Reverse bytes of a Hexadecimal Number
Given an unsigned integer N. The task is to reverse all bytes of N without using a temporary variable and print the reversed number.
Examples:
Input: N = 0xaabbccdd
Output: 0xddccbbaaInput: N = 0xa912cbd4
Output: 0xd4cb12a9
The naive approach is to extract the appropriate byte is to use mask(&) with shift operators.
#define REV(x) ( ((x&0xff000000)>>24) | (((x&0x00ff0000)<<8)>>16) | (((x&0x0000ff00)>>8)<<16) |
((x&0x000000ff) << 24) )
Efficient approach:
The idea is to use shift operators only.
- Move the position of the last byte to the first byte using left shift operator(<<).
- Move the position of the first byte to the last byte using right shift operator(>>).
- Move the middle bytes using the combination of left shift and right shift operator.
- Apply logical OR (|) to the output of all the above expression, to get the desired output.
Below is the implementation of the above approach :
// C program to reverse bytes of a hexadecimal number#include <stdio.h>// macro which reverse the hexadecimal integer#define REV(n) ((n << 24) | (((n>>16)<<24)>>16) | \ (((n<<16)>>24)<<16) | (n>>24)) // Driver codeint main(){ unsigned int n = 0xa912cbd4; // n = 0xaabbccdd // (n >> 24) - 0x000000aa // (n << 24) - 0xdd000000 // (((n >> 16) << 24) >> 16) - 0xbb00 // (((n >> 8) << 24) >> 8) - 0xcc0000 // If output of all the above expression is // OR'ed then it results in 0xddccbbaa printf("%x is reversed to %x", n, REV(n)); return 0;} |
a912cbd4 is reversed to d4cb12a9
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