I'm studying assembly language from the book "Assembly Language Step-by-Step: Programming with Linux" by Jeff Dunteman, and have come across an interesting paragraph in the book which I'm most likely misunderstanding, hence would appreciate some clarification on:
"The stack must be clean before we destroy the stack frame and return control. This simply means that any temporary values that we may have pushed onto the stack during the program’s run must be gone. All that is left on the stack should be the caller’s EBP, EBX, ESI, and EDI values.
...
Once the stack is clean, to destroy the stack frame we must first pop the caller’s register values back into their registers, ensuring that the pops are in the correct order.
...
We restore the caller’s ESP by moving the value from EBP into ESP, and finally pop the caller’s EBP value off the stack."
Consider the following code generated from Visual Studio 2008:
int myFuncSum( int a, int b)
{
001B1020 push ebp
001B1021 mov ebp,esp
001B1023 push ecx <------------------
int c;
c = a + b;
001B1024 mov eax,dword ptr [ebp+8]
001B1027 add eax,dword ptr [ebp+0Ch]
001B102A mov dword ptr [ebp-4],eax
return c;
001B102D mov eax,dword ptr [ebp-4]
}
001B1030 mov esp,ebp
001B1032 pop ebp
001B1033 ret
The value of ecx (indicated), pushed to make space on the stack for my variable c, is, as far as I can see, only gone from the stack when we reset ESP; however, as quoted, the book states that the stack must be clean before we reset ESP. Can someone please clarify whether or not I am missing something?
The example from Visual Studio 2008 doesn't contradict the book. The book is covering the most elaborate case of a call. See the x86-32 Calling Convention as a cross-reference which spells it out with pictures.
In your example, there were no caller registers saved on the stack, so there are no pop instructions to be performed. This is part of the "clean up" that must occur before mov esp, ebp that the book is referring to. So more specifically, let's suppose the callee is saving si and di for the caller, then the prelude and postlude for the function might look like this:
push ebp ; save base pointer
mov ebp, esp ; setup stack frame in base pointer
sub esp, 4 ; reserve 4 bytes of local data
push si ; save caller's registers
push di
; do some stuff, reference 32-bit local variable -4(%ebp), etc
; Use si and di for our own purposes...
; clean up
pop di ; do the stack clean up
pop si ; restoring the caller's values
mov esp, ebp ; restore the stack pointer
pop ebp
ret
In your simple example, there were no saved caller registers, so no final pop instructions needed at the end.
Perhaps because it's simpler or faster, the compiler elected to do the following instruction in place of sub esp, 4:
push ecx
But the effect is the same: reserve 4 bytes for a local variable.
Notice the instruction:
push ebp
mov ebp,esp ; <<<<=== saves the stack base pointer
and the instruction:
mov esp,ebp ; <<<<<== restore the stack base pointer
pop ebp
So after this sequence the stack is clean again
Related
Hi heads up this is a homework. I'm given an assembly generated by MSVC 32-bit Release with optimizations on, and I'm supposed to decode it back into C++. I've included the top of the function to the line I'm having problems with. The comments are mine, which I'm wrote while trying to understand this.
Note: Code is supposedly generated from C++. Not traditional ASM.
Note 2: There is one area of undefined behavior in the code.
Here are the lines I'm stuck with
TheFunction: ; TheFunction(int* a, int s);
0F2D4670 push ebp ; Push/clear/save ebp
0F2D4671 mov ebp,esp ; ebp now points to top of stack
0F2D4673 push ecx ; Push/clear/save ecx
0F2D4674 push ebx ; Push/clear/save ebs
0F2D4675 push esi ; Push/clear/save esi
0F2D4676 mov ebx,edx ; ebx = int s
0F2D4678 mov esi,1 ; esi = 1
0F2D467D push edi ; calling convention ; Push/clear/save edi
0F2D467E mov edi,dword ptr [a (0F2D95E8h)] ; edi = a[0]
0F2D4684 cmp ebx,esi ; if(s < 1)
0F2D4686 jl SomeFunction+3Ch (0F2D46ACh) ; Jump to return
0F2D4688 nop dword ptr [eax+eax] ; !! <-- No op involving dereferencing? What does this do?
0F2D4690 mov eax,dword ptr [edi+esi*4-4] ; !! <-- edi is *a, while esi is 1. There is no address
here!
