为什么在转到下一个堆栈帧时存储旧的帧指针

Question

I study the x86 assembly recently, and not understand why we have to do push ebp when entering a new function.

From the survey, I can see the need for ebp is because that it can easily access the arguments of function call and the local variables in this new callee.

But I don't know why we have to store the old frame pointer when go to the new stack frame?

Is that because doing so will make stack trace easier for debugging?

The following is my test code:

foobar:
.LFB0:
    .cfi_startproc
    push ebp                 #Why do this here??
    ....
    ....
    mov ebp, esp

Thanks in advance

Answer 1

You don't even have to use a frame pointer, the calling convention doesn't mandate that and optimized code frequently doesn't. What the calling convention does prescribe though is that some registers are callee-saved, that is they must be preserved for the caller. This normally includes ebp . This requirement may be fulfilled by pushing/popping it.

As a side effect, if you do use frame pointers and you do know the structure of the caller, you may use it to access its frame. This is however not typically used.