Why stack frame aligned by 24 bytes?

Question

I am reading Computer Systems: A Programmer's Perspective 3rd Edition by Randal E. Bryant and David R. O'Hallaron.

In chapter 3 and section 7.5, there is figure demonstrating how stack frames are allocated as follows:

I couldn't understand that why are line 4 and 12 needed. It seems that those lines are not required because the surplus 8 bytes of stack memory are not used at all.

As comment indicates, IMHO, it seems that it is inevitably allocated to align stack frame by 24 bytes:

• additional 16 bytes for pushq %rbp and pushq %rbx respectively
• , and 8 bytes for subq $8, %rsp

So, my question can be summarized to "Why stack frames are aligned by 24 bytes?"

Answer 1

24 bytes is not a power of 2 and hence cannot be an alignment. The real alignment is to 16 bytes and you forgot to account for the call instruction also pushing 8 bytes on the stack. So in total, the stack pointer is moved by 32 bytes, preserving alignment to 16 bytes.