如何将字符串作为参数传递给“调用”函数内联汇编指令？

发布于 2024-08-22 06:01:37 字数 281 浏览 5 评论 0原文

本质上，我希望能够做这样的事情：

//assume myFunction is defined and takes one argument that is an int
char * functionName = "myFunction";
int arg = 5;

__asm{
    push a
    call functionName
}

基本上我想调用一个名称存储在字符串中的函数。执行此操作的正确语法是什么？

编辑：我们正在谈论 x86 汇编

原文

Essentially, I'd like to be able to do something like this:

//assume myFunction is defined and takes one argument that is an int
char * functionName = "myFunction";
int arg = 5;

__asm{
    push a
    call functionName
}

Basically I want to call a function whose name is stored in a string. What would be the proper syntax for doing this?

Edit:
We are talking about x86 assembly

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清风夜微凉 2024-08-29 06:01:37

你不能，至少不能直接。

call 将地址作为参数。即使您编写“call functionName”，链接器也会将 functionname 替换为函数的实际地址。您需要首先将该字符串映射到其地址。一般来说，C 和 C++ 不支持任何类型的有关函数名称映射的运行时元数据，从而实现这一点。如果函数是从 DLL 导出的，则可以使用 GetProcAddress 查找其地址。

如果函数列表是静态的，您可以自己提前创建映射。

类似的东西：

std:map<string, PVOID> functionMappings;
functionMappings["MyFunction"] = MyFunction;


// Later


PVOID function = functionMappings["MyFunction"];

__asm
{
    push a;
    call [function]
}

一些注释：

我相信标准说函数指针可能比 PVOID 更大。这应该适用于 Windows x86/x64 平台。

您没有说明您使用的调用约定 - 这段代码假定 stdcall。

这是一件非常非常奇怪的事情——你想解决什么问题？

You can't, at least not directly.

call takes an address as a parameter. Even though you write "call functionName", the linker replaces functionname with the actual address of the function. You'd need to first map that string to its address. In general, C and C++ don't support any sort of runtime metadata about function name mappings that would allow for this. If the function is exported from a DLL, you can use GetProcAddress to find its address.

If the list of functions is static, you can create the mapping ahead of time yourself.

Something like:

std:map<string, PVOID> functionMappings;
functionMappings["MyFunction"] = MyFunction;


// Later


PVOID function = functionMappings["MyFunction"];

__asm
{
    push a;
    call [function]
}

Some notes:

I believe the standard says that a function pointer may be larger than a PVOID. This should work on Windows x86/x64 platforms.

You didn't say what calling convention you were using - this code presumes stdcall.

This is a very, very odd thing to want to accomplish - what problem are you trying to solve?

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