如何获得“简单”函数指针从成员函数 [英] How to get a "simple" function pointer from a member function

问题描述

我有一个函数指针的问题，我在网上找到的没有帮助我解决这个问题。

I'm having a problem with function pointers and nothing I found on the net helped me to solve this problem.

我有一个来自C API的函数取一个void函数的指针：

I have a function from a C API which take a pointer of a void function :

extern int APIFunction（int，void（* func）（int））;

我有一个类，当我调用API函数时，我想要的函数。

I have a class with the function I would like to put when I call the API function.

class MyClass
{
   public:
      void myFunction(int status, otherAPi arguments...);
};

然后，我创建了一个指向我的成员函数的指针，并创建了我的类的一个新实例

Then, I created a pointer to my member function and created a new instance of my class

typedef  void (MyClass::*MyClassFunctionPointer)(int stat, otherAPi arguments...);
MyClassFunctionPointer fctPointer= &MyClass::myFunction; 
LicenseSecurity instance;  

当我尝试使用我创建的函数指针调用APi函数时出现错误： p>

I get an error when I try to call my APi function with the function pointer I created:

int stat = APIFunction(5, fctPointer ); // -> error 1
int stat = APIFunction(5, instance.*fctPointer ); // -> error 2

我在第一种和第二种情况下分别有错误：

I got errors respectively in the first and second case:

E2034 Impossible to convert 'void (MyClass::*)(int, otherAPITypes...)' into 'void (*) (int, otherAPITypes...)'
E2342 Bad type correspondence in the parameter 'func' ('void (*)(int, otherAPITypes...)' desired, 'void(int, otherAPITypes...)' obtained)

我无法访问API函数，因此我无法修改它。总结问题：如何获得一个简单C函数指针，以便从我的类的成员函数中插入一个函数的参数？

I don't have access to the API function so I can't modify it. To summary the problem: how How to get a "simple" C function pointer to put in argument of a function from a member function of my class?

感谢

推荐答案

不幸的是，你不能。很抱歉。

Unfortunately, you can't. Sorry.

理想情况下，您的API会接受类似 std :: function 或成员函数。但是如果你不能修改API，那么你别无选择，只能提供一个免费的函数。

Ideally, your API would accept something like std::function that would allow you to wrap free functions or member functions. But if you can't modify the API, then you have no choice but to provide a free function.

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