如何传递成员函数指针到std :: function [英] how to pass member function pointer to std::function

问题描述

如何通过函数传递成员函数指针到 std :: function 。我将通过比较解释（实时测试）：

How can I pass member function pointer to std::function through a function. I am going to explain it by comparison (Live Test):

template<class R, class... FArgs, class... Args>
    void easy_bind(std::function<R(FArgs...)> f, Args&&... args){ 
}

int main() {
    fx::easy_bind(&Class::test_function, new Class);
    return 0;
}

我收到错误讯息：

no matching function for call to ‘easy_bind(void (Class::*)(int, float, std::string), Class*)’

我不明白为什么函数指针不能传递给 std :: function 当它通过一个函数参数。如何传递该函数？我愿意将 easy_bind 函数参数从 std :: function 更改为函数指针，但我真的不知道如何。

I just don't understand why a function pointer cannot be passed to std::function when its being passed through a function parameter. How can I pass that function? I am willing to change the easy_bind function parameter from std::function into a function pointer but I really don't know how.

编辑：问题简化。

编辑：感谢@remyabel，我需要： http://ideone.com/FtkVBg

Thanks to @remyabel, I was able to get what I needed: http://ideone.com/FtkVBg

template <typename R, typename T, typename... FArgs, typename... Args>
auto easy_bind(R (T::*mf)(FArgs...), Args&&... args)
-> decltype(fx::easy_bind(std::function<R(T*,FArgs...)>(mf), args...)) {
    return fx::easy_bind(std::function<R(T*,FArgs...)>(mf), args...);
}

推荐答案

缩小到此：

template<class R, class... FArgs>
void test(std::function<R(FArgs...)> f)
{
}

int main() {
  test(&SomeStruct::function);
}

错误消息非常类似， > easy_bind stuff：

The error message is pretty similar without the rest of the easy_bind stuff:

main.cpp: In function 'int main()':
main.cpp:63:31: error: no matching function for call to 
'test(void (SomeStruct::*)(int, float, std::string))'
     test(&SomeStruct::function);
main.cpp:63:31: note: candidate is:
main.cpp:49:10: note: template<class R, class ... FArgs> 
void test(std::function<_Res(_ArgTypes ...)>)
     void test(std::function<R(FArgs...)> f)
          ^
main.cpp:49:10: note:   template argument deduction/substitution failed:
main.cpp:63:31: note:   'void (SomeStruct::*)(int, float, std::string) 
{aka void (SomeStruct::*)(int, float, std::basic_string<char>)}' 
is not derived from 'std::function<_Res(_ArgTypes ...)>'
     test(&SomeStruct::function);

本质上，它不能神奇地创建一个 std :: function 给你。您需要类似 Functor 别名。

Essentially, it can't magically create an std::function for you. You need something like your Functor alias.

对于通用成员函数指针作为模板参数提供的答案，您可以执行以下操作：

So thanks to the answer provided in generic member function pointer as a template parameter, here's what you can do:

//Test Case:
struct SomeStruct {
public:
 int function(int x, float y, std::string str) {
   std::cout << x << " " << y << " " << str << std::endl;
   return 42;
 }
};

template <typename Ret, typename Struct, typename ...Args>
std::function<Ret (Struct*,Args...)> proxycall(Ret (Struct::*mf)(Args...))
{
    return std::function<Ret (Struct*,Args...)>(mf);
}

int main() {
    auto func3 = fx::easy_bind(proxycall(&SomeStruct::function), new SomeStruct);
    int ret = func3(5, 2.5, "Test3");
    std::cout << ret << "\n";

    return 0;
}

现在可以自动工作。

实例

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