Question

if I have the following code:

int i = 5;
void * ptr = &i;
printf("%p", ptr);

Will I get the LSB address of i, or the MSB?
Will it act differently between platforms?
Is there a difference here between C and C++?

Answer 1

Consider the size of int is 4 bytes. Always &i will gives you the first address of those 4 bytes.

If the architecture is little endian, then the lower address will have the LSB like below.

        +------+------+------+------+
Address | 1000 | 1001 | 1002 | 1003 |
        +------+------+------+------+
Value   |   5  |    0 |    0 |    0 |
        +------+------+------+------+

If the architecture is big endian, then the lower address will have the MSB like below.

        +------+------+------+------+
Address | 1000 | 1001 | 1002 | 1003 |
        +------+------+------+------+
Value   |   0  |    0 |    0 |    5 |
        +------+------+------+------+

So &i will give LSB address of i if little endian or it will give MSB address of i if big endian

In mixed endian mode also, either little or big endian will be chosen for each task dynamically.

Below logic will tells you the endianess

int i = 5; 
void * ptr = &i; 
char * ch = (char *) ptr;

printf("%p", ptr); 
if (5 == (*ch))
    printf("\nlittle endian\n");
else
    printf("\nbig endian\n");

This behaviour will be same for both c and c++