..... More code but I've figured these out ....
I've more or less got the gist of the function. Its a function that takes a pointer to an int, with an underlying array, and a size. It then goes through each element in the array from last to first, adding to each subsequent one and printing it out. However, I still haven't got the details down and need help
Two questions, both at the end of the code snippet. What does no op on a dereference pointer do, and am I reading the last line in that its attempting to dereference something not in memory?
The nop dword ptr [eax+eax] instruciton does nothing. It doesn't even access the memory location given by the operand. It literally performs no operation.
It's just there so the next instruction is aligned to a 16-byte boundary. You'll notice that next instruction address is 0F2D4690 which ends with 0 which means it's 16-byte aligned. This can improve the performance of loops. Somewhere there will be an instruction that jumps back to 0F2D4690 as part of a loop. This particular form of a NOP instruction is used because it encodes a single NOP instruction in 8 bytes.
There is no corresponding C++ code for this instruction. You shouldn't try to represent it in your C++ code, just ignore it.
Also note that your comment for mov edi,dword ptr [a (0F2D95E8h)] is incorrect. Instead of being edi = a[0] it's simply edi = a. The variable a isn't a parameter at all, instead it's a global (or file level static) variable located at memory location 0F2D95E8h. This instruction just loads the value from memory.
I have a function which takes 3 arguments, dest, src0, src1, each a pointer to data of size 12. I made two versions. One is written in C and optimized by the compiler, the other one is fully written in _asm. So yeah. 3 arguments? I naturally do something like:
mov ecx, [src0]
mov edx, [src1]
mov eax, [dest]
I am a bit confused by the compiler, as it saw fit to add the following:
_src0$ = -8 ; size = 4
_dest$ = -4 ; size = 4
_src1$ = 8 ; size = 4
?vm_vec_add_scalar_asm##YAXPAUvec3d##PBU1#1#Z PROC ; vm_vec_add_scalar_asm
; _dest$ = ecx
; _src0$ = edx
; 20 : {
sub esp, 8
mov DWORD PTR _src0$[esp+8], edx
mov DWORD PTR _dest$[esp+8], ecx
; 21 : _asm
; 22 : {
; 23 : mov ecx, [src0]
mov ecx, DWORD PTR _src0$[esp+8]
; 24 : mov edx, [src1]
mov edx, DWORD PTR _src1$[esp+4]
; 25 : mov eax, [dest]
mov eax, DWORD PTR _dest$[esp+8]
Function body etc.
add esp, 8
ret 0
What does the _src0$[esp+8] etc. even means? Why does it do all this stuff before my code? Why does it try to [apparently]stack anything so badly?
In comparison, the C++ version has only the following before its body, which is pretty similar:
_src1$ = 8 ; size = 4
?vm_vec_add##YAXPAUvec3d##PBU1#1#Z PROC ; vm_vec_add
; _dest$ = ecx
; _src0$ = edx
mov eax, DWORD PTR _src1$[esp-4]
Why is this little sufficient?
The answer of Mats Petersson explained __fastcall. But I guess that is not exactly what you're asking ...
Actually _src0$[esp+8] just means [_src0$ + esp + 8], and _src0$ is defined above:
_src0$ = -8 ; size = 4
So, the whole expression _src0$[esp+8] is nothing but [esp] ...
To see why it does all these stuff, you should probably first understand what Mats Petersson said in his post, the __fastcall, or more generally, what is a calling convention. See the link in his post for detailed informations.
Assuming that you have understood __fastcall, now let's see what happens to your codes. The compiler is using __fastcall. Your callee function is f(dst, src0, src1), which requires 3 parameters, so according to the calling convention, when a caller calls f, it does the following:
Move dst to ecx and src0 to edx
Push src1 onto the stack
Push the 4 bytes return address onto the stack
Go to the starting address of the function f
And the callee f, when its code begins, then knows where the parameters are: dst and src0 are in the registers ecx and edx, respectively; esp is pointing to the 4 bytes return address, but the 4 bytes below it (i.e. DWORD PTR[esp+4]) is exactly src1.
So, in your "C++ version", the function f just does what it should do:
mov eax, DWORD PTR _src1$[esp-4]
Here _src1$ = 8, so _src1$[esp-4] is exactly [esp+4]. See, it just retrieves the parameter src1 and stores it in eax.
