Reverse bytes for 64-bit value

byte, c, endianness, reverse, swap

Solution

You cannot use `char x` for a pointer!!!! A `char` is only a single byte long.

You need at the very least

unsigned long int swapPtr(unsigned long int x) {

Or better, use the type of the pointer

void* swapPtr(void* x) {

Quite likely your compiler will complain when you start bit shifting pointers; in that case you're better off explicitly casting your argument to an unsigned 64 bit integer:

#include <stdint.h>
uint64_t x;

Note also that you have to call with the address of a variable, so you call with

result = swapLong(&loc);

not `*loc` (which looks at the place where `loc` is pointing - the value, not the address).

Complete program:

#include <stdio.h>
#include <stdint.h>

uint64_t swapLong(void *X) {
  uint64_t x = (uint64_t) X;
x = (x & 0x00000000FFFFFFFF) << 32 | (x & 0xFFFFFFFF00000000) >> 32;
x = (x & 0x0000FFFF0000FFFF) << 16 | (x & 0xFFFF0000FFFF0000) >> 16;
x = (x & 0x00FF00FF00FF00FF) << 8  | (x & 0xFF00FF00FF00FF00) >> 8;
return x;
}

int main(void) {
  char a;
  printf("the address of a is 0x%016llx\n", (uint64_t)(&a));
  printf("swapping all the bytes gives 0x%016llx\n",(uint64_t)swapLong(&a));
}

Output:

the address of a is 0x00007fff6b133b1b
swapping all the bytes gives 0x1b3b136bff7f0000

EDIT you could use something like

#include <inttypes.h>

printf("the address of a is 0x%016" PRIx64 "\n", (uint64_t)(&a));

where the macro `PRIx64` expands into "the format string you need to print a 64 bit number in hex". It is a little cleaner than the above.

Problem

Original source