Why does first parameter in x86 assembly starts from offset 8?

Question

I'm trying to develop understanding of Assembly language. I understand that when function creates stack frame, it pushes current EBP, than copies stack pointer value to the EBP. First (and only) function parameter is accessed by EBP + 8. But why 8? Next value after pushed EBP is logically offset 4. I read many webpages, but it seems I don't understand this part.

Faunia

Answer 1

The "missing" DWORD is the return address. The call stack looks like:

ebp     : saved ebp
ebp + 4 : return address
ebp + 8 : pushed parameter

And then if the function uses local variables, since stack space is (typically) reserved for those after the stack frame, they are referenced as ebp - xx:

ebp - 8 : second local
ebp - 4 : first local
ebp     : saved ebp
ebp + 4 : return address
ebp + 8 : pushed parameter

Yerkovich

Thank you. I focused on the wrong part (read about calling conventions, but forgot basic thing) – Faunia

Is worth mentioning that the stack in this question grows downward that's why you substrate local variables from ebp which were pushed after arguments and return address. – Sebi