Why is the size of this struct 24?

Question

I have a structure of which I want to calculate its size:

#pragma pack(push,4)
struct  MyStruct
{  
    uint32_t i1;    /* size=4, offset=0. */
    uint32_t i2;    /* size =4 offset =4 */
    uint16_t s1;    /* size =2 offset=8 */
    unsigned char c[8]; /* size=8 offset=12*/
    uint16_t s2;    /* size=2 offset=20. */
    uint16_t s3;    /* size=2 offset=24. */

} ; // total size is 26 

static_assert(sizeof(MyStruct) == 24, "size of MyStruct incorrect");
#pragma pack(pop)

The static assert shows that the size is 24, but my calculation shows that it should be 26.

Why is the size 24?

I am working on windows 7, 32 bit application using visual studio 2012

Answer 1

The alignment of uint16_t is only 2, hence the offsets are:

#pragma pack(push,4)
struct  MyStruct
{  
    uint32_t i1;        /* offset=0  size=4 */
    uint32_t i2;        /* offset=4  size=4 */
    uint16_t s1;        /* offset=8  size=2 */
    unsigned char c[8]; /* offset=10 size=8 */
    uint16_t s2;        /* offset=18 size=2 */
    uint16_t s3;        /* offset=20 size=2 */
                        /* offset=22 padding=2 (needed to align MyStruct) */
} ; // total size is 24

Edit The padding at the end is necessary to ensure that all elements of

MyStruct A[10]; // or
MyStruct*B = new MyStruct[10];

are aligned appropriately. This requires that sizeof(MyStruct) is a multiple of alignof(MyStruct). Here, sizeof(MyStruct)=6*alignof(MyStruct).

Any struct/class type is always padded to the next multiple of its alignment.

Answer 2

In addition to Walter's answer, consider catching this fish yourself. All you need is the printf function and simple arithmetic:

  struct MyStruct ms;

  printf("sizeof(ms): %zd\n", sizeof(ms));

  printf("i1\t%td\n", (uint8_t*)&ms.i1 - (uint8_t*)&ms);
  printf("i2\t%td\n", (uint8_t*)&ms.i2 - (uint8_t*)&ms);
  printf("s1\t%td\n", (uint8_t*)&ms.s1 - (uint8_t*)&ms);
  printf("c \t%td\n", (uint8_t*)&ms.c  - (uint8_t*)&ms);
  printf("s2\t%td\n", (uint8_t*)&ms.s2 - (uint8_t*)&ms);
  printf("s3\t%td\n", (uint8_t*)&ms.s3 - (uint8_t*)&ms);

(%zd is for printing size_t, %td for printing ptrdiff_t. A plain %d will probably work just fine on most systems.)

Output:

sizeof(ms): 24
i1      0
i2      4
s1      8
c       10
s2      18
s3      20