试图理解 gcc 在复制返回地址的 main 顶部的复杂堆栈对齐 [英] Trying to understand gcc's complicated stack-alignment at the top of main that copies the return address

问题描述

我已经反汇编了一些我写的程序(linux)以更好地理解它是如何工作的，我注意到主要功能总是以:

hi I have disassembled some programs (linux) I wrote to understand better how it works, and I noticed that the main function always begins with:

lea    ecx,[esp+0x4] ; I assume this is for getting the adress of the first argument of the main...why ?
and    esp,0xfffffff0 ; ??? is the compiler trying to align the stack pointer on 16 bytes ???
push   DWORD PTR [ecx-0x4] ; I understand the assembler is pushing the return adress....why ?
push   ebp                
mov    ebp,esp
push   ecx  ;why is ecx pushed too ??

所以我的问题是:为什么所有这些工作都完成了??我只了解使用:

so my question is: why all this work is done ?? I only understand the use of:

push   ebp                
mov    ebp,esp

其余的对我来说似乎没用...

the rest seems useless to me...

推荐答案

我已经试过了:

;# As you have already noticed, the compiler wants to align the stack
;# pointer on a 16 byte boundary before it pushes anything. That's
;# because certain instructions' memory access needs to be aligned
;# that way.
;# So in order to first save the original offset of esp (+4), it
;# executes the first instruction:
lea    ecx,[esp+0x4]

;# Now alignment can happen. Without the previous insn the next one
;# would have made the original esp unrecoverable:
and    esp,0xfffffff0

;# Next it pushes the return addresss and creates a stack frame. I
;# assume it now wants to make the stack look like a normal
;# subroutine call:
push   DWORD PTR [ecx-0x4]
push   ebp
mov    ebp,esp

;# Remember that ecx is still the only value that can restore the
;# original esp. Since ecx may be garbled by any subroutine calls,
;# it has to save it somewhere:
push   ecx

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