以下汇编代码到底做了什么？

Question

我不确定以下应该做什么，但这是我到目前为止所做的。

mov eax, 5       (move 5 into register eax)   
add eax, ebx     (add 5 from eax to 0 from ebx and store in eax)
nop              (no operation)
nop              (no operation)
push ebx         (push 0 onto the hardware stack)
nop              (no operation)
pop ebx          (pop the 0 from off the stack and store in ebx)
call [eax]       (get the 5 from eax)

Answer 1

片段A的更正说明是：

var self = this;

mov eax, 5 ; move 5 into register eax add eax, ebx ; add contents of ebx to eax, changing eax nop ; no operation nop ; no operation push ebx ; push contents of ebx onto the stack nop ; no operation pop ebx ; pop top of the stack into ebx call [eax] ; call the subroutine pointed to at location [eax] 后跟nop后跟push ebx后跟nop的{​​{1}}说明不会改变任何内容（除了保留pop ebx的前值之外在堆栈空间中的可用位置）。因此在功能上（尽管消耗的CPU周期数和代码空间减少），这相当于：

ebx

片段B是：

mov eax, 5       ; move 5 into register eax
add eax, ebx     ; add contents of ebx to eax, changing eax
call [eax]       ; call the subroutine pointed to at location [eax]

连续两次交换两个寄存器没有净效应，除了消耗CPU周期和代码空间。因此片段B在功能上归结为：

mov eax, 5       ; move 5 into register eax
push ecx         ; push contents of ecx onto the stack
pop ecx          ; pop top of the stack into ecx
add eax, ebx     ; add contents of ebx to eax, changing eax
swap eax, ebx    ; swap the contents of eax and ebx
swap ebx, eax    ; swap the contents of eax and ebx
call [eax]       ; call the subroutine pointed to at location [eax]
nop              ; no operation

功能与片段A相同。

Answer 2

两个代码片段通过EAX进行间接调用。如果两个片段中的EBX值相同，则会调用相同的代码，因为两个片段都将ebx添加到eax。