C language: function stack frame

1. What is a function stack frame

stay C In language , Every running function has a stack frame corresponding to it , The return address and local variables of the function are stored in the stack frame . Logically speaking , Stack frame is a function execution environment ： Function parameter 、 A local variable of a function 、 Where to return after the function is executed, and so on .

2. Function stack frame creation and destruction

When the function is called , It will open up a space on the stack for this function , During the operation of this function , register ebp Save the address at the bottom of the stack , register esp Save the stack top address . And it should be clear ,esp and ebp Registers can store only one address at a time , therefore , anytime , This pair of pointers will all point to the stack frame structure of the same function . also ebp Generally, the system changes its value , and esp It will move with the data in and out of the stack , in other words esp Always point to the top of the stack . Let's take an example ：

#include<stdio.h>

int Add(int x, int y)
{
    
	int z = 0;
	z = x + y;
	return z;
}

int main()
{
    
	int a = 10;
	int b = 20;
	int c = 0;
	c = Add(a, b);
	printf("%d\n", c);

	return 0;
}

The following code is main The corresponding disassembly code in the function ：

int main()
{
    
00B118B0  push        ebp  //ebp Pressing stack 
00B118B1  mov         ebp,esp  // take esp The value is assigned to ebp
00B118B3  sub         esp,0E4h  // take esp The value of minus 0E4h, to main Function stack frame allocation space 
00B118B9  push        ebx  // Pressing stack 
00B118BA  push        esi  // Pressing stack 
00B118BB  push        edi  // Pressing stack 
00B118BC  lea         edi,[ebp-24h]  
00B118BF  mov         ecx,9  
00B118C4  mov         eax,0CCCCCCCCh  
00B118C9  rep stos    dword ptr es:[edi]   // from edi From start to down 9 The values of all spaces are changed to eax, namely 0CCCCCCCCh
00B118CB  mov         ecx,0B1C003h  
00B118D0  call        00B1131B  // Get into main function 
	int a = 10;
00B118D5  mov         dword ptr [ebp-8],0Ah  //ebp-8 by a The location of , take a The value of is assigned to 10
	int b = 20;
00B118DC  mov         dword ptr [ebp-14h],14h  //ebp-14h by b The location of , take b The value of is assigned to 20
	int c = 0;
00B118E3  mov         dword ptr [ebp-20h],0  // ditto 
	c = Add(a, b);
00B118EA  mov         eax,dword ptr [ebp-14h]  // The process of transferring parameters , First the b The value of is passed to eax
00B118ED  push        eax  // then eax Pressing stack 
00B118EE  mov         ecx,dword ptr [ebp-8]  // The ginseng , First the a The value of is passed to ecx
00B118F1  push        ecx  // then ecx Pressing stack 
00B118F2  call        00B110B4  
00B118F7  add         esp,8  
00B118FA  mov         dword ptr [ebp-20h],eax  
	printf("%d\n", c);
00B118FD  mov         eax,dword ptr [ebp-20h]  
00B11900  push        eax  
00B11901  push        0B17B30h  
00B11906  call        00B110D2  
00B1190B  add         esp,8  

	return 0;
00B1190E  xor         eax,eax  
}

Here is Add Disassembly code of function ：

int Add(int x, int y)
{
    
00B11770  push        ebp  // Record the last ebp Value 
00B11771  mov         ebp,esp  // assignment 
00B11773  sub         esp,0CCh  
00B11779  push        ebx  
00B1177A  push        esi  
00B1177B  push        edi  
00B1177C  lea         edi,[ebp-0Ch]  
00B1177F  mov         ecx,3  
00B11784  mov         eax,0CCCCCCCCh  
00B11789  rep stos    dword ptr es:[edi]  // And main The function is the same , from edi Start down 3 All units are assigned CCC
00B1178B  mov         ecx,0B1C003h  
00B11790  call        00B1131B  
	int z = 0;
00B11795  mov         dword ptr [ebp-8],0  
	z = x + y;
00B1179C  mov         eax,dword ptr [ebp+8]  
00B1179F  add         eax,dword ptr [ebp+0Ch]  // Do addition calculation 
00B117A2  mov         dword ptr [ebp-8],eax  
	return z;
00B117A5  mov         eax,dword ptr [ebp-8]  
}

The stack frame destruction process is as follows ：

00B117A8  pop         edi  // Out of the stack 
00B117A9  pop         esi  // Out of the stack 
00B117AA  pop         ebx  // Out of the stack 
00B117AB  add         esp,0CCh  // The destruction Add function 
00B117B1  cmp         ebp,esp  
00B117B3  call        00B11244  // go back to main function 
00B117B8  mov         esp,ebp  
00B117BA  pop         ebp  // Out of the stack 
00B117BB  ret  

Tell the truth , My expression ability is limited , In addition, there are too many underlying contents involved , Is difficult , There will inevitably be mistakes and inappropriate places , Welcome to correct ~~