typedef struct structc_tag
{
char c;
double d;
int s;
} structc_t;
I read in a blog that this will take 24 bytes of data:
sizeof(char) + 7 byte padding + sizeof(double) + sizeof(int) + 4 byte padding = 1 + 7 + 8 + 4 + 4 = 24 bytes.
My question is why the 7 byte padding, why can't we use 3bytes of padding there and utilise next 8 bytes for double? And what is the need for last 4 bytes?
You need to consider what the happens if you allocate an array of these structures with malloc():
structc_t *p = malloc(2 * sizeof *p);
Consider a platform where sizeof(double) == 8, sizeof(int) == 4 and the required alignment of double is 8. malloc() always returns an address correctly aligned for storing any C type - so in this case a will be 8 byte aligned. The padding requirements then naturally fall out:
In order for a[0].d to be 8-byte aligned, there must therefore be 7 bytes of padding after a[0].c;
In order for a[1].d to be 8-byte aligned, the overall struct size must be a multiple of 8, so there must therefore be 4 bytes of padding after a[0].s.
If you re-order the struct from largest to smallest:
typedef struct structc_tag
{
double d;
int s;
char c;
} structc_t;
...then the only padding required is 3 bytes after .c, to make the structure size a multiple of 8. This results in the total size of the struct being 16, rather than 24.
It's platform dependent, but it depends on what double is aligned to. If it's aligned to 8 bytes, which appears to be this case, 3 bytes of padding won't cut it.
If double was aligned to 4 bytes, you'd be right and 3 bytes of padding would be used.
