Size of struct is more than expected [duplicate]

Question

I already read this question: struct padding in c++ and this one Why isn't sizeof for a struct equal to the sum of sizeof of each member?

and I know this isn't standardized but still I believe it's a legit question.

Why is the size of this struct 16 on a x64 system?

struct foo { char* b; char a;};

The effective size would be 8 + 1 = 9, but I know there's padding involved. Anyway I thought a would only be padded to reach the size of an int, i.e. with other 3 bytes giving a total of 12 bytes.

Is there any reason why the specific compiler (gcc) thought it should have 16 bytes as a size?

Wild guess: is it possible that the biggest type (e.g. double or in this case x64 pointer) will dictate the padding to use?

Answer 1

Likely the compiler is aligning the struct on an 8-byte word boundary to improve access speed. A struct size of 9 is probably going to slow down the CPU quite a bit with unaligned accesses (plus the stack pointer should never be on an odd address). A size of 12 (3 padding bytes), would work, but some operations, like the FPU operations, prefer an alignment of 8 or 16 bytes.