如何创建类似std :: function的包装器？

Question

Attempted this:

template <class R, class... Ts>
class MyFunction
{
public:
    using func_type = R(*)(Ts...);

    MyFunction(func_type f)
      : m_func(f) 
    {
    }

    R operator()(Ts ... args) 
    {
        return m_func(args...);
    }

private:
    func_type m_func;
};

int Testfn(int a)
{
    std::cout << "value is " << a;
    return 42;
}

void Testing()
{
    MyFunction<int(int)> func(Testfn);
    std::cout << "Ret is " << func(1) << std::endl;
}

But fails with:

 error C2064: term does not evaluate to a function taking 1
 C2091: function returns function   
 C2091: function returns
 C2664: 'MyFunction<int (int),>::MyFunction(const MyFunction<int
 (int),> &)' : cannot convert argument 1 from 'int (__cdecl *)(int)' to
 'int (__cdecl *(__cdecl
 *)(void))'

Compiler is MSVC2013.

Answer 1

It should be like this:

template <typename T>
class MyFunction;

template<typename R, class... Ts>
class MyFunction<R(Ts...)>
{
public:
    using func_type = R(*)(Ts...);

    MyFunction(func_type f)
      : m_func(f) 
    {
    }

    R operator()(Ts ... args) 
    {
        return m_func(args...);
    }

private:
    func_type m_func;
};

MyFunction should be specialized for function signature type. Note: std::function is really more complicated.