C / C++ / MFC

strange 4 assembly instructions for a function

21-Jul-08 1:23

akira32 wrote:
Add/Remove operation is impossible, becuase the code element CDialogCommon is read only

The message itself it is clear that you cannot added new handler because the file is read only. So just take the DialogCommon.h and CDialogCommon.cpp and remove the readonly attribute D'Oh! | :doh:

nave

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George_George20-Jul-08 22:56

20-Jul-08 22:56

Hello everyone,

I am confused what is the function of the 4 assembly instructions? I posted the 4 assembly instructions and related whole source code/assembly code below.

Any ideas?

00BA13CC  lea         edi,[ebp-0C0h] 
00BA13D2  mov         ecx,30h 
00BA13D7  mov         eax,0CCCCCCCCh 
00BA13DC  rep stos    dword ptr es:[edi] 


int sumExample (int a, int b)
{
00BA13C0  push        ebp  
00BA13C1  mov         ebp,esp 
00BA13C3  sub         esp,0C0h 
00BA13C9  push        ebx  
00BA13CA  push        esi  
00BA13CB  push        edi  
00BA13CC  lea         edi,[ebp-0C0h] 
00BA13D2  mov         ecx,30h 
00BA13D7  mov         eax,0CCCCCCCCh 
00BA13DC  rep stos    dword ptr es:[edi] 
    return a + b;
00BA13DE  mov         eax,dword ptr [a] 
00BA13E1  add         eax,dword ptr [b] 
}

#include <iostream>

using namespace std;

int sumExample (int a, int b)
{
    return a + b;
}

int main()
{
	int c = sumExample (100, 200);

	cout << c << endl;

	return 0;
}

thanks in advance,
George

Naveen20-Jul-08 23:22

20-Jul-08 23:22

George_George wrote:
Any ideas?

I think you are running a debug version. In the release version it wont come.

The purpose of that instructions is to initialize all the memory location of from ebp-0C0h to edp with value 0CCCCCCCCh. The address "ebp - 0C0h" to ebp is the stack range of the function sumExample().

nave

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George_George20-Jul-08 23:28

20-Jul-08 23:28

Thanks Naveen,

1.

You are correct, I am using debug version. So, only debug version will initialize stack used in current function to zero, and release version will not (I have tried in my sample, it is yes. Here I want to ask for a general answer.) Smile | :)

2.

How do you calculate the memory used will be 0x0C0 bytes?

regards,
George

Naveen20-Jul-08 23:38

20-Jul-08 23:38

George_George wrote:
So, only debug version will initialize stack used in current function to zero

No to 0 but to 0x0CCCCCCCC Smile | :)

George_George wrote:
How do you calculate the memory used will be 0x0C0 bytes?

Compiler does this at compile time itself. It calculate this value by adding the size of variables, flags to push etc.

nave

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George_George20-Jul-08 23:44

20-Jul-08 23:44

Cool, nave!

From the code I posted in my 1st post, do you have any ideas why the stack used is 30h? I am very confused why the address begins with ebp-0C0h?

regards,
George

Naveen20-Jul-08 23:51

20-Jul-08 23:51

George_George wrote:
do you have any ideas why the stack used is 30h?

Stack size is actually C0h. An that 30h is a counter ( C0h/4 = 30h ). I dont know why the stack size became C0h in your case. May be the compiler allocates atleast that much stack for a function and I have found it depends on the compiler version.

nave

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George_George21-Jul-08 0:58

21-Jul-08 0:58

Thanks nave,

I found why I am confused by reading your reply, the counter is in the basis of 4-byte, not 1-byte, correct?

regards,
George

Rajkumar R21-Jul-08 1:17

Rajkumar R

21-Jul-08 1:17

4 is the sizeof doubleword and the count is the count of doublewords as it uses "store string instruction" (rep stosd).[^]. if it uses rep stosb[^] it would be 1 byte

George_George21-Jul-08 2:43

Re: strange 4 assembly instructions for a function [modified]

21-Jul-08 2:43

1.
Cool, Rajkumar. I think the 4 assembly instruction is doing stack initialization work for the stack space used for foo function, correct?

2.

But from the 4 assembly instruction, how do you know it ends from address ebp-0C0h and is 30 * 4 bytes length?

regards,
George

Rajkumar R21-Jul-08 3:05

Rajkumar R

21-Jul-08 3:05

George_George wrote:
I think the 4 assembly instruction is doing stack initialization work for the stack space used for foo function, correct?

i think so (i admit), to facilitate ease debug VC++ compiler fills stack variables with 0cch and heap with 0cdh(if my knowledge is correct Smile | :)

George_George wrote:
But from the 4 assembly instruction, how do you know it ends from address ebp-0C0h and is 30 * 4 bytes length?

as naveen said, it is the stack frame size calculated by compiler, refer fixed-allocation-size
in [Prolog and Epilog^] and [Stack Allocation^] for paramters considered for that.

modified on Monday, July 21, 2008 9:28 AM

George_George21-Jul-08 19:50

21-Jul-08 19:50

Thanks Rajkumar,

1.

Rajkumar R wrote:
it is the stack frame size calculated by compiler, refer fixed-allocation-size

Do you mean stack will be initialized just according to how much data will be used actually? Or no matter how much data will be used, a fixed size will be initialized?

2.

My confusion is, from the assembly code I posted, more stack space than I acually used is initialized. Any comments?

regards,
George

Luc Pattyn21-Jul-08 2:53

Luc Pattyn

21-Jul-08 2:53

Hi George,

I believe the whole idea is twofold:

1. to initialize your function's local variables to a strange value,
increasing the probability that you will discover potential "variable value used before
being set" mistakes. Your function does not have local variabled, so this does not apply here.

2. allocate more stack space than needed, and fill it with something that springs out
in a hex dump, so you can easily recognize the stack frames of each of your functions.

Both of these reasons don't hold true for release code, since they cost both CPU cycles
and stack space.

Smile | :)

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George_George21-Jul-08 19:45

21-Jul-08 19:45

Thanks Luc,

Luc Pattyn wrote:
allocate more stack space than needed

It is ok. But how to calculate the actual stack space used? For example, in my sample?

regards,
George

Luc Pattyn22-Jul-08 1:12

Luc Pattyn

22-Jul-08 1:12

The compiler is responsible for stack allocation. It knows what it wants
to store on stack, and it can round it up to whatever number it chooses,
as long as it emits correct code (e.g. it should not initialize more memory
than it has allocated for each and every function).

Smile | :)

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George_George10-Aug-08 21:54