为什么printf函数会导致序言的变化?
C code_1:
#include <cstdio>
int main(){
int a = 11;
printf("%d", a);
}
GCC -m32生成一个:
.LC0:
.string "%d"
main:
lea ecx, [esp+4] // What's purpose of this three
and esp, -16 // lines?
push DWORD PTR [ecx-4] //
push ebp
mov ebp, esp
push ecx
sub esp, 20 // why sub 20?
mov DWORD PTR [ebp-12], 11
sub esp, 8
push DWORD PTR [ebp-12]
push OFFSET FLAT:.LC0
call printf
add esp, 16
mov eax, 0
mov ecx, DWORD PTR [ebp-4]
leave
lea esp, [ecx-4]
ret
C code_2:
#include <cstdio>
int main(){
int a = 11;
}
GCC -m32:
main:
push ebp
mov ebp, esp
sub esp, 16
mov DWORD PTR [ebp-4], 11
mov eax, 0
leave
ret
在第一个代码中添加前三行的目的是什么?如果可以,请解释第一个汇编代码。
编辑:
64位模式:
.LC0:
.string "%d"
main:
push rbp
mov rbp, rsp
sub rsp, 16
mov DWORD PTR [rbp-4], 11
mov eax, DWORD PTR [rbp-4]
mov esi, eax
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
mov eax, 0
leave
ret
洞察力是编译器在函数调用时保持堆栈对齐。 对齐是16字节。
lea ecx, [esp+4] ;Save original ESP to ECX (ESP+4 actually)
and esp, -16 ;Align stack on 16 bytes (Lower esp)
push DWORD PTR [ecx-4] ;Push main return address (Stack at 16B + 4)
;My guess is to aid debugging tools that expect the RA
;to be at [ebp+04h]
push ebp
mov ebp, esp ;Prolog (Stack at 16B+8)
push ecx ;Save ECX (Original stack pointer) (Stack at 16B+12)
sub esp, 20 ;Reserve 20 bytes (Stack at 16B+0, ALIGNED AGAIN)
;4 for alignment + 1x16 for a variable (variable space is
;allocated in multiple of 16)
mov DWORD PTR [ebp-12], 11 ;a = 11
sub esp, 8 ;Stack at 16B+8 for later alignment
push DWORD PTR [ebp-12] ;a
push OFFSET FLAT:.LC0 ;"%d" (Stack at 16B)
call printf
add esp, 16 ;Remove args+pad from the stack (Stack at 16B)
mov eax, 0 ;Return 0
mov ecx, DWORD PTR [ebp-4] ;Restore ECX without the need to add to esp
leave ;Restore EBP
lea esp, [ecx-4] ;Restore original ESP
ret
我不知道为什么编译器在esp+4
而不是ecx
中保存esp
(esp+4
是main
的第一个参数的地址)。