英文:
cpp: permit base member access only via derived by consumer
问题
我有一个应用程序，其中Base是一个工厂类，用于根据模板对创建的对象进行不同的创建。然而，只有非常特定（可计数的无限）数量的模板对是有意义的。对于每个有意义的模板对，我都有一个（有时是有模板的）派生工厂类。由于需要选择可行的正确模板对，使用Base Factory容易出错。
我的问题是：如何防止消费者直接使用Base Factory？
template<typename T1, typename T2>
class Base{
Base()=delete;
public:
// 许多公共静态例程
T1 static foo(){ return 1; }
T2 static bar(){ return 2; }
};
template<typename T1>
class Derived: public Base<T1,float>{
// 希望很少的代码，或者根本没有，就像现在一样。
};
int main(){ // main是消费者
Base<int,float>::foo(); // 阻止
Derived<int> ::foo(); // 允许
}
我希望编译器在main中拒绝/抛出/报错带有注释“prevent”的行。我希望Base<..>::foo和Base<..>::bar不直接可访问消费者，除非通过Derived<..>的访问。我曾希望“虚拟”概念可以用于此目的；但它不自然地从非静态到静态情况转移。
英文:
I have an application where Base is a factory class for objects that shall be created differently depending on a template pair. However, only very particular (countable infinite) number of template pairs make sense. For each of sensible pair, I have a (sometimes) templated Derived Factory class. The use of Base Factory is error prone due to the need for feasible correct choice of the template pair.
My question is: How can I prevent a consumer from using the Base Factory directly?
template<typename T1, typename T2>
class Base{
Base()=delete;
public:
// many public static routines
T1 static foo(){ return 1; }
T2 static bar(){ return 2; }
};
template<typename T1>
class Derived: public Base<T1,float>{
// hopefully very little code, or none at all, as of now.
};
int main(){ // main is the consumer
Base<int,float>::foo(); // prevent
Derived<int> ::foo(); // permit
}
I want the compiler to reject/thow/err the line with comment "prevent" in main. I want Base<..>::foo and Base<..>::bar not directly accessible by a consumer unless via access through Derived<..>. I had hoped that the virtual concept could be used for that purpose; however it does not transfer naturally from the non-static to the static case.
答案1
得分: 1
你可以将foo方法设为protected，然后通过Derived类使用public的using语句来暴露它：
template<typename T1, typename T2>
class Base {
Base() = delete;
protected:
T1 static foo() { return 1; }
T2 static bar() { return 2; }
};
template<typename T1>
class Derived : public Base<T1, float> {
public:
using Base<T1, float>::foo;
};
int main() { // main is the consumer
Base<int, float>::foo(); // 阻止
Derived<int>::foo(); // 允许
}
英文:
You can make the foo method protected, and expose it via the Derived class with a public using statement:
template<typename T1, typename T2>
class Base {
Base() = delete;
protected:
T1 static foo() { return 1; }
T2 static bar() { return 2; }
};
template<typename T1>
class Derived : public Base<T1, float> {
public:
using Base<T1, float>::foo;
};
int main() { // main is the consumer
Base<int, float>::foo(); // prevent
Derived<int> ::foo(); // permit
}
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