About the data alignment in c

Question

And I define a struct :

#include <stdint.h>
#include <stdio.h>
#define O(type, field) (size_t)(&(((type *)0)->field))
struct byname {
int16_t int16;
int32_t int32;
int64_t int64;};

Then I use sizeof(struct byname) and it return 16 which I can understand.

However when I define the like adding a int8_t:

#include <stdint.h>
#include <stdio.h>
#define O(type, field) (size_t)(&(((type *)0)->field))
struct byname {
int16_t int16;
int32_t int32;
int64_t int64;
int8_t int8;};

It just return 24, I think a int8 only takes 1 by and there are 3 bys padding according to data alignment, so I think the answer should be 20.

Anyone can kindly explain to me how the 24 comes?

this will help you undersant https://en.wikipedia.org/wiki/Data_structure_alignment — imbr, Aug 13 '21 at 10:35
The "duplicate question" is not the same as OP's question... — Daniel Kleinstein, Aug 13 '21 at 10:43

score 3 · Answer 1 · answered Aug 13 '21 at 10:31

The structure contains int64_t. If the compiler thinks that int64_t should be aligned to 8-byte boundary, it is reasonable to make the size of the structure multiple of 8 (therefore 24 bytes instead of 20) to align every int64_t int64; in an array of struct byname to 8-byte boundary.

About the data alignment in c

1 Answers1