Modify a bit at a given position

Last Updated : 18 Sep, 2023

Given a number n, a position p and a binary value b, we need to change the bit at position p in n to value b.

Examples :

Input : n = 7, p = 2, b = 0
Output : 3
7 is 00000111 after clearing bit at 
2nd position, it becomes 0000011.

Input : n = 7, p = 3, b = 1
Output : 15
7 is 00000111 after setting bit at 
3rd position it becomes 00001111.

We first create a mask that has set bit only 
at given position using bit wise shift.
      mask = 1 << position

Then to change value of bit to b, we first
make it 0 using below operation
      value & ~mask

After changing it 0, we change it to b by
doing or of above expression with following
(b << p) & mask, i.e., we return
      ((n & ~mask) | (b << p))

Below is the implementation of above steps :

C++

// CPP program to modify a bit at position
// p in n to b.
#include <bits/stdc++.h>
using namespace std;

// Returns modified n.
int modifyBit(int n, int p, int b)
{
    int mask = 1 << p;
    return ((n & ~mask) | (b << p));
}

// Driver code
int main()
{
    cout << modifyBit(6, 2, 0) << endl;
    cout << modifyBit(6, 5, 1) << endl;
    return 0;
}

// C program to modify a bit at position
// p in n to b.
#include <stdio.h>

// Returns modified n.
int modifyBit(int n, int p, int b)
{
    int mask = 1 << p;
    return ((n & ~mask) | (b << p));
}

// Driver code
int main()
{
    printf("%d\n",modifyBit(6, 2, 0));
    printf("%d\n",modifyBit(6, 5, 1));
    return 0;
}

// This code is contributed by kothvvsaakash.

Java

// Java program to modify a bit
// at position p in n to b.
import java.io.*;

class GFG 
{
    
// Returns modified n.
public static int modifyBit(int n, 
                            int p, 
                            int b)
{
    int mask = 1 << p;
    return (n & ~mask) | 
           ((b << p) & mask);
}
    // Driver Code
    public static void main (String[] args)
    {
        System.out.println(modifyBit(6, 2, 0));
        System.out.println (modifyBit(6, 5, 1));
    }
}

// This code is contributed by m_kit

Python3

# Python3 program to modify a bit at position
# p in n to b.

# Returns modified n.
def modifyBit( n,  p,  b):
    mask = 1 << p
    return (n & ~mask) | ((b << p) & mask)
 
# Driver code
def main():
    print(modifyBit(6, 2, 0))
    print(modifyBit(6, 5, 1))
    
if __name__ == '__main__':
    main()
# This code is contributed by PrinciRaj1992

// C# program to modify a bit
// at position p in n to b.
using System;

class GFG
{
// Returns modified n.
public static int modifyBit(int n, 
                            int p, 
                            int b)
{
    int mask = 1 << p;
    return (n & ~mask) | 
           ((b << p) & mask);
}

    // Driver Code
    static public void Main ()
    {
    
        Console.WriteLine(modifyBit(6, 2, 0));
        Console.WriteLine(modifyBit(6, 5, 1));
    }
}

// This code is contributed by ajit

PHP

<?php
// PHP program to modify a bit
// at position p in n to b.

// Returns modified n.
function modifyBit($n, $p, $b)
{
    $mask = 1 << $p;
    return ($n & ~$mask) | 
           (($b << $p) & $mask);
}

    // Driver code
    echo modifyBit(6, 2, 0),"\n";
    echo modifyBit(6, 5, 1) ,"\n";
    
// This code is contributed by ajit
?>

JavaScript

<script>

// Javascript program to modify a bit
// at position p in n to b.

// Returns modified n.
function modifyBit(n, p, b)
{
    let mask = 1 << p;
    return (n & ~mask) |
           ((b << p) & mask);
}

// Driver code
document.write(modifyBit(6, 2, 0) + "<br/>");
document.write(modifyBit(6, 5, 1));

// This code is contributed by susmitakundugoaldanga

</script>

Output :

 2
 38

Time Complexity : O(1)

Auxiliary Space : O(1)

Binary representation of a given number

Pawan Asipu

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Bit Magic

Modify a bit at a given position

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