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assemblyx86-64

Why stack frame aligned by 24 bytes?

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:

figure

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?"

Solution

  • This is consistent with x64 ABI (both Microsoft and SystemV). Stack must be aligned on 16 bytes boundary before calling a function ("call Q" in your example). Let's assume that the stack is initially at a 16 bytes boundary. When the program reaches "P" label (due to a "call P" instruction), RSP points 8 byte below, because "call" subtracts 8 to RSP and saves RIP (8 bytes) at *RSP. Then, there are two "pushq" (line 2 and 3), each decreases RSP by 8, so RSP is still misaligned. That's why The compiler must subtract 8 to align RSP before executing "call Q" at line 8 and 10.

    This article describes x86/x64 ABIs quite well: https://en.wikipedia.org/wiki/X86_calling_conventions