Why does GCC allocate more space than necessary on the stack, beyond what's needed for alignment?

问题

I'm reading a textbook which shows assembly code based on C code:

C code:

void echo()
{
   char buf[8];
   otherFunction(buf);
}

assembly code:

echo:
   subq $24, %rsp      //Allocate 24 bytes on stack, but why allocate 24 instead of 8 bytes?
   movq %rsp, %rdi     //Compute buf as %rsp
   call otherFunction

I don't understand why stack pointer %rsp is decremented by 24 bytes. I only assign 8 bytes' buffer as char buf[8];, and there no callee saved registers to push on stack, shouldn't the instruction be

subq $8, %rsp

回答1:

Allocating an extra 16 bytes of stack space is a GCC missed optimization that pops up occasionally. I don't know why it happens, but it's reproducible with GCC10.1 -O3. Clang doesn't do it, it just reserves 8 bytes (with a dummy push). Example on Godbolt, where -fno-stack-protector -fno-pie is the default, unlike GCC in many GNU/Linux distros.

Even int buf; / foo(&buf) results in over-allocation.

My wild guess is that there's something GCC doesn't optimize away until after it's already decided it needs more than 8 bytes of space (and thus needs 24). Hopefully this good MCVE will let GCC devs find an fix that bug, if it's easily fixable.

Feel free to report this as a GCC missed-optimization bug (https://gcc.gnu.org/bugzilla/); I looked recently but didn't find an existing one.

You're correct that allocating 8 bytes would be enough for char buf[8] and re-align RSP by 16 before the call, as required by the x86-64 System V ABI (Why does System V / AMD64 ABI mandate a 16 byte stack alignment?).

GCC is not trying to maintain 32-byte stack alignment or anything. The default for -mpreferred-stack-boundary is the minimum allowed by the ABI, 4 (2^4 = 16).