Add 1 to a given number - GeeksforGeeks

Write a program to add one to a given number. The use of operators like ‘+’, ‘-‘, ‘*’, ‘/’, ‘++’, ‘–‘ …etc are not allowed.
Examples:

Input:  12
Output: 13

Input:  6
Output: 7

This question can be approached by using some bit magic. Following are different methods to achieve the same using bitwise operators.

Method 1
To add 1 to a number x (say 0011000111), flip all the bits after the rightmost 0 bit (we get 0011000000). Finally, flip the rightmost 0 bit also (we get 0011001000) to get the answer.

C++

// C++ code to add add  
// one to a given number  
#include <bits/stdc++.h> 
using namespace std; 
  
int addOne(int x)  
{  
    int m = 1;  
      
    // Flip all the set bits  
    // until we find a 0  
    while( x & m )  
    {  
        x = x ^ m;  
        m <<= 1;  
    }  
      
    // flip the rightmost 0 bit  
    x = x ^ m;  
    return x;  
}  
  
/* Driver program to test above functions*/
int main()  
{  
    cout<<addOne(13);  
    return 0;  
}  
  
// This code is contributed by rathbhupendra 

C

// C++ code to add add 
// one to a given number  
#include <stdio.h> 
  
int addOne(int x) 
{ 
    int m = 1; 
      
    // Flip all the set bits  
    // until we find a 0  
    while( x & m ) 
    { 
        x = x ^ m; 
        m <<= 1; 
    } 
      
    // flip the rightmost 0 bit  
    x = x ^ m; 
    return x; 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
    printf("%d", addOne(13)); 
    getchar(); 
    return 0; 
} 

Java

// Java code to add add 
// one to a given number  
class GFG { 
  
    static int addOne(int x) 
    { 
        int m = 1; 
          
        // Flip all the set bits  
        // until we find a 0  
        while( (int)(x & m) == 1) 
        { 
            x = x ^ m; 
            m <<= 1; 
        } 
      
        // flip the rightmost 0 bit  
        x = x ^ m; 
        return x; 
    } 
      
    /* Driver program to test above functions*/
    public static void main(String[] args) 
    { 
        System.out.println(addOne(13)); 
    } 
} 
  
// This code is contributed by prerna saini. 

Python3

# Python3 code to add 1 
# one to a given number  
def addOne(x) : 
      
    m = 1; 
    # Flip all the set bits 
    # until we find a 0  
    while(x & m): 
        x = x ^ m 
        m <<= 1
      
    # flip the rightmost  
    # 0 bit  
    x = x ^ m 
    return x 
  
# Driver program 
n = 13
print addOne(n) 
  
# This code is contributed by Prerna Saini. 

C#

// C# code to add one 
// to a given number  
using System; 
  
class GFG { 
  
    static int addOne(int x) 
    { 
        int m = 1; 
          
        // Flip all the set bits  
        // until we find a 0  
        while( (int)(x & m) == 1) 
        { 
            x = x ^ m; 
            m <<= 1; 
        } 
      
        // flip the rightmost 0 bit  
        x = x ^ m; 
        return x; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        Console.WriteLine(addOne(13)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP code to add add 
// one to a given number  
  
  
function addOne($x) 
{ 
    $m = 1; 
      
    // Flip all the set bits  
    // until we find a 0  
    while( $x & $m ) 
    { 
        $x = $x ^ $m; 
        $m <<= 1; 
    } 
      
    // flip the rightmost 0 bit  
    $x = $x ^ $m; 
    return $x; 
} 
  
// Driver Code 
echo addOne(13); 
  
// This code is contributed by vt_m. 
?> 

Output:

Method 2
We know that the negative number is represented in 2’s complement form on most of the architectures. We have the following lemma hold for 2’s complement representation of signed numbers.

Say, x is numerical value of a number, then

~x = -(x+1) [ ~ is for bitwise complement ]

(x + 1) is due to addition of 1 in 2’s complement conversion

To get (x + 1) apply negation once again. So, the final expression becomes (-(~x)).

C

#include<stdio.h> 
  
int addOne(int x) 
{ 
   return (-(~x)); 
} 
  
/* Driver program to test above functions*/
int main() 
{ 
  printf("%d", addOne(13)); 
  getchar(); 
  return 0; 
} 

Java

// Java code to Add 1 to a given number 
class GFG 
{ 
    static int addOne(int x) 
    { 
         return (-(~x)); 
    } 
      
    // Driver program 
    public static void main(String[] args) 
    { 
        System.out.printf("%d", addOne(13)); 
    } 
} 
  
// This code is contributed  
// by Smitha Dinesh Semwal 

Python3

# Python3 code to add 1 to a given number 
  
def addOne(x): 
    return (-(~x)); 
  
  
# Driver program  
print(addOne(13)) 
  
# This code is contributed by Smitha Dinesh Semwal 

C#

// C# code to Add 1  
// to a given number 
using System; 
  
class GFG 
{ 
    static int addOne(int x) 
    { 
        return (-(~x)); 
    } 
      
    // Driver program 
    public static void Main() 
    { 
        Console.WriteLine(addOne(13)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP Code to Add 1  
// to a given number 
  
function addOne($x) 
{ 
return (-(~$x)); 
} 
  
// Driver Code 
echo addOne(13); 
  
// This code is contributed by vt_m. 
?> 

Output:

Example :

Assume the machine word length is one *nibble* for simplicity.
And x = 2 (0010),
~x = ~2 = 1101 (13 numerical)
-~x = -1101

Interpreting bits 1101 in 2’s complement form yields numerical value as -(2^4 – 13) = -3. Applying ‘-‘ on the result leaves 3. Same analogy holds for decrement. Note that this method works only if the numbers are stored in 2’s complement form.