下面的汇编代码究竟做了什么？
What does the following assembly code really do?
我不确定下面应该做什么，但这是我目前得到的结果。
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)
两个代码片段通过EAX做间接调用。如果两个片段中的 EBX 值相同，则将调用相同的代码，因为两个片段将 ebx 添加到 eax。
片段 A 的更正描述是：
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]
nop 指令后跟 push ebx 后跟 nop 后跟 pop ebx 再次不改变任何东西（除了保留 [=18 的前值=] 在堆栈中的可用位置 space)。所以在功能上（尽管减少了消耗的 CPU 周期数和代码 space），这相当于：
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]
片段 B 是：
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
除了消耗 CPU 个周期和代码 space 之外，连续两次交换两个寄存器没有任何实际效果。所以片段 B 在功能上归结为：
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]
功能与片段 A 相同。
我不确定下面应该做什么，但这是我目前得到的结果。
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)
两个代码片段通过EAX做间接调用。如果两个片段中的 EBX 值相同，则将调用相同的代码，因为两个片段将 ebx 添加到 eax。
片段 A 的更正描述是：
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]
nop 指令后跟 push ebx 后跟 nop 后跟 pop ebx 再次不改变任何东西（除了保留 [=18 的前值=] 在堆栈中的可用位置 space)。所以在功能上（尽管减少了消耗的 CPU 周期数和代码 space），这相当于：
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]
片段 B 是：
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
除了消耗 CPU 个周期和代码 space 之外，连续两次交换两个寄存器没有任何实际效果。所以片段 B 在功能上归结为：
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]
功能与片段 A 相同。