There is however a tricky point here. In the code of f, if you want to use the parameter src1 multiple times, you can certainly do that, because it's always stored in the stack, right below the return address; but what if you want to use dst and src0 multiple times? They are in the registers, and can be destroyed at any time.
So in that case, the compiler should do the following: right after entering the function f, it should remember the current values of ecx and edx (by pushing them onto the stack). These 8 bytes are the so-called "shadow space". It is not done in your "C++ version", probably because the compiler knows for sure that these two parameters will not be used multiple times, or that it can handle it properly some other way.
Now, what happens to your _asm version? The problem here is that you are using inline assembly. The compiler then loses its control to the registers, and it cannot assume that the registers ecx and edx are safe in your _asm block (they are actually not, since you used them in the _asm block). Thus it is forced to save them at the beginning of the function.
The saving goes as follows: it first raises esp by 8 bytes (sub esp, 8), then move edx and ecx to [esp] and [esp+4] respectively.
And then it can enter safely your _asm block. Now in its mind (if it has one), the picture is that [esp] is src0, [esp+4] is dst, [esp+8] is the 4 byte return address, and [esp+12] is src1. It no longer thinks about ecx and edx.
Thus your first instruction in the _asm block, mov ecx, [src0], should be interpreted as mov ecx, [esp], which is the same as
mov ecx, DWORD PTR _src0$[esp+8]
and the same for the other two instructions.
At this point, you might say, aha it's doing stupid things, I don't want it to waste time and space on that, is there a way?
Well there is a way - do not use inline assembly... it's convenient, but there is a compromise.
You can write the assembly function f in a .asm source file and public it. In the C/C++ code, declare it as extern 'C' f(...). Then, when you begin your assembly function f, you can play directly with your ecx and edx.
The compiler has decided to use a calling convention that uses "pass arguments in registers" aka __fastcall. This allows the compiler to pass some of the arguments in registers, instead of pushing onto stack, and this can reduce the overhead in the call, because moving from a variable to a register is faster than pushing onto the stack, and it's now already in a register when we get to the callee function, so no need to read it from the stack.
There is a lot more information about how calling conventions work on the web. The wikipedia article on x86 calling conventions is a good starting point.
I'm compiling this C++ code with the VC compiler. I'm trying to call a function that takes two WORD (aka unsigned short) parameters using the __asm statement, like this:
__declspec(naked) void __stdcall CallFunction(WORD a, WORD b)
{
__asm {
PUSH EBP
MOV EBP, ESP
PUSH a
PUSH b
CALL functionAddress
LEAVE
RETN
}
}
The function at functionAddress simply outputs the result of doing a + b. Then calling CallFuncion(5, 5); prints "64351" or something like that. The problem is when using the a and b variables inside the __asm statement because this works:
PUSH EBP
MOV EBP, ESP
PUSH 5
PUSH 5
CALL functionAddress
LEAVE
This is the function at functionAddress:
void __stdcall Add(WORD a, WORD b)
{
WORD c;
c = a + b;
printf("The result is %d\n", c);
}
How can I do this the right way? So the __asm statement interpretate the a and b values?
Since you're using __declspec(naked) and setting up your own stack frame, I don't believe the compiler will let you refer to a and b by name. Using __declspec(naked) basically means you're responsible for dealing with the stack frame, parameters, etc., on your own.
You probably want code more on this general order:
__asm {
PUSH EBP
MOV EBP, ESP
mov eax, [ebp+8]
mov ebx, [ebp+12]
push eax
push ebx
CALL functionAddress
LEAVE
RETN
}
I'ts been a while since I've handled things like this by hand, so you might want to re-check those offsets, but if I recall correctly, the return address should be at [ebp+4]. Parameters are (usually) pushed from right to left, so the the left-most parameter should be next at [ebp+8], and the next parameter at [ebp+12] (keeping in mind that the stack grows downward).
Edit: [I should have looked more carefully at the function heading.]
