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

问题

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.

回答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

来源：https://stackoverflow.com/questions/42771550/why-does-first-parameter-in-x86-assembly-starts-from-offset-8