Weird endianness related bug

So, apparently when casting an array of bytes, C automatically reverses the endianness:

Code:

---

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

int main(){
    uint8_t a[4] = {0x52,0x4a,0x75,0xd8};
    uint32_t x = *(uint32_t*)a;
    uint32_t y = 0x524a75d8;

    printf("x=%x, y=%x, a=%02x%02x%02x%02x\n",x,y,a[0],a[1],a[2],a[3]);
}

---

It outputs x=d8754a52, y=524a75d8, a=524a75d8, which means the bytes (but not the bits!) of x are reversed with respect to y.

1. Why does this happen automatically?
2. How to make it not happen?
3. Why is endiannes split into bits and bytes, and the order is big bit, small byte?

For those wondering, I am using GCC 11.1.0 without any extra options.

Nothing is happening.
Your machine (our common Intel/AMD machines) are little endian.
So the value 0xd8754a52 is stored in memory with the "little end" (least significant byte) "first" (lowest) in memory, i.e., 0x52, 0x4a, 0x75, 0xd8.

Look at y's bytes:

Code:

---

#include <stdio.h>
#include <stdint.h>
 
int main(){
    uint32_t y = 0x524a75d8;
    uint8_t *p = (uint8_t*)&y;
    printf("%02x %02x %02x %02x\n", p[0], p[1], p[2], p[3]);  // prints d8754a52
    return 0;
}

---

I don't understand what you mean about "endianness split into bits and bytes...".

Incidentally, you can detect endianness very easily at runtime.

Code:

---

#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>

bool little_endian(void) {
  static bool state = false;
  static bool known = false;
  if (!known) {
    unsigned long sample = 0xff;
    uint8_t* bytes = (uint8_t*)&sample;
    if (bytes[0] == 0xff)
      state = true;
    known = true;
  }
  return state;
}

#define big_endian() !little_endian()

void reverse_bytes(void* data, size_t size) {
  uint8_t* left = (uint8_t*)data;
  uint8_t* right = left + size;
  while (left < --right) {
    uint8_t copy = *left;
    *left++ = *right;
    *right = copy;
  }
}

void to_little_endian(void* data, size_t size) {
  if (!little_endian())
    reverse_bytes(data, size);
}

void to_big_endian(void* data, size_t size) {
  if (!big_endian())
    reverse_bytes(data, size);
}

#include <stdio.h>

int main(void) {
  uint8_t a1[4] = {0x52, 0x4a, 0x75, 0xd8};
  uint32_t x1 = *(uint32_t*)a1;
  uint32_t y1 = 0x524a75d8;

  printf("CPU is %s-endian\n", big_endian() ? "big" : "little");
  printf("x=%x, y=%x, a=%02x%02x%02x%02x\n", x1, y1, a1[0], a1[1], a1[2],
        a1[3]);

  puts("Switching to big-endian format");
  to_big_endian(a1, 4);
  printf("x=%x, y=%x, a=%02x%02x%02x%02x\n", x1, y1, a1[0], a1[1], a1[2],
        a1[3]);

  uint8_t a2[4] = {0x52, 0x4a, 0x75, 0xd8};
  uint32_t x2 = *(uint32_t*)a2;
  uint32_t y2 = 0x524a75d8;

  puts("Switching to little-endian format");
  to_little_endian(a2, 4);
  printf("x=%x, y=%x, a=%02x%02x%02x%02x\n", x2, y2, a2[0], a2[1], a2[2],
        a2[3]);
}

---

Output on my machine:

Quote:

---

CPU is little-endian
x=d8754a52, y=524a75d8, a=524a75d8
Switching to big-endian format
x=d8754a52, y=524a75d8, a=d8754a52
Switching to little-endian format
x=d8754a52, y=524a75d8, a=524a75d8

---

Big and little are not your only choices - Question 20.9
Nor is it necessarily something cast in stone - Endianness - Wikipedia

The only guaranteed way to do this, without assuming anything about the current machine, is

Code:

---

int main(){
    uint8_t a[4] = {0x52,0x4a,0x75,0xd8};
    uint32_t x = (uint32_t)a[0] << 24 |
                (uint32_t)a[1] << 16 |
                (uint32_t)a[2] <<  8 |
                (uint32_t)a[3];
    uint32_t y = 0x524a75d8;
 
    printf("x=%x, y=%x, a=%02x%02x%02x%02x\n",x,y,a[0],a[1],a[2],a[3]);
}

---