You've marked CallFunction as using the __stdcall calling convention. That means it's required to clean up the parameters that were passed to it. So, since it receives 8 bytes of parameters, it needs to remove 8 bytes from the stack as it returns:
PUSH EBP
MOV EBP, ESP
mov eax, [ebp+8]
mov ebx, [ebp+12]
push eax
push ebx
CALL Add_f
LEAVE
RET 8
From http://www.learncpp.com/cpp-tutorial/79-the-stack-and-the-heap/
Here is the sequence of steps that takes place when a function is
called:
The address of the instruction beyond the function call is pushed onto the stack. This is how the CPU remembers where to go after the
function returns.
Room is made on the stack for the function’s return type. This is just a placeholder for now.
The CPU jumps to the function’s code.
The current top of the stack is held in a special pointer called the stack frame.
Everything added to the stack after this point is considered “local” to the function.
All function arguments are placed on the stack.
The instructions inside of the function begin executing.
Local variables are pushed onto the stack as they are defined.
I am not sure how point #6 works. If all function arguments are placed on the stack, how are they accessed?
If for example, there are three arguments a, b and c and are placed on stack like this from top
| a |
| b |
| c |
| |
...
|___|
Now what happens when the function wants to access c? Are a and b popped out?
The stack is a metaphoric stack. Remember it is still a RAM, so you can access each address without popping the rest, if you know what you are looking for.
Since the automatic variable's size is known at compile time - the compiler marks offset for each variable, the offset is determined from where the automatic variables section on stack start [or the stack's head, both are valid and the specific implementation depends might depend on architecture], and it access them by merely: start + offset for each variable's offset.
No they are not. The stack pointer (typically the esp registry) points to a, esp+8h points to b, esp+16h points to c and so on. There's no need for a to be popped.
Note that this is an implementation detail. You shouldn't worry about these. The number I've given is purely theoretical, on some architectures descending addresses are given to latter parameters, on others the other way around. There's no guarantee this happens.
EDIT: It seems to me like that's not a very reliable source of information. It speaks of stack and heap, but these are implementation details, and might not even be there.
There's no constraint in the standard for anything to be implemented via a stack either. For example, I have the following code generated:
void foo(int x, int y, int z)
{
01241380 push ebp
01241381 mov ebp,esp
01241383 sub esp,0CCh
01241389 push ebx
0124138A push esi
0124138B push edi
0124138C lea edi,[ebp-0CCh]
01241392 mov ecx,33h
01241397 mov eax,0CCCCCCCCh
0124139C rep stos dword ptr es:[edi]
int c = x;
0124139E mov eax,dword ptr [x]
012413A1 mov dword ptr [c],eax
c = y;
012413A4 mov eax,dword ptr [y]
012413A7 mov dword ptr [c],eax
c = z;
012413AA mov eax,dword ptr [z]
012413AD mov dword ptr [c],eax
}
012413B0 pop edi
012413B1 pop esi
012413B2 pop ebx
012413B3 mov esp,ebp
012413B5 pop ebp
So you see, there's no stack there. The runtime has direct access to the elements: dword ptr [x], etc.
It uses the stack pointer and a relative adress to point out c.
I got the following simple C++ code:
#include <stdio.h>
int main(void)
{
::printf("\nHello,debugger!\n");
}
And from WinDbg, I got the following disassembly code:
SimpleDemo!main:
01111380 55 push ebp
01111381 8bec mov ebp,esp
01111383 81ecc0000000 sub esp,0C0h
01111389 53 push ebx
0111138a 56 push esi
0111138b 57 push edi
0111138c 8dbd40ffffff lea edi,[ebp-0C0h]
01111392 b930000000 mov ecx,30h
01111397 b8cccccccc mov eax,0CCCCCCCCh
0111139c f3ab rep stos dword ptr es:[edi]
0111139e 8bf4 mov esi,esp
011113a0 683c571101 push offset SimpleDemo!`string' (0111573c)
011113a5 ff15b0821101 call dword ptr [SimpleDemo!_imp__printf (011182b0)]
011113ab 83c404 add esp,4
011113ae 3bf4 cmp esi,esp
011113b0 e877fdffff call SimpleDemo!ILT+295(__RTC_CheckEsp) (0111112c)
011113b5 33c0 xor eax,eax
011113b7 5f pop edi
011113b8 5e pop esi
011113b9 5b pop ebx
011113ba 81c4c0000000 add esp,0C0h
011113c0 3bec cmp ebp,esp
011113c2 e865fdffff call SimpleDemo!ILT+295(__RTC_CheckEsp) (0111112c)
011113c7 8be5 mov esp,ebp
011113c9 5d pop ebp
011113ca c3 ret
I have some difficulties to fully understand it. What is the SimpleDemo!ILT things doing here?
