# 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: 0xddccbbaa

Input: N = 0xa912cbd4
Output: 0xd4cb12a9

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `// macro which reverse the hexadecimal integer ` `#define REV(n) ((n << 24) | (((n>>16)<<24)>>16) | \ ` `                    ``(((n<<16)>>24)<<16) | (n>>24)) ` ` `  `// Driver code ` `int` `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; ` `} `

Output:

```a912cbd4 is reversed to d4cb12a9
```

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