# Swap two nibbles in a byte

• Difficulty Level : Easy
• Last Updated : 10 Jun, 2022

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.

## 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.

Another Approach:

Using binary instead of hexadecimal values. It is much clearer to beginners.

Step 1: Take & of 00001111 with number to get right nibble i.e. 0b00001111 & N

Step 2: Take & of 11110000 with number to get left nibble i.e. 0b11110000 & N

Step 3: Left shift the right nibble obtained in step 1 by 4 positions to get it as left nibble in the final answer i.e. <<4

Step 4: Right shift the left nibble obtained in step 2 by 4 positions to get it as right nibble in final answer >>4

Step 5: Do or( | ) operation between values obtained in step 3 & 4 to get the answer

## C++

 `// C++ program to swap two``// nibbles in a byte``#include ``using` `namespace` `std;` ` ``int` `swapNibbles(``int` `N) {``      ``// Step 1``        ``int` `right = (N & 0b00001111);``      ``// Step 3``        ``right= (right<<4);``      ``// Step 2``        ``int` `left = (N & 0b11110000);``      ``// Step 4``        ``left = (left>>4);``      ``// Step 5``        ``return` `(right | left);``    ``}` `// Driver code``int` `main()``{``    ``int` `n = 100;``    ``cout << ``"Original: "` `<< n << ``" Swapped: "` `<< swapNibbles(n);``        ` `    ``return` `0;``}` `// This code is contributed by sanjoy_62.`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;``class` `Solution{``    ``static` `int` `swapNibbles(``int` `N) {``      ``// Step 1``        ``int` `right = (N & 0b00001111);``      ``// Step 3``        ``right= (right<<``4``);``      ``// Step 2``        ``int` `left = (N & 0b11110000);``      ``// Step 4``        ``left = (left>>``4``);``      ``// Step 5``        ``return` `(right | left);``    ``}``}``class` `GFG {``    ``public` `static` `void` `main (String[] args) {``      ``Solution s = ``new` `Solution();``      ``int` `n = ``100``;``        ``System.out.println(``"Original: "``+ n + ``" Swapped: "` `+ s.swapNibbles(n));``    ``}``}`

## Python3

 `# Python code for the above approach``from` `math ``import` `ceil, sqrt` `def` `swapNibbles(N) :``    ` `    ``# Step 1``    ``right ``=` `(N & ``0b00001111``)``    ` `    ``# Step 3``    ``right``=` `(right<<``4``)``    ` `    ``# Step 2``    ``left ``=` `(N & ``0b11110000``)``      ` `    ``# Step 4``    ``left ``=` `(left>>``4``)``    ` `    ``# Step 5``    ``return` `(right | left)` `# Driver Code``n ``=` `100``;``print``(``"Original: "``, n, end ``=` `" "``)``print``(``" Swapped: "` `, swapNibbles(n))` `# This code is contributed by code_hunt.`

## C#

 `// C# program to swap two``// nibbles in a byte``using` `System;` `public` `class` `GFG{` `  ``static` `int` `swapNibbles(``int` `N) {``    ``// Step 1``    ``int` `right = (N & 0b00001111);``    ``// Step 3``    ``right= (right<<4);``    ``// Step 2``    ``int` `left = (N & 0b11110000);``    ``// Step 4``    ``left = (left>>4);``    ``// Step 5``    ``return` `(right | left);``  ``}` `  ``// Driver Code``  ``static` `public` `void` `Main (){``    ``int` `n = 100;``    ``Console.Write(``"Original: "``+ n + ``" Swapped: "` `+ swapNibbles(n));``  ``}``}` `// This code is contributed by shruti456rawal`

## Javascript

 `// JavaScript code for the above approach``function` `swapNibbles(N)``{``    ` `    ``// Step 1``    ``var` `right = (N & 0b00001111);``    ` `    ``// Step 3``    ``var` `right= (right<<4);``    ` `    ``// Step 2``    ``var` `left = (N & 0b11110000);``      ` `    ``// Step 4``    ``var` `left = (left>>4);``    ` `    ``// Step 5``    ``return` `(right | left);` `}` `// Driver Code``var` `n = 100;``console.log(``"Original:"``, n, ``" Swapped:"``, swapNibbles(n));` `// This code is contributed by phasing17`

Output

`Original: 100 Swapped: 70`