

Saída: 

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