How to turn on a particular bit in a number?

05 May, 2021
Given a number n and a value k, turn on the k’th bit in n.Examples:  
Input:  n = 4, k = 2
Output: 6

Input:  n = 3, k = 3
Output: 7

Input:  n = 64, k = 4
Output: 72

Input:  n = 64, k = 5
Output: 80
 
The idea is to use bitwise << and | operators. Using expression “(1 << (k – 1))“, we get a number which has all bits unset, except the k’th bit. If we do bitwise | of this expression with n, we get a number which has all bits same as n except the k’th bit which is 1.Below is the implementation of above idea. 
C++
Java
Python 3
C#
PHP
Javascript
// CPP program to turn on a particular bit#include <iostream>using namespace std; // Returns a number that has all bits same as n// except the k'th bit which is made 1int turnOnK(int n, int k){    // k must be greater than 0    if (k <= 0)        return n;     // Do | of n with a number with all    // unset bits except the k'th bit    return (n | (1 << (k - 1)));} // Driver program to test above functionint main(){    int n = 4;    int k = 2;    cout << turnOnK(n, k);    return 0;}
// Java program to turn on a particular// bitclass GFG {         // Returns a number that has all    // bits same as n except the k'th    // bit which is made 1    static int turnOnK(int n, int k)    {                 // k must be greater than 0        if (k <= 0)            return n;             // Do | of n with a number with        // all unset bits except the        // k'th bit        return (n | (1 << (k - 1)));    }         // Driver program to test above    // function    public static void main(String [] args)    {        int n = 4;        int k = 2;        System.out.print(turnOnK(n, k));    }} // This code is contributed by Smitha
# Python 3 program to turn on a# particular bit # Returns a number that has all# bits same as n except the k'th# bit which is made 1def turnOnK(n, k):     # k must be greater than 0    if (k <= 0):        return n     # Do | of n with a number    # with all unset bits except    # the k'th bit    return (n | (1 << (k - 1))) # Driver program to test above# functionn = 4k = 2print(turnOnK(n, k)) # This code is contributed by# Smitha
// C# program to turn on a particular// bitusing System; class GFG {         // Returns a number that has all    // bits same as n except the k'th    // bit which is made 1    static int turnOnK(int n, int k)    {                 // k must be greater than 0        if (k <= 0)            return n;             // Do | of n with a number        // with all unset bits except        // the k'th bit        return (n | (1 << (k - 1)));    }         // Driver program to test above    // function    public static void Main()    {        int n = 4;        int k = 2;        Console.Write(turnOnK(n, k));    }} // This code is contributed by Smitha
<?php// PHP program to turn on a particular bit // Returns a number that has// all bits same as n except// the k'th bit which is made 1function turnOnK($n, $k){         // k must be greater than 0    if ($k <= 0)        return $n;     // Do | of n with a number with all    // unset bits except the k'th bit    return ($n | (1 << ($k - 1)));}     // Driver Code    $n = 4;    $k = 2;    echo turnOnK($n, $k); // This code is contributed by m_kit?>
<script>    // Javascript program to turn on a particular bit         // Returns a number that has all    // bits same as n except the k'th    // bit which is made 1    function turnOnK(n, k)    {                   // k must be greater than 0        if (k <= 0)            return n;               // Do | of n with a number        // with all unset bits except        // the k'th bit        return (n | (1 << (k - 1)));    }         let n = 4;    let k = 2;    document.write(turnOnK(n, k)); // This code is contributed by suresh07.</script>
6
 
jit_t
Smitha Dinesh Semwal
suresh07
Bit Magic
Bit Magic
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