模板化已模板化的 Base class 的 Derived class

Question

我收到编译器错误“使用未声明的标识符‘_storage’”

#include <iostream>

struct Storage{
    int t;
};

struct DerivedStorage : Storage{
    int s;
};

struct DerivedStorage2 : DerivedStorage{
    int r;
};

template<typename DS = Storage>
class Base{
public:
    DS* _storage = nullptr;
    Base(){
        _storage = new DS();
    }
    void m(){
        _storage->t++;
    }
};

template<typename DS = DerivedStorage>
class Derived : public Base<DS>{
public:
    void m2(){
        _storage->s++; //error here
    }
};

template<typename DS = DerivedStorage2>
class Derived2 : public Derived<DS>{
public:

    void m3(){
        _storage->r++; //error here
    }
};

int main(int argc, const char * argv[]) {
    Derived2<DerivedStorage2> t;
    for(int i = 0;i<3;i++){
        t.m3();
    }
    
    return 0;
}

知道问题出在哪里吗？

Answer 1

因为Derived本身没有名为_storage的成员变量，所以使用this->_storage来访问Base的_storage。

void m2(){
  this->_storage->s++;
}

Demo

模板化已模板化的 Base class 的 Derived class

Templating a Derived class of a Base class that is already templated

c++

c++11