c++ - 模板参数接受指向类成员函数的指针

Question

#include <utility>

class Base {
public:
    virtual ~Base() {}
    virtual void base() {}
};

class Derived : public Base {
public:
    virtual void derived() {}
};

template<typename... Params>
using MemberFuncPtr = void(Derived::*)(Params...);

template<typename... Params, typename... Args>
void wrapper(MemberFuncPtr<Params...> ptr, Args&&... args)
{
    Derived* d = new Derived();

    (d->*ptr)(std::forward<Args>(args)...);

    delete d;
}

int main()
{
    wrapper(&Derived::derived);
    wrapper(&Derived::base);
    return 0;
}

尝试运行此代码 (GCC 7.0) 时出现以下错误:

prog.cc: In function 'int main()':
prog.cc:33:27: error: no matching function for call to 'wrapper(void (Base::*)())'
     wrapper(&Derived::base);
                           ^
prog.cc:18:6: note: candidate: template<class ... Params, class ... Args> void wrapper(MemberFuncPtr<Params ...>, Args&& ...)
 void wrapper(MemberFuncPtr<Params...> ptr, Args&&... args)
      ^~~~~~~
prog.cc:18:6: note:   template argument deduction/substitution failed:
prog.cc:33:27: note:   mismatched types 'Derived' and 'Base'
     wrapper(&Derived::base);
                           ^

我真的不明白为什么来自基类的方法是个问题？它也是派生类的一种方法。我做了简单的测试，我将 Derived::base 分配给 Derived::*ptr 类型并且有效。

Answer 1

I don't really understand why method from the base class is an issue?

这只是类型的问题。 &Derived::derived的类型是void (Derived::*)()，但是&Derived::base的类型是void (Base::*)()。这与 void (Derived::*)(Args...) 不匹配，因此推导失败。模板推导不允许转换，即使在这种情况下存在有效的转换也是如此。

I've done simple test where I assigned Derived::base to Derived::*ptr type and that worked.

这个:

MemberFuncPtr<> d = &Derived::base;
wrapper(d);

之所以有效，是因为现在 d 是正确的类型(它现在 确实 匹配 void (Derived::*)(Args...))，并且第一行很好，因为将指向基类成员的指针转换为指向派生类成员的指针是有效的。