Swap two nibbles in a byte

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 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.

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

Below is the implementation of the above idea. 

C++

// C++ program to swap two
// nibbles in a byte
#include <bits/stdc++.h>
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 <stdio.h>

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

<?php
// PHP program to swap two
// nibbles in a byte

// function to Swap two nibbles
// in a byte in php program
function swapNibbles($x)
{
    return ( ($x & 0x0F) << 4 |
           ($x & 0xF0) >> 4 );
}

    // Driver Code
    $x = 100;
    echo swapNibbles($x);

// This Code is Contributed by Ajit
?>

Javascript

<script>
// javascript program to swap two
// nibbles in a byte
function swapNibbles(x)
{
    return ((x & 0x0F) << 4 | (x & 0xF0) >> 4);
}

// Driver code
var x = 100;
document.write(swapNibbles(x));

// This code is contributed by Princi Singh
</script>

Output

70

Time Complexity: O(1)
Auxiliary Space: O(1)

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Another Approach:

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

• Take & of 00001111 with number to get right nibble i.e. 0b00001111 & N
• ake & of 11110000 with number to get left nibble i.e. 0b11110000 & N
• 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
• Right shift the left nibble obtained in step 2 by 4 positions to get it as right nibble in final answer >>4
• Do or( | ) operation between values obtained in step 3 & 4 to get the answer

Below is the implementation of the above approach:

C++

// C++ program to swap two
// nibbles in a byte
#include <bits/stdc++.h>
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

Time Complexity: O(1)
Auxiliary Space: O(1)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.