试图了解gcc在main顶部复制返回地址的复杂堆栈对齐方式 [英] Trying to understand gcc's complicated stack-alignment at the top of main that copies the return address
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问题描述
我已经分解了一些程序(Linux),我为了更好地理解它是如何工作的而编写的,我注意到主要功能总是以以下内容开头:
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? 我只了解以下用途:
so my question is: why all this work is done ?? I only understand the use of:
push ebp
mov ebp,esp
其余的东西对我来说似乎没用...
the rest seems useless to me...
推荐答案
我已经尝试过了:
;# 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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