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Home/ Questions/Q 6964855
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T15:58:39+00:00 2026-05-27T15:58:39+00:00

Here is the code in question #include <stdio.h> struct test { unsigned char t;

  • 0

Here is the code in question

#include <stdio.h>

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


int main ()
{
    struct test  * a = (void *) 0x1000;

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

    return 0;
}

The sizeof(struct test) prints 6, so I would expect to see:

6 0xffa 0x1006

Instead I get

6 0x1024 0xfdc

Last time I checked, 0x24, or 36, was not equal to 6. It’s not even aligned to anything that I can tell. I am at a complete loss.

Can someone please explain to me why I’m getting these values?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T15:58:40+00:00Added an answer on May 27, 2026 at 3:58 pm

    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);
    
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