Printing using hexadecimal format specifier

Question

Why does the given code give the following output.

#include <stdio.h>
int main(void) 
{
    int x = 0x12345678;
    unsigned short *p = &x;
    unsigned char *q = &x;
    printf("%x %x\n",*p++,*p++);
    printf("%x %x %x %x\n",q[0],q[1],q[2],q[3]);
    return 0;
}

Output:

1234 5678
78 56 34 12

and not:

1234 5678
12 34 56 78

The thing which I feel could be the answer is the endianness of the architecture must be causing it. But I can not comprehend how, because the whole 4-byte must be stored in a contiguous manner.

Also don't *q++ and *(q+1) point to the same address?

Answer 1

*q++ and *(q + 1) do not point to the same memory location during evaluation of their values in expressions. *q++ will point to the next memory location after q after the expression has been evaluated.

The reason you got 78 56 34 12 instead of 12 34 56 78 has to do with endianness; in memory your int is likely stored as 78563412. Try replacing q[0] thru q[3] in your printf statement with x & 0xff, (x >> 8) & 0xff, (x >> 16) & 0xff, (x >> 24) & 0xff and see if you get what you expected.