C++ 将派生 class 的成员函数指针作为参数传递时选择了错误的模板专业化

Question

正如标题所说，我在将从基础 class 继承的成员函数指针传递给专门的模板函数时遇到了一些麻烦。

如果以继承方法作为参数调用模板函数时不指定类型，编译器会select错误的(Base)一个：

struct Base {
  void method() {
  }
};

struct Derived: public Base {
  void other_method() {
  }
};

template <typename T> void print_class_name(void (T::*func)()) {
  std::cout << "Unknown" << std::endl;
}

template <> void print_class_name<Base>(void (Base::*func)()) {
  std::cout << "Base" << std::endl;
}

template <> void print_class_name<Derived>(void (Derived::*func)()) {
  std::cout << "Derived" << std::endl;
}

int main(int argc, char** argv) {
  print_class_name(&Base::method);    // output: "Base"
  print_class_name(&Derived::method); // output: "Base"???
  print_class_name(&Derived::other_method); // output: "Derived"
  print_class_name<Derived>(&Derived::method); // output: "Derived"
  return 0;
}

在这种情况下是否需要调用指定派生的专用模板函数 class 还是我做错了什么？

Answer 1

method 从 Derived 可见，但实际上是 Base 的成员，因此 &Derived::method 的类型是 void (Base::*)().

这些语句都打印相同的值：

std::cout << typeid(&Base::method).name() << "\n";
std::cout << typeid(&Derived::method).name() << "\n";

C++ 将派生 class 的成员函数指针作为参数传递时选择了错误的模板专业化

C++ Wrong template specialization selected when passing member function pointer of derived class as a parameter

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derived-class

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