# C Program-Swap two nibbles in a byte

C Program Swap two nibbles in a byte - Bit Algorithm - A nibble is a four-bit aggregation, or half an octet. There are 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.
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To swap the nibbles, we can use bitwise &, bitwise ‘<<‘ and ‘>>’ operators. A byte can be represented using a unsigned char in C as size of char is 1 byte in a typical C compiler. Following is C program to swap the two nibbles in a byte.

C programming
``````#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;
}``````

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.