在IA-32和x86-64 POP指令中组装

Question

在pop ebp（返回）代码指令之前的末尾，ret在IA-32和x86-64机器中意味着什么？所以我确实有新旧ebp，并且通过调用函数将新的ebp推入堆栈。那么这个ebp是pop'ed？ pop指令如何更改ebp的值？

Answer 1

PUSH EAX

基本上意味着：

SUB ESP,4
MOV [ESP],EAX

并且

POP EAX

意思是：

MOV EAX,[ESP]
ADD ESP,4

---

当你谈论新旧EBP时，我猜你指的是功能序言和结语？

PUSH EBP     ; Store caller's EBP on the stack
MOV EBP,ESP  ; Set EBP to the current stack pointer

; Here we can do things like:
MOV EAX,[EBP+8]
PUSH EAX
MOV EBX,[EBP+12]
POP EAX
; ..to access the stack. Since we've got a fixed reference point for
; the stack in EBP we don't have to worry about the stack pointer
; changing.

; For example, we could do this instead to access the same stack
; elements as above:
MOV EAX,[ESP+8]
PUSH EAX
MOV EBX,[ESP+16]
POP EAX
; But notice that we had to change the second offset since the push
; instruction changed the stack pointer. It's obviously easier to deal
; with a base address that doesn't change every time we execute
; certain instructions.

MOV ESP,EBP  ; Restore the stack pointer
POP EBP      ; Restore the caller's EBP before returning