memory allocation for structures elements

Question

Hi I am having difficulties in understanding about how the memory is allocated to the structure elements.

For example if i have the below structure and the size of c

Answer 1

Every structure will also have alignment requirements

for example :

typedef struct structc_tag { char c;`` double d; int s; } structc_t;

Applying same analysis, structc_t needs sizeof(char) + 7 byte padding + sizeof(double) + sizeof(int) = 1 + 7 + 8 + 4 = 20 bytes. However, the sizeof(structc_t) will be 24 bytes. It is because, along with structure members, structure type variables will also have natural alignment. Let us understand it by an example. Say, we declared an array of structc_t as shown below structc_t structc_array[3];

Assume, the base address of structc_array is 0×0000 for easy calculations. If the structc_t occupies 20 (0×14) bytes as we calculated, the second structc_t array element (indexed at 1) will be at 0×0000 + 0×0014 = 0×0014. It is the start address of index 1 element of array. The double member of this structc_t will be allocated on 0×0014 + 0×1 + 0×7 = 0x001C (decimal 28) which is not multiple of 8 and conflicting with the alignment requirements of double. As we mentioned on the top, the alignment requirement of double is 8 bytes. In order to avoid such misalignment, compiler will introduce alignment requirement to every structure. It will be as that of the largest member of the structure. In our case alignment of structa_t is 2, structb_t is 4 and structc_t is 8. If we need nested structures, the size of largest inner structure will be the alignment of immediate larger structure.

In structc_t of the above program, there will be padding of 4 bytes after int member to make the structure size multiple of its alignment. Thus the sizeof (structc_t) is 24 bytes. It guarantees correct alignment even in arrays. You can cross check