c++11 - C++ CRTP : How to make only one (some) function of the base class as a friend to to the derived class?

Question

我只想做 Base<DerivedImpl>::fct1()可以访问类(class) DerivedImpl成员。

基地看起来像:

template < typename Derived>
class Base<Derived>{

protected:
void fct1(){
static_cast<Derived*>(this)->topfunc();
}

void fct2(){
...
}

};

派生类如下所示:

class DerivedImpl: public Base<DerivedImpl>{

void callbase(){fct1();}
void topfunc(){std::cout << "topfunc" <<std::endl;}

friend Base<DerivedImpl>; //this works
//friend void Base<DerivedImpl>::fct1(); //does not work!!
};

主要 C++:

int main(){
DerivedImpl obj;
obj.callbase();
}

Answer 1

免责声明:这回答了所问的问题，但在我看来，不同的设计方法可能更可取，所以我不建议您在生产中这样做，除非您绝对必须这样做。

您可以通过滥用允许派生类访问 protected 的事实来解决此问题。 静态 其父类的成员:

#include <iostream>

template<typename Derived>
class Base {
protected:
  static void fct1(Base* self){
    static_cast<Derived*>(self)->topfunc();
  }

  void fct2() {}
};

class DerivedImpl: public Base<DerivedImpl> {

  void callbase() { fct1(this); }
  void topfunc() { std::cout << "topfunc" << std::endl; }

  friend void Base<DerivedImpl>::fct1(Base*); // works fine now!
};