C Pointer Arithmetic sizeof(struct)

Question

Here is the code in question

#include 

struct test {
    unsigned char t;
    unsigned short u;
    unsigned char v;
};


int main ()
{
    s         


        

Answer 1

You have a typed pointer.

So when you increment it my 1 (i.e. a + 1) it means a + sizeof(type).

So a + sizeof(type) = a + sizeof(type) * sizeof(type) = a + 6 * 6 (in your case as sizeof(test) = 6)

That's where you are getting 0x24 or 36 from.

Answer 2

The problem is that when you do pointer arithmetic, it increments by a multiple of the size of the datatype.

So what you're effectively doing is adding by the square of sizeof(struct test).

Since sizeof(struct test) = 6, you are incrementing the address by 6 * 6 = 36. Hence why you get 0x1024 and 0xfdc instead of 0x1006 and 0xffa. (You also switched the + and -, but that's a small thing.)

Instead, just do this:

printf("%x %p %p\n",
       sizeof(struct test),
       a + 1,
       a - 1);

Answer 3

When you do pointer arithmetic like this, you move forward or back by that number of elements, as though that variable were in an array. So you really want to just use a + 1 and a - 1, which should advance by 6 bytes each time.

IMPORTANT: Keep in mind that the compiler can add padding in your struct to help with alignment. Don't just assume that because you have two one-byte chars and a two-byte short that your struct will be 4 bytes in size---this isn't the case here. (In fact, don't assume that you know the size of char or short; I've seen 2-byte chars before).

Answer 4

I think you are looking for a + 1 and a - 1.

(a + x) is the same is &a[x].