Question

I always assume, as they said here http://en.wikipedia.org/wiki/Data_structure_alignment, “It is important to note that the last member is padded with the number of bytes required so that the total size of the structure should be a multiple of the largest alignment of any structure member”

So for the struct like this, its size should be 16 at a 32 processor

typedef struct
{
   double   s;  /* 8 bytes */
   char     c;  /* 7 bytes padding at the end of make the total size 8*2 */
} structa_t;  

So I was quite surprised to the size is 12 instead of 16!! Why is that ? Can someone cast some light on it ?

sizeof(double) = 8
sizeof(structa_t) = 12

BTW, so system info

$ uname -a
Linux 2.6.18-8.el5xen #1 SMP Thu Mar 15 21:02:53 EDT 2007 i686 i686 i386 GNU/Linux
$ gcc --version
gcc (GCC) 4.1.1 20070105 (Red Hat 4.1.1-52)

Answer 1

The key wording here is:

…the total size of the structure should be a multiple of the largest alignment of any structure member…

On your system, the aligment of a double is 4, not 8.

If you wait for C1x, you can use the _Alignof operator (similar to sizeof). On your system,

sizeof(double) == 8
_Alignof(double) == 4

You can test the alignment in a more primitive way in C89,

#include <stdlib.h>
struct char_double { char x; double y; };
#define DOUBLE_ALIGNMENT offsetof(struct char_double, y)

Or with a macro,

#define alignof(x) offsetof(struct { char a; x b; }, b)