为什么GCC在堆栈上分配的空间超出了对齐所需的空间? [英] Why does GCC allocate more space than necessary on the stack, beyond what's needed for alignment?

问题描述

我正在读一本教科书，其中显示了基于C代码的汇编代码:

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

C代码:

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

汇编代码:

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  

我不明白为什么堆栈指针%rsp会减少24个字节.我只将8个字节的缓冲区分配为char buf[8];，并且没有被调用方保存的寄存器要压入堆栈，指令不应该

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

推荐答案

分配额外的16个字节的堆栈空间是GCC遗漏的优化，偶尔会弹出.我不知道为什么会发生，但是可以用GCC10.1 -O3重现. Clang不会这样做，它只保留8个字节(带有伪push).上 Godbolt ，其中在许多GNU/Linux发行版的默认，不像GCC.

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.

即使int buf;/foo(&buf)也会导致过度分配.

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

我的疯狂猜测是，直到之后它已经确定需要超过8个字节的空间(因此需要24个字节)，GCC才会对其进行优化.希望这个好的MCVE可以让GCC开发人员找到一个可以轻松修复的bug.

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.

随时将其报告为GCC missed-optimization错误( https://gcc.gnu.org/bugzilla/);我最近看了一下，但是没有找到现有的.

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.

您是正确的，按照x86-64 System V ABI的要求，分配8个字节足以使char buf[8] 和在call之前将RSP对齐16. a href ="https://stackoverflow.com/q/49391001">为什么System V/AMD64 ABI要求16字节堆栈对齐?).

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不是试图保持32字节堆栈对齐或其他任何内容. -mpreferred-stack-boundary的默认值是ABI允许的最小值4(2 ^ 4 = 16).

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).

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