Creation and destruction of function stack frames

The creation and destruction of function stack frames are under the environment of different compilers , The principle is basically the same .

1. register  

eax
ebx
ecx
edx
The following two registers store addresses , These two addresses are used to maintain the function stack frame
ebp	Pointer at the bottom of the stack
esp	Top pointer of stack

Every function call , Create a space in the stack area

  We write the following code as an example ：

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;
}

After executing the above code , Our stack space will be ours main Open up a space , among ebp To record our main The high address of the occupied space ,esp To record our main Low address occupied by space  

The usage habit of stack area is to use high address first , Then use low address , That is to use our space below first , Then use the space above

stay VS2013 in ,main Functions are also called by other functions ：

mainCRTStartup       Called      __tmainCRTStartup     Called main function

Stack pressing is to put an element from the top of the stack push

Out of stack is to pop an element from the top of the stack pop

  In our main Function before , our  __tmainCRTStartup The function is already on the stack , our ebp and esp The pointer records  __tmainCRTStartup High address and low address of function （ At the bottom of this picture is the high address , It's a low address ）

First, we perform stack pressing ：push ebp, Take us ebp Press into the top of our stack , Now our ebp We are stored in  __tmainCRTStartup High address of

 

When we ebp After entering the stack , Our stack top pointer esp At the same time, it also moves up  

 

 mov ebp,esp

take esp The value is assigned to ebp, It is equivalent to ebp The pointer of moves up to the following figure

 

sub esp,0E4h 

take esp subtract 0E4h,0E4h It's an octal number 228

It's equivalent to us esp Move the pointer up , As shown in the figure below , Among them, we found our esp and ebp The pointer no longer maintains the original __tmainCRTStartup Space , At this time, the purple space is actually for us main The space opened up by the function

push ebx

push esi

push edi

  Put our ebx,esi,edi Three are pushed into the stack , But at the same time , We need to notice our stack top pointer esp With each stack , Point to the space at the top of the stack in time

 

 lea edi,[ebp-0E4h]

lea yes load effective address, It means loading valid addresses . So the above code is to load the following valid address into edi Go inside

Here we give ebp Minus the 0E4h, Be careful , We have 0E4h It has appeared before , It is our main Function stack frame space , Give us the ebp subtract 0E4h Just move up to main Top of function stack frame

 mov ecx,39h

take 39h Put in ecx in

mov eax,0CCCCCCCCh

take 0CCCCCCCCh Put in eax in

rep stos dword ptr es:[edi]

Will just from edi At the beginning 39h Time of dword(double word Two words , Four bytes ) All the data of are changed to 0CCCCCCCCh, That is to say, we just named main All the contents of the function are changed to CCCCCCCC

 

 int a=10;

mov dword ptr [ebp-8]0Ah

take 0Ah This hexadecimal number , It's the decimal system 10 Put it in ebp-8 The location of

 int b=20;

dword ptr [ebp-14h],14h

int c=0;

dword ptr [ebp-20h],0

Empathy

 c=Add(a,b);

mov eax,dword ptr [ebp-14h]

take ebp-14h, That is to say b Put the value of eax In the middle , Also is to 20 Put in eax in

push eax

take eax Pressure into the stack

 

 mov ecx,dword ptr [ebp-8]

take ebp-8 Put the value in ecx in , That is to put our a,10, Put in ecx in

push ecx

take ecx Push

The same as the above step , Are parameters of functions

 

 call 00C210E1

Call function （call The address for 00C2144B）, And will call The address of the next instruction of the instruction is pushed onto the stack .

This means call After the instruction is executed, it will jump to the address of the function , When you jump back from the address of the function , Direct execution call Next instruction for

 

  Then we enter add Internal function

The following code is similar to our main Function creation is very similar

push ebp

take ebp Pressure into the stack

 

 

mov ebp,esp

take esp The value is assigned to ebp

That means ours ebp Yes esp The location of

 

sub esp,0CCh

to sub subtract 0CCh

That's for us add Function opens up space ​​​​​​​

 ​​​​​​​

 

push ebx

push esi

push edi

Separately ebx,esi,edi Pressure into the stack

 ​​​​​​​

 lea edi,[ebp+FFFFFF34h]

Load valid address , take ebp+FFFFFF34h The address of is loaded into our edi in

mov ecx,33h

take 33h Deposit our ecx in

mov eax,0CCCCCCCCh

take 0CCCCCCCCh Assign a value to eax This register

rep stos dword ptr es:[edi]

Let's start with edi This position starts down to ebp All the middle values are assigned 0CCCCCCCCh

 

 int z=0;

mov dword ptr [ebp-8],0

take 0 Put in ebp-8 The location of , That's our z​​​​​​​

 ​​​​​​​

mov eax,dword ptr [ebp+8]

take ebp+8 Put the value of eax in , That's our a

add eax,dword ptr [ebp+0Ch]

Will we ebp+0Ch The value at is added to our eax In the middle , That is to put our b give a, This is a , our eax by 30

dword ptr [ebp-8],eax

After the addition , Then we eax Put the value of into us ebp-8 The location of , That's our z The address of

Here we can notice that we press the stack first b Push the , Recompression a, So we pass from right to left

That's our add(a,b) First of all b Push the , then a Press in

So formal parameters are not what we are add Created inside the function , Instead, go back to find the parameters when the parameters are transferred after the parameters are generated

 

return z;

mov  eax,dword ptr [ebp-8]

So here return z The order of , That is to put our ebp-8 The value of is put into our eax in , That is now our 30 Put in our eax in , because z It will be destroyed after going out later , So we need to put the result of our calculation into our register for storage .（）

pop edi

Pop up the elements at the top of the stack and put them into our edi In the register , At this time, the original data at the top of the stack is edi, That is to say, it will just edi It's just data reading , The same is true for the following two

​​​​​​​

pop esi

pop ebx

 

 

mov esp,ebp

take ebp Assign a value to esp, That means ours esp It points to us ebp The same place

 

 pop ebp

Pop up the element at the top of the stack and assign it to ebp, That is our original ebp-main That is to say, now our ebp and esp Re maintain our main Function space .

ret 

Before that we will call The address of the next instruction of the instruction is saved , Now we ret once , Now we have successfully found our former call Next instruction of instruction

 add esp,8

esp+8 That is, let our stack top pointer move down to us b Below , That is to put the two formal parameters we have used before x,y Space destruction of

 

mov dword ptr [ebp-20h],eax 

take eax Put the value of ebp-20h The location of , That's for us c assignment

 

  thus , We add The creation and destruction of function stack frames are completed