Using this example coming from wikipedia, in which DrawSquare() calls DrawLine(),
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(Note that this diagram has high addresses at the bottom and low addresses at the top.)
Could anyone explain me what ebp and esp are in this context?
From what I see, I'd say the stack pointer points always to the top of the stack, and the base pointer to the beginning of the the current function? Or what?
edit: I mean this in the context of windows programs
edit2: And how does eip work, too?
edit3: I have the following code from MSVC++:
var_C= dword ptr -0Ch
var_8= dword ptr -8
var_4= dword ptr -4
hInstance= dword ptr 8
hPrevInstance= dword ptr 0Ch
lpCmdLine= dword ptr 10h
nShowCmd= dword ptr 14h
All of them seem to be dwords, thus taking 4 bytes each. So I can see there is a gap from hInstance to var_4 of 4 bytes. What are they? I assume it is the return address, as can be seen in wikipedia's picture?
(editor's note: removed a long quote from Michael's answer, which doesn't belong in the question, but a followup question was edited in):
This is because the flow of the function call is:
* Push parameters (hInstance, etc.)
* Call function, which pushes return address
* Push ebp
* Allocate space for locals
My question (last, i hope!) now is, what is exactly what happens from the instant I pop the arguments of the function i want to call up to the end of the prolog? I want to know how the ebp, esp evolve during those moments(I already understood how the prolog works, I just want to know what is happening after i pushed the arguments on the stack and before the prolog).