Output of this C code

Question

#include<stdio.h>
int main()
{
    int i, j;
    int *pi,*pj;
    pi=&i;
    pj=&j;

    printf("pi-pj=%d\n",pi-pj);
    return 0;
}

I tried this code on different compilers, but each time I am getting the same result, can anybody please help me understand why it is the same?

Ouput:

pi -pj = 3

I am confused, as the memory normally would be contiguously allocated. So, if let's say, our system stack is growing downwards and we have &i = 0xA, then the address of j(&j) = 0x6 (since integers are 4 bytes). Now as we are printing the difference between these two int pointer values, output should be "1". But it is coming as "3". Why is that?

Answer 1

I was not able to replicate your experience. With gcc on Linux x86:

[wally@lenovotower ~]$ cat t.c
#include<stdio.h>
int main()
{
    int i, j;
    int *pi,*pj;
    pi=&i;
    pj=&j;

    printf("pi-pj=%d\n",pi-pj);
    return 0;
}

[wally@lenovotower ~]$ gcc -o t t.c
[wally@lenovotower ~]$ ./t
pi-pj=1
[wally@lenovotower ~]$ 

This means that i and j are adjacent. Pointer subtraction returns the number of items between the pointers, not the address difference. To get your result, there would have to be two items-worth of padding in between. I can't explain how that could be.