assembly registers esp and ebp

Question

I am currently learning assembly for Intel processors. Since the stack 'grows down', why do we have to add in order to access a specific element

[ebp + 8] ;; This will access the first param

I konw we have to skip the old ebp value and the return address and that's why we use 8 ( because each are 4 bytes long ). This is kinda strange.

Also, if ebp is a backup copy for esp, what is ebp's value in the main function Ex:

_start:

;; what's ebp value here ?

push eax
push ebx
call someFunction ;; this will create a stack frame

+-----------+
|   param1  |
+-----------+
|   param0  | <--- this is [ebp + 8] ( assuming mov ebp, esp )
+-----------+
|ret_address|
+-----------+
|    ebp    |
+-----------+ <--- esp

And also when we allocate memory for locals, we have to substract from ebp ... Please give a non-ambiguous answer. Thank you!

A main() function is no different from a regular function, it has a prologue too. You didn't show one in your assembly snippet, not sure why. Space for locals is created by adjusting esp, not ebp. — Hans Passant
There are same questions: stackoverflow.com/questions/1395591/… , and other linked to it — Abyx

ninjalj ninjalj · Accepted Answer · 2011-12-02T20:55:17

The initial value for %ebp is usually 0. This is so debuggers know when to end following the link chain in a backtrace.

You should think of %ebp as a reference point. For convenience, it is placed between the function arguments and local variables. That way, you access arguments with a positive offset, and variables with a negative offset, so it's easy to tell whether you are accessing a variable or an argument.

assembly registers esp and ebp

2 Answers