Question

The Code used

#include<stdio.h>

struct st
{
 char a;
 short c;
 int b;
};

struct st s1;
int main()
{
        printf("%p %p \n",(&s1.b)-1, &s1);
}

If I print the address of &s1.b it prints 0x804a01c and &s1.b-2 prints 0x804a018
why it is printing same address 0x804a01c if i select &s1.b-1 ?

Answer 1

Thank you for posting your code. Now I see the issue. It is because of padding. To wit:

printf("sizeof(char): %d\n", sizeof(char));
printf("sizeof(short): %d\n", sizeof(short));
printf("sizeof(int): %d\n", sizeof(int));
printf("sizeof(struct st): %d\n", sizeof(struct st));

On my machine this prints

1
2
4
8

You might think, shouldn’t sizeof(struct st) be 1 + 2 + 4 = 7? That’s certainly a reasonable thought but because of alignment issues there is padding between a and c. Therefore, in memory, the struct looks like the following (relative to the first byte of the struct):

0x00000000: char                 a
0x00000001: padding
0x00000002: first byte of short  c
0x00000003: second byte of short c
0x00000004: first byte of int    b
0x00000005: second byte of int   b
0x00000006: third byte of int    b
0x00000007: fourth byte of int   b

Consequently (relative to &s1):

&s1.b - 1 is ((long)&s1.b) - sizeof(int) = 4 - 4 = 0 = &s1

This is why both &s1 and &s1.b - 1 will print the same address. In particular if

&s1 = 0x804a01c

then

&s1.b = 0x804a01c + 0x00000004 = 0x804a020

and

&s1.b - 1 = 0x804a020 - 0x00000004 = 0x804a01c

and

&s1.b - 2 = 0x804a020 - 0x00000008 = 0x804a018

Note, finally, that this is implementation-specific behavior. This is not portable!