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 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

#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
?>


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.

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



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Improved By : nitin mittal, jit_t


 
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