函数调用原理–参数篇

4 Replies

1、汇编模拟函数调用的基本过程：

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assume cs:code, ds:data, ss:stack
 
stack segment   
    db 20h dup(1)
stack ends    
 
data segment    
    db 20h dup(2)
data ends
 
code segment
start:
;---初始化----              
    mov ax, stack;加载栈
    mov ds, ax
    mov sp, 20h
 
    mov ax, data;加载数据
    mov ds, ax  
;---业务代码---
    push 1122h
    push 3344h;
    call sum;系统会push下一行代码地址到栈
    add sp, 4;人为压栈的数据执行出栈操作(否则会造成内存泄漏)
 
;---退出-----
    mov ah, 4ch
    int 21h     
;---函数实现
sum:           
    mov bp, sp;获取栈顶位置
    mov ax, ss:[bp+2];取出栈最上面数据(3344h)
    add ax, ss:[bp+4];取出下一条数据(1122h)累加到ax
    ret;返回时系统会取出上一次压栈的位置   
code ends
end start

2、C++中函数调用时的反汇编代码
2.1、C++源码：

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// VCTest.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
 
int sum(int a, int b)
{
	return a + b;
}
int main(int argc, char* argv[])
{
	sum(2, 3);
	return 0;
}

2.2、反汇编部分代码

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--- E:\Project\Assemble\VCTest\VCTest.cpp  --------------------
10:   int main(int argc, char* argv[])
11:   {
00401050   push        ebp
00401051   mov         ebp,esp
00401053   sub         esp,40h
00401056   push        ebx
00401057   push        esi
00401058   push        edi
00401059   lea         edi,[ebp-40h]
0040105C   mov         ecx,10h
00401061   mov         eax,0CCCCCCCCh
00401066   rep stos    dword ptr [edi]
12:       sum(2, 3);
00401068   push        3
0040106A   push        2
0040106C   call        @ILT+15(print) (00401014)
00401071   add         esp,8
13:       return 0;
00401074   xor         eax,eax
14:   }
00401076   pop         edi
00401077   pop         esi
00401078   pop         ebx
00401079   add         esp,40h
0040107C   cmp         ebp,esp
0040107E   call        __chkesp (00401120)
00401083   mov         esp,ebp
00401085   pop         ebp
00401086   ret
--- No source file  --------------------------------------------