程序调用在汇编器中如何工作? [英] how do procedure calls work in assembler?

问题描述

我刚刚开始修改ASM，但不确定我对过程调用的理解是否正确.

I just started tinkering with ASM and I'm not sure if my understanding of procedure calls is correct.

说在代码的某个地方有一个过程调用

say at some point in the code there is a procedure call

call dword ptr[123]

，该过程仅包含一个命令，请重新输入:

and the procedure consists of only one command, ret:

ret 0004

此过程调用将产生什么结果，返回值将存储在何处?我读到某处将在AX中存储2字节的返回值，但是当我用以下方式替换过程调用时:

what would be the effect of this procedure call, and where would the return value be stored? I read somewhere that a return value of 2 bytes would be stored in AX, but when I replace the procedure call by

mov AX, 0004

(连同必要的NOP一起)，程序崩溃.

(together with the necessary NOPs) the program crashes.

推荐答案

在x86汇编程序中，ret指令的参数含义:

in x86 assembler the parameter to the ret instruction means:

RET immediate

返回调用过程，并从堆栈中弹出立即个字节.

(引自英特尔®64和IA-32体系结构软件开发人员手册 第2B卷)

因此，当您键入:

ret 0004

您要告诉CPU在call之后立即返回指令，并从堆栈中弹出4个字节.如果您在调用之前将推入 4个字节到堆栈中，那就太好了.

You're telling the CPU to return to the instruction immediately after the call, and to pop 4 bytes off the stack. This is great if you pushed 4 bytes onto the stack before the call.

push eax
call dword ptr[123]

请注意，这与返回值无关.实际上，Assembly中的过程无法指定值是 return 值.这一切都是按照惯例完成的.据我所知，大多数编译器都将使用EAX来保存返回值，但这是正确的，仅是因为 calling 函数将在那里期望结果.

Note that this has nothing to do with the return value. In fact, a procedure in Assembly has no way of specifying that a value is a return value. This is all done by convention. Most compilers of which I am aware will use EAX to hold the return value, but this is true only because the calling function will expect the result there.

因此您的呼叫代码为:

call dword ptr [123]
mov dword ptr [result], eax

，返回值4的函数将是:

and your function that returns the value 4 would be:

mov eax, 4
ret

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