How offset of structure is calculated? Explain printf statement here?

Question

#include<stdio.h>
main()
{
        unsigned char c;
        typedef struct name {
                long  a;
                int b;
                long c;
        }r;
        r re = {3,4,5};
        r *na = &re;
        printf("%d", *(int*) ((char*)na + (unsigned int) & ( (struct name *)0 )->b));
}

OUTPUT :

4

I would be thankful if somebody explain in detail what printf statement is doing in this c program ? According to me int is casted . then char pointer is casted to na. But I cannot understand what whole statement overall doing ? What casting doing here ? b is not pointer but it seems it is used as pointer. I have found this problem one of the c programming book.It seems be it calculate structure offset but I still not able to know how. Please help me to understand program in detail.

score 3 · Accepted Answer · edited May 23 '17 at 12:15

The code &((struct name *)0)->b is indeed known as the "offsetof" operation and is in fact a well-known C macro.

The idea is that if a is a structure of type A then &a.b is equal to the address of a plus the offset of b in a. Since here NULL is used instead of the address of a, the result is just the offset of b in structures of type A.

According to Wikipedia,

This works by casting a null pointer into a pointer to structure st, and then obtaining the address of member m within said structure. While this implementation works correctly in many compilers, it has undefined behavior according to the C standard,2 since it involves a dereference of a null pointer (although, one might argue that no dereferencing takes place, because the whole expression is calculated at compile time)

So let's look at the expression in printf step-by-step.

The expression (unsigned int) & ( (struct name *)0 )->b)) calculates offset of b in struct name as described above and casts the result to an integer. The result should be equal to sizeof(long) on most platforms.

The (char*)na in the code casts na, which is a pointer to struct name to a char pointer. This is required because sizeof(char) can be assumed to be 1 while sizeof(*na) is larger than 1. What we want to do is to use the address of *na as a raw numerical address instead of doing pointer arithmetic, so that if for example na==0x1234, then the result of ((char*)na + 4) is equal to 0x1234 + 4 = 0x1238.

The sum results in a memory address. That address is equal to the address of member variable b in the object na, and is of type char *. Knowing that, the last step is to cast the address back to int * (because the type of b is int) and then to dereference the result (once again, we know that it points to b). The final outcome is the value of b, which is then printed.

My actual question is to understand that printf statement dear. Still I am upvoting your answer as you tried — Punit Vara, Jan 05 '16 at 19:23
@PunitVara: which parts are you still missing? I can expand on the answer. Once you understand the offsetof stuff, the rest is plain pointer arithmetic. — kfx, Jan 05 '16 at 19:26
I am really thankful to you if you expand step by step that statement. — Punit Vara, Jan 05 '16 at 19:29
Good breakdown of the pointer arithmetic. Basically, it's a convoluted way of saying `printf("%d", na->b);` — dbush, Jan 05 '16 at 19:46
Thanks @dbush Why ending with unsigned int what is purpose of that one? — Punit Vara, Jan 05 '16 at 19:51
The `unsigned int` cast treats the offset address as a number instead of a pointer. The updated answer describes it pretty well. — dbush, Jan 05 '16 at 19:54

How offset of structure is calculated? Explain printf statement here?

1 Answers1