What's the point of the instruction comparing ebp and esp at 011113c0?
Since I don't have any local variables in main() function, why there's still a sub esp,0C0h at the loacation of 01111383?
Many thanks.
Update 1
Though I still don't know what ILT means, but the __RTC_CheckESP is for runtime checks. These code can be elimiated by placing the following pragma before the main() function.
#pragma runtime_checks( "su", off )
Reference:
http://msdn.microsoft.com/en-us/library/8wtf2dfz.aspx
http://msdn.microsoft.com/en-us/library/6kasb93x.aspx
Update 2
The sub esp,0C0h instruction allocate another 0C0h bytes extra space on the stack. Then EAX is filled with 0xCCCCCCCC, this is 4 bytes, since ECX=30h, 4*30h=0C0h, so the instruction rep stos dword ptr es:[edi] fill exactly the extra spaces with 0xCC. But what is this extra space on stack for? Is this some kind of safe belt? Also I notice that if I turn off the runtime check as Update 1 shows, there's still such extra space on stack, though much smaller. And this space is not filled with 0xCC.
The assembly code without runtime check is like below:
SimpleDemo!main:
00231250 55 push ebp
00231251 8bec mov ebp,esp
00231253 83ec40 sub esp,40h <-- Still extra space allocated from stack, but smaller
00231256 53 push ebx
00231257 56 push esi
00231258 57 push edi
00231259 683c472300 push offset SimpleDemo!`string' (0023473c)
0023125e ff1538722300 call dword ptr [SimpleDemo!_imp__printf (00237238)]
00231264 83c404 add esp,4
00231267 33c0 xor eax,eax
00231269 5f pop edi
0023126a 5e pop esi
0023126b 5b pop ebx
0023126c 8be5 mov esp,ebp
0023126e 5d pop ebp
0023126f c3 ret
Most of the instructions are part of MSVC runtime checking, enabled by default for debug builds. Just calling printf and returning 0 in an optimized build takes much less code. (Godbolt compiler explorer). Other compilers (like GCC and clang) don't do as much stuff like stack-pointer comparison after calls, or poisoning stack memory with a recognizable 0xCC pattern to detect use-uninitialized, so their debug builds are like MSVC debug mode without its extra runtime checks.
I've annotated the assembler, hopefully that will help you a bit. Lines starting 'd' are debug code lines, lines starting 'r' are run time check code lines. I've also put in what I think a debug with no runtime checks version and release version would look like.
; The ebp register is used to access local variables that are stored on the stack,
; this is known as a stack frame. Before we start doing anything, we need to save
; the stack frame of the calling function so it can be restored when we finish.
push ebp
; These two instructions create our stack frame, in this case, 192 bytes
; This space, although not used in this case, is useful for edit-and-continue. If you
; break the program and add code which requires a local variable, the space is
; available for it. This is much simpler than trying to relocate stack variables,
; especially if you have pointers to stack variables.
mov ebp,esp
d sub esp,0C0h
; C/C++ functions shouldn't alter these three registers in 32-bit calling conventions,
; so save them. These are stored below our stack frame (the stack moves down in memory)
r push ebx
r push esi
r push edi
; This puts the address of the stack frame bottom (lowest address) into edi...
d lea edi,[ebp-0C0h]
; ...and then fill the stack frame with the uninitialised data value (ecx = number of
; dwords, eax = value to store)
d mov ecx,30h
d mov eax,0CCCCCCCCh
d rep stos dword ptr es:[edi]
; Stack checking code: the stack pointer is stored in esi
r mov esi,esp
; This is the first parameter to printf. Parameters are pushed onto the stack
; in reverse order (i.e. last parameter pushed first) before calling the function.
push offset SimpleDemo!`string'
; This is the call to printf. Note the call is indirect, the target address is
; specified in the memory address SimpleDemo!_imp__printf, which is filled in when
; the executable is loaded into RAM.
call dword ptr [SimpleDemo!_imp__printf]
; In C/C++, the caller is responsible for removing the parameters. This is because
; the caller is the only code that knows how many parameters were put on the stack
; (thanks to the '...' parameter type)
add esp,4
; More stack checking code - this sets the zero flag if the stack pointer is pointing
; where we expect it to be pointing.
