Author Topic: I have some questions about returns.  (Read 5271 times)

Offline ben321

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I have some questions about returns.
« on: March 23, 2015, 02:26:56 AM »
To get an idea of how programs look in assembly language, I've written a few simple programs in VB6 and then disassembled them with OllyDbg.

One of the things I notice is that at the end of a sub routine there are typicially two types of returns, both with the mnemonic RET, but with different op-codes. The two op-codes I've seen are C2 and C3. C3 takes no parameter. C2 takes a two byte "word" parameter. What I can't figure out is what you are supposed to pass to that parameter. Is it some relative memory location? Is it supposed to be a value that gets put on some register, or somewhere in the stack, for later use by the program? And in what situation would I most often use a C2 return, versus when should I use a C3 return?

I also notice that just prior to the RET, there is usually a LEAVE command. This has the op-code C9. According to this documentation http://ref.x86asm.net/coder32.html a C9 is a "High Level Procedure Exit". What exactly does this mean? In what situations is a LEAVE needed? In cases where LEAVE is not used, the RET is always preceded by a POP EBP. But I can't figure out, in what situations should a POP EBP be used instead of a LEAVE? I also notice that at the start of any subroutine, without fail, and regardless of what commands are used near the end of the routine, the first command is always PUSH EBP. Why is that used?

Please help me out here. Thanks.

Offline Frank Kotler

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Re: I have some questions about returns.
« Reply #1 on: March 23, 2015, 08:11:21 AM »
It might be easier to understand the two kinds of "ret" by looking at how they're called, first. As you probably know, there are different "calling conventions" - agreed upon rules for interfaces with library (and other) code. In "cdecl", used in the interface with the C library and other places, it is the caller's responsibility to "clean up the stack" or "balance the stack" by "removing" the parameters that were pushed there.
Code: [Select]
    push 42
    push fmt_string ; pretend it's db "%d", 0
    call _printf
    add esp, 8 ; "clean up stack" - 2 parameters, 4 bytes each
We don't need to see the _printf code to know it ends with a plain "ret". As you can see, nothing is actually "removed" from the stack - the parameters are still there - we've just moved esp (the "stack pointer") above them. The next time the stack is used, they'll be overwritten.

In the "stdcall" convention, used by Windows APIs (and other places), it is the callee's responsibility to "clean up the stack".
Code: [Select]
    push 0 ; msgbox style
    push caption
    push message
    push 0 ; hWin
    call MesssageBoxA
    ; we don't need to "clean up stack"
    ; the parameters have been "removed" for us
We don't get to see the code for MessageBoxA, but we know it ends in "ret 16". The operand is the number of bytes to "remove" from the stack - not the number of parameters but the number of bytes (same as what we added in the cdecl example). As you can imagine, this doesn't work with functions with a variable number of parameters - like _printf.

Which one you should use to end your subroutines depends on which convention you used to call 'em.

Besides who gets to "clean up stack", the calling conventions specify which registers need to be preserved (ebp, ebx, esi, and edi) and which ones can be altered (ecx and edx - eax is the return value). Back in the good old days of 16-bit code, addressing modes were quite limited - [bx] and [bp] were the only two "base registers" that could be used - [sp] was not a valid effective address. bp was also special in that it defaulted to using ss as a segment register to form the complete address. So if we wanted to access parameters passed on the stack, we didn't have much choice but to  create a "stack frame" using bp.
Code: [Select]
myfunc:
; create a stack frame
    push bp ; save caller's register
    mov bp, sp ; set the "frame pointer" to current sp
    sub sp, 16 ; make room for "local" variables

    mov ax, [bp + 4] ; get last parameter pushed
    mov [bp - 6], ax ; store it in a local variable (for no reason)

; destroy the stack frame
    mov sp, bp ; "free" local variables and restore sp
    pop bp ; restore caller's reg
    ret

In 32-bit code, any register can be a "base" register, and all the segment registers point to the same memory, so ebp isn't as "special" as it was, but it it still used as the "frame pointer". We don't even need to set up a stack frame at all.
Code: [Select]
myfunc:
    mov eax, [esp + 4] ; get last parameter pushed
...
But it is still convenient to use a stack frame. It allows a debugger to "back trace" through a call chain, and it makes local variables easier to use (we could use locals indexed off esp, but it would soon drive us crazy keeping track of 'em). Compilers usually do it unless asked not to.
Code: [Select]
myfunc:
    push ebp ; our caller was using this
    mov ebp, esp ; set our frame pointer
    sub esp, 4 ; room for one local variable

    mov eax, [ebp + 8] ; last-pushed parameter

    push esi ; convention expects us to preserve this
    mov esi, [ebp + 12] ; next parameter
    add eax, esi
    mov [ebp - 4], eax ; store in local variable
    pop esi ; restore caller's esi, per convention

    xor eax, eax ; pretend we did something useful with eax
    mov eax, [ebp - 4] ; get eax back from local
; note that parameters and locals are at the
; same offset from ebp, whether we've put
; something on the stack or not.

; destroy stack frame
    leave
; does the same thing as:
    ; mov esp, ebp ; restore esp to where it was
    ; pop ebp ; restore caller's ebp
    ret
As shown, "leave" is a shorter way of doing "mov esp, ebp" and "pop ebp". You might see either. If you haven't got local variables (and didn't butcher the stack), you don't need the "mov esp, ebp" - esp was there anyway.

There's a matching "enter" instruction:
Code: [Select]
myfunc:
    enter 4, 0
; does the same as:
    ; push ebp
    ; mov ebp, esp
    ; sub esp, 4
...
You won't often see it, because while it's smaller, it's slower. What's that second operand? Don't ask! It's a "lex level" - allows a function to access its caller's variables (and its caller's caller's variables, etc.). I think Pascal uses it. I never have. Just make it zero.

Best,
Frank