**Difficulty Level:** Rookie

Given a number n and a value k, turn of the k’th bit in n.

Examples:

Input: n = 15, k = 1 Output: 14 Input: n = 15, k = 2 Output: 13 Input: n = 15, k = 3 Output: 11 Input: n = 15, k = 4 Output: 7 Input: n = 15, k >= 5 Output: 15

The idea is to use bitwise <<, & and ~ operators. Using expression "** ~(1 << (k - 1))**“, we get a number which has all bits set, 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 0.

Following is C++ implementation of this.

## C/C++

#include <iostream> using namespace std; // Returns a number that has all bits same as n // except the k'th bit which is made 0 int turnOffK(int n, int k) { // k must be greater than 0 if (k <= 0) return n; // Do & of n with a number with all set bits except // the k'th bit return (n & ~(1 << (k - 1))); } // Driver program to test above function int main() { int n = 15; int k = 4; cout << turnOffK(n, k); return 0; }

## Java

// Java program to turn off a particular bit in a number import java.io.*; class TurnOff { // Function to returns a number that has all bits same as n // except the k'th bit which is made 0 static int turnOffK(int n, int k) { // k must be greater than 0 if (k <= 0) return n; // Do & of n with a number with all set bits except // the k'th bit return (n & ~(1 << (k - 1))); } // Driver program public static void main (String[] args) { int n = 15; int k = 4; System.out.println(turnOffK(n, k)); } } // Contributed by Pramod Kumar

Output:

7

**Exercise:** Write a function turnOnK() that turns the k’th bit on.

This article is contributed by **Rahul Jain**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above