c++ - 混合 typedef 和 CRTP？

Question

考虑以下示例:

#include <iostream>
#include <iostream>
#include <type_traits>

template<typename Type, template<typename> class Crtp>
class Base
{
    public:
        typedef int value;

        // f1: OK
        // Expected result: casts 4.2 to Base<Type, Crtp>::value
        value f1() {return 4.2;}

        // f2: NOT OK
        // Expected result: casts 4.2 to Crtp<Type>::value
        // But f2 does not compile: no type named 'value' 
        // in 'class Derived<double>'
        typename Crtp<Type>::value f2() {return 4.2;} 
};

template<typename Type>
class Derived : public Base<Type, Derived>
{
    public:
        typedef Type value;
};

int main()
{
    Derived<double> a;
    std::cout<<a.f1()<<std::endl;
    std::cout<<a.f2()<<std::endl;
    return 0;
}

如何解决这个问题(Derived typedef unknown from the Base class)？

编辑:我发现了一个非常简单的技巧。有人可以向我解释为什么以下版本有效而​​以前的版本无效吗？这个技巧是否适用于标准 C++11，还是因为编译器的工作方式(此处为 g++ 4.7.1)而起作用？

#include <iostream>
#include <iostream>
#include <type_traits>

template<typename Type, template<typename> class Crtp>
class Base
{
    public:
        typedef int value;
        value f1() {return 4.2;}
        template<typename T = Crtp<Type>> typename T::value f2() {return 4.2;}
};

template<typename Type>
class Derived : public Base<Type, Derived>
{
    public:
        typedef Type value;
};

int main()
{
    Derived<double> a;
    std::cout<<a.f1()<<std::endl;
    std::cout<<a.f2()<<std::endl;
    return 0;
}

Answer 1

您必须使用在定义Base 之前声明的包装类(这里是value_getter)。然后，您可以在定义 Derived 之前对其进行特殊化:

template<typename T>
struct value_getter;

template<typename Type, template<typename> class Crtp>
class Base
{
    public:
        typedef int value;

        value f1() {return 4.2;}

        // in 'class Derived<double>'
        typename value_getter<Crtp<Type> >::value f2() {return 4.