How to reverse the 4 bytes of an unsigned integer?

Question

I am trying to reverse the unsigned integer by using the '«' and '»', and bitwise 'AND' and 'OR' (& and |), but can't figure out how to do this.

What I already have;

int main(int argc, char** argv) {
    unsigned int getal;
    scanf("%i", &getal);
    printf("%X\n", getal);
    return 0;
}

User input: 0xaabbccdd, output now: AABBCCDD, what it should output DDCCBBAA

I also tried;

int main(int argc, char** argv) {
    unsigned int getal;
    unsigned int x;
    scanf("%i", &getal);
    int i;
    for (i = 0; i < 32; ++i) {
        getal <<= 1;
        getal |= (x & 1);
        x >>= 1;
        printf("%X\n", getal);
    }
    return 0;
}

but the result was completely different.

First, initialize your x to 0 before the loop...and don´t confuse getal and x in the loop. Other than that, you´re on the right way — deviantfan, Feb 25 '14 at 11:18
@Gone what do you mean by show the code which reverse the integer? That's the part I think I am missing and asking for help. — MOTIVECODEX, Feb 25 '14 at 11:18
There are 4 bytes in an int that need to be reversed. So don't reverse the bits. — Hans Passant, Feb 25 '14 at 11:19
Is this what you looking for ?: http://stackoverflow.com/questions/2182002/convert-big-endian-to-little-endian-in-c-without-using-provided-func — qwark, Feb 25 '14 at 11:19
Should 87654321 become 12345678, or should it become 84C2A6E1? (or even 21436587) — user253751, Feb 25 '14 at 11:20
If you're not going to give any other examples, then: `unsigned int reverse(unsigned int x) {return 0xDDCCBBAA;}`. What does 87654321 become? — user253751, Feb 25 '14 at 11:22
each bit or 4 bit or byte? and `scanf("%i", &getal);` --> `scanf("%i", &x);` — BLUEPIXY, Feb 25 '14 at 11:33

frogatto · Accepted Answer · 2019-10-02T16:32:03.287

3

Try this:

int getal;     // original number
int reversed;  // 4 bytes reversed of getal

/* get number form console here */

uint8_t *n1, *n2;
n1 = (uint8_t *) &getal;
n2 = (uint8_t *) &reversed;

n2[0] = n1[3];
n2[1] = n1[2];
n2[2] = n1[1]; 
n2[3] = n1[0];

/* print reversed here */

edited Oct 02 '19 at 16:32

answered Feb 25 '14 at 11:24

frogatto

28,539
11
83
129

Thank you, your code worked, but I do not really think it's a correct way to reverse bytes?.. Anyway thanks – MOTIVECODEX Feb 25 '14 at 11:33
@F4LLCON This solution only works with 4byte integers, also you can extend it to support other integers, As I know this solution is correct and there is no problem with it – frogatto Feb 25 '14 at 11:37
1

this may violate aliasing rules, and may not work correctly. Using simple bit shifts is the best – phuclv Mar 19 '14 at 07:19

unwind · Answer 2 · 2014-02-25T11:25:47.560

2

Your code seems like it's trying to reverse the bits, but your indicated desired outcome is a reversal of the 8-bit groups that make up each pair of hexadecimal digits. These are not the same.

You need something like:

unsigned int reverse_nibbles(unsigned int x)
{
  unsigned int out = 0, i;
  for(i = 0; i < 4; ++i)
  {
    const unsigned int byte = (x >> 8 * i) & 0xff;
    out |= byte << (24 - 8 * i);
  }
  return out;
}

The above (untested) code assumes unsigned int is 32 bits; generalizing it is trivial but I left it out for simplicity's sake.

It simply extracts one byte (8-bit chunk) at a time from one direction, and uses bitwise or to merge it into the result from the other direction.

edited Feb 25 '14 at 11:25

answered Feb 25 '14 at 11:18

unwind

391,730
64
469
606

It says that nibble is unassigned? – MOTIVECODEX Feb 25 '14 at 11:23
Thank you, but I could not get it working with the code I had, but the other answer from ABFORCE worked, tho it does not seem to be a correct way to do.. – MOTIVECODEX Feb 25 '14 at 11:33

How to reverse the 4 bytes of an unsigned integer?

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