r cmp esi,esp
; ILT - Import Lookup Table? This is a statically linked function which throws an
; exception/error if the zero flag is cleared (i.e. the stack pointer is pointing
; somewhere unexpected)
r call SimpleDemo!ILT+295(__RTC_CheckEsp))
; The return value is stored in eax by convention
xor eax,eax
; Restore the values we shouldn't have altered
r pop edi
r pop esi
r pop ebx
; Destroy the stack frame
r add esp,0C0h
; More stack checking code - this sets the zero flag if the stack pointer is pointing
; where we expect it to be pointing.
r cmp ebp,esp
; see above
r call SimpleDemo!ILT+295(__RTC_CheckEsp)
; This is the usual way to destroy the stack frame, but here it's not really necessary
; since ebp==esp
mov esp,ebp
; Restore the caller's stack frame
pop ebp
; And exit
ret
; Debug only, no runtime checks
push ebp
mov ebp,esp
d sub esp,0C0h
d lea edi,[ebp-0C0h]
d mov ecx,30h
d mov eax,0CCCCCCCCh
d rep stos dword ptr es:[edi]
push offset SimpleDemo!`string'
call dword ptr [SimpleDemo!_imp__printf]
add esp,4
xor eax,eax
mov esp,ebp
pop ebp
ret
; Release mode (The optimiser is clever enough to drop the frame pointer setup with no VLAs or other complications)
push offset SimpleDemo!`string'
call dword ptr [SimpleDemo!_imp__printf]
add esp,4
xor eax,eax
ret
Number one your code's main() is improperly formed. It doesn't return the int you promised it would return. Correcting this defect, we get:
#include
int main(int argc, char *argv[])
{
::printf("\nHello,debugger!\n");
return 0;
}
Additionally, any more, it is very strange to see #include <stdio.h> in a C++ program. I believe you want #include <cstdio>
In all cases, space must be made on the stack for arguments and for return values. main()'s return value requires stack space. main()s context to be saved during the call to printf() requires stack space. printf()'s arguments require stack space. printf()'s return value requires stack space. That's what the 0c0h byte stack frame is doing.
The first thing that happens is the incoming bas pointer is copied to the top of the stack. Then the new stack pointer is copied into the base pointer. We'll be checking later to be sure that the stack winds up back where it started from (because you have runtime checking turned on). Then we build the (0C0h bytes long) stack frame to hold our context and printf()'s arguments during the call to printf(). We jump to printf(). When we get back, we hop over the return value which you didn't check in your code (the only thing left on its frame) and make sure the stack after the call is in the same place it was before the call. We pop our context back off the stack. We then check that the final stack pointer matches the value we saved way up at the front. Then we pop the prior value of the base pointer off the very top of the stack and return.
That is code that is inserted by the compiler when you build with runtime checking (/RTC). Disable those options and it should be clearer. /GZ could also be causing this depending on your VS version.
For the record, I suspect that ILT means "Incremental Linking Thunk".
The way incremental linking (and Edit&Continue) works is the following: the linker adds a layer of indirection for every call via thunks which are grouped at the beginning of executable, and adds a huge reserved space after them. This way, when you're relinking the updated executable it can just put any new/changed code into the reserved area and patch only the affected thunks, without changing the rest of the code.
The 40 bytes is the worst case stack allocation for any called or subsequently called function. This is explained in glorious detail here.
What is this space reserved on the top of the stack for? First, space is created for any local variables. In this case, FunctionWith6Params() has two. However, those two local variables only account for 0x10 bytes. What’s the deal with the rest of the space created on the top of the stack?
On the x64 platform, when code prepares the stack for calling another function, it does not use push instructions to put the parameters on the stack as is commonly the case in x86 code. Instead, the stack pointer typically remains fixed for a particular function. The compiler looks at all of the functions the code in the current function calls, it finds the one with the maximum number of parameters, and then creates enough space on the stack to accommodate those parameters. In this example, FunctionWith6Params() calls printf() passing it 8 parameters. Since that is the called function with the maximum number of parameters, the compiler creates 8 slots on the stack. The top four slots on the stack will then be the home space used by any functions FunctionWith6Params() calls.