# How to swap two bits in a given integer?

Given an integer n and two bit positions p1 and p2 inside it, swap bits at the given positions. The given positions are from least significant bit (lsb). For example, the position for lsb is 0.

Examples:

```Input: n = 28, p1 = 0, p2 = 3
Output: 21
28 in binary is 11100.  If we swap 0'th and 3rd digits,
we get 10101 which is 21 in decimal.

Input: n = 20, p1 = 2, p2 = 3
Output: 24
```

We strongly recommend you to minimize your browser and try this yourself first.

The idea is to first find the bits, then use XOR based swapping concept, i..e., to swap two numbers ‘x’ and ‘y’, we do x = x ^ y,  y = y ^ x and x = x ^ y.

Below is the implementation of the above idea

## C++

 `// C program to swap bits in an intger ` `#include ` ` `  `// This function swaps bit at positions p1 and p2 in an integer n ` `int` `swapBits(unsigned ``int` `n, unsigned ``int` `p1, unsigned ``int` `p2) ` `{ ` `    ``/* Move p1'th to rightmost side */` `    ``unsigned ``int` `bit1 =  (n >> p1) & 1; ` ` `  `    ``/* Move p2'th to rightmost side */` `    ``unsigned ``int` `bit2 =  (n >> p2) & 1; ` ` `  `    ``/* XOR the two bits */` `    ``unsigned ``int` `x = (bit1 ^ bit2); ` ` `  `    ``/* Put the xor bit back to their original positions */` `    ``x = (x << p1) | (x << p2); ` ` `  `    ``/* XOR 'x' with the original number so that the ` `       ``two sets are swapped */` `    ``unsigned ``int` `result = n ^ x; ` `} ` ` `  `/* Drier program to test above function*/` `int` `main() ` `{ ` `    ``int` `res =  swapBits(28, 0, 3); ` `    ``printf``(``"Result = %d "``, res); ` `    ``return` `0; ` `} `

## Java

 `// Java program to swap bits in an intger ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// This function swaps bit at  ` `// positions p1 and p2 in an integer n ` `static` `int` `swapBits( ``int` `n, ``int` `p1, ``int` `p2) ` `{ ` `    ``/* Move p1'th to rightmost side */` `    ``int` `bit1 = (n >> p1) & ``1``; ` ` `  `    ``/* Move p2'th to rightmost side */` `    ``int` `bit2 = (n >> p2) & ``1``; ` ` `  `    ``/* XOR the two bits */` `    ``int` `x = (bit1 ^ bit2); ` ` `  `    ``/* Put the xor bit back to ` `    ``their original positions */` `    ``x = (x << p1) | (x << p2); ` ` `  `    ``/* XOR 'x' with the original ` `    ``number so that the ` `    ``two sets are swapped */` `    ``int` `result = n ^ x; ` `    ``return` `result; ` `} ` ` `  `    ``/* Driver code*/` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `res = swapBits(``28``, ``0``, ``3``); ` `        ``System.out.println (``"Result = "` `+ res); ` `    ``} ` `} ` ` `  `// This code is contributed by ajit.. `

/div>

## C#

 `// C# program to swap bits in an intger ` `using` `System; ` ` `  `class` `GFG ` `{ ` `         `  `// This function swaps bit at positions  ` `// p1 and p2 in an integer n ` `static` `int` `swapBits(``int` `n, ``int` `p1, ``int` `p2) ` `{ ` `    ``/* Move p1'th to rightmost side */` `    ``int` `bit1 = (n >> p1) & 1; ` ` `  `    ``/* Move p2'th to rightmost side */` `    ``int` `bit2 = (n >> p2) & 1; ` ` `  `    ``/* XOR the two bits */` `    ``int` `x = (bit1 ^ bit2); ` ` `  `    ``/* Put the xor bit back to ` `    ``their original positions */` `    ``x = (x << p1) | (x << p2); ` ` `  `    ``/* XOR 'x' with the original ` `    ``number so that the ` `    ``two sets are swapped */` `    ``int` `result = n ^ x; ` `    ``return` `result; ` `} ` ` `  `/* Driver code*/` `static` `public` `void` `Main () ` `{ ` `    ``int` `res = swapBits(28, 0, 3); ` `    ``Console.WriteLine(``"Result = "` `+ res); ` `} ` `} ` ` `  `// This code is contributed by akt_mit `

Output:

`Result = 21`

## tags:

Bit Magic Bit Magic