# Swap two nibbles in a byte

• Difficulty Level : Easy
• Last Updated : 07 Apr, 2021

A nibble is a four-bit aggregation, or half an octet. There are two nibbles in a byte.
Given a byte, swap the two nibbles in it. For example 100 is be represented as 01100100 in a byte (or 8 bits). The two nibbles are (0110) and (0100). If we swap the two nibbles, we get 01000110 which is 70 in decimal.

To swap the nibbles, we can use bitwise &, bitwise ” operators. A byte can be represented using a unsigned char in C as size of char is 1 byte in a typical C compiler.
Below is the implementation of above idea.

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## C++

 `// C++ program to swap two``// nibbles in a byte``#include ``using` `namespace` `std;` `int` `swapNibbles(``int` `x)``{``    ``return` `( (x & 0x0F) << 4 | (x & 0xF0) >> 4 );``}` `// Driver code``int` `main()``{``    ``int` `x = 100;``    ``cout << swapNibbles(x);``    ``return` `0;``}` `//This code is contributed by Shivi_Aggarwal`

## C

 `#include ` `unsigned ``char` `swapNibbles(unsigned ``char` `x)``{``    ``return` `( (x & 0x0F)<<4 | (x & 0xF0)>>4 );``}` `int` `main()``{``    ``unsigned ``char` `x = 100;``    ``printf``(``"%u"``, swapNibbles(x));``    ``return` `0;``}`

## Java

 `// Java program to swap two``// nibbles in a byte` `class` `GFG {``    ` `static` `int` `swapNibbles(``int` `x)``{``    ``return` `((x & ``0x0F``) << ``4` `| (x & ``0xF0``) >> ``4``);``}` `// Driver code``public` `static` `void` `main(String arg[])``{``    ``int` `x = ``100``;``    ``System.out.print(swapNibbles(x));``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# python program Swap``# two nibbles in a byte` `def` `swapNibbles(x):``    ``return` `( (x & ``0x0F``)<<``4` `| (x & ``0xF0``)>>``4` `)` `# Driver code` `x ``=` `100``print``(swapNibbles(x))` `# This code is contributed``# by Anant Agarwal.`

## C#

 `// C# program to swap two``// nibbles in a byte``using` `System;` `class` `GFG {` `// Function for swapping   ``static` `int` `swapNibbles(``int` `x)``{``    ``return` `((x & 0x0F) << 4 |``            ``(x & 0xF0) >> 4);``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `x = 100;``    ``Console.Write(swapNibbles(x));``}``}` `// This code is contributed by Nitin Mittal.`

## PHP

 `> 4 );``}` `    ``// Driver Code``    ``\$x` `= 100;``    ``echo` `swapNibbles(``\$x``);` `// This Code is Contributed by Ajit``?>`

## Javascript

 ``

Output:

`70`

Explanation:
100 is 01100100 in binary. The operation can be split mainly in two parts
1) The expression “x & 0x0F” gives us last 4 bits of x. For x = 100, the result is 00000100. Using bitwise ‘<<‘ operator, we shift the last four bits to the left 4 times and make the new last four bits as 0. The result after shift is 01000000.
2) The expression “x & 0xF0” gives us first four bits of x. For x = 100, the result is 01100000. Using bitwise ‘>>’ operator, we shift the digit to the right 4 times and make the first four bits as 0. The result after shift is 00000110.
At the end we use the bitwise OR ‘|’ operation of the two expressions explained above. The OR operator places first nibble to the end and last nibble to first. For x = 100, the value of (01000000) OR (00000110) gives the result 01000110 which is equal to 70 in decimal.