struct representation in memory on 64 bit machine

Question

For my curiosity I have written a program which was to show each byte of my struct. Here is the code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
#include <limits.h>

#define MAX_INT 2147483647
#define MAX_LONG 9223372036854775807

typedef struct _serialize_test{
   char a;
   unsigned int b;
   char ab;
   unsigned long long int c;
}serialize_test_t;


int main(int argc, char**argv){
   serialize_test_t *t;
   t = malloc(sizeof(serialize_test_t));
   t->a = 'A';
   t->ab = 'N';
   t->b = MAX_INT;
   t->c = MAX_LONG;

   printf("%x %x %x %x %d %d\n", t->a, t->b, t->ab, t->c, sizeof(serialize_test_t), sizeof(unsigned long long int));

   char *ptr = (char *)t;

   int i;
   for (i=0; i < sizeof(serialize_test_t) - 1; i++){
      printf("%x = %x\n", ptr + i, *(ptr + i));
   }

   return 0;
}

and here is the output:

41 7fffffff 4e ffffffff 24 8
26b2010 = 41
26b2011 = 0
26b2012 = 0
26b2013 = 0
26b2014 = ffffffff
26b2015 = ffffffff
26b2016 = ffffffff
26b2017 = 7f
26b2018 = 4e
26b2019 = 0
26b201a = 0
26b201b = 0
26b201c = 0
26b201d = 0
26b201e = 0
26b201f = 0
26b2020 = ffffffff
26b2021 = ffffffff
26b2022 = ffffffff
26b2023 = ffffffff
26b2024 = ffffffff
26b2025 = ffffffff
26b2026 = ffffffff

And here is the question: if sizeof(long long int) is 8, then why sizeof(serialize_test_t) is 24 instead of 32 - I always thought that size of struct is rounded to largest type and multiplied by number of fields like here for example: 8(bytes)*4(fields) = 32(bytes) — by default, with no pragma pack directives?

Also when I cast that struct to char * I can see from the output that the offset between values in memory is not 8 bytes. Can you give me a clue? Or maybe this is just some compiler optimizations?

Answer 1

On modern 32-bit machines like the SPARC or the Intel [34]86, or any Motorola chip from the 68020 up, each data iten must usually be ``self-aligned'', beginning on an address that is a multiple of its type size. Thus, 32-bit types must begin on a 32-bit boundary, 16-bit types on a 16-bit boundary, 8-bit types may begin anywhere, struct/array/union types have the alignment of their most restrictive member.

The total size of the structure will depend on the packing.In your case it's going as 8 byte so final structure will look like

typedef struct _serialize_test{

   char a;//size 1 byte

   padding for 3 Byte;

   unsigned int b;//size 4 Byte

   char ab;//size 1 Byte again

   padding of 7 byte;

   unsigned long long int c;//size 8 byte

}serialize_test_t;

in this manner first two and last two are aligned properly and total size reaches upto 24.