Known that sizeof(std::string) is 8, sizeof(int) is 4.
class Test1
{
int i; // 4
std::string str; // 8
};
sizeof(Test1) is 16, because 8 = N1 * 4 and 16 = N2 * 8. (N1 and N2 are natural numbers)
However, if I replace std::string with char [8] as below:
class Test2
{
int i; // 4
char ch[8]; // 8
};
I don't know why sizeof(Test2) gives me 12.
4 + 8 = 12, so 12 bytes is the minimum amount of memory to fit a 4 and a 8 byte object.
However, there is another consideration besides the total number of bytes of the subobjects. Each subobject has a particular alignment requirement.
The alignment of std::string depends on its complete subobjects, which are implementation defined. It appears, that std::string on your implementation has to be aligned to an 8 byte boundary. As a consequence, the alignment requirement of Test1 is also 8, and there must be 4 bytes of padding after the member i. So, the size of Test1 is the sum of the sizes of the subobjects + the pad bytes, which totals 4 + 8 + 4 = 16.
char and consequently char[8] have alignment requirement of 1. Therefore the alignment requirement of Test2 is 4 (because of int member) and no padding is required after the member i. Since there is no padding, the size of Test2 matches the sum of the sizes of the subobjects.
Another point of view is that the size must be a multiple of the alignment requirement. 16 is the smallest multiple of 8 (alignment of Test1) that is greater than or equal to 12. 12 is the smallest multiple of 4 (alignment of Test2) that is greater than or equal to 12.
The alignment of char is 1, and this holds for arrays of characters too. So no padding is added between the class members. Thus you get the size as 4 + 8 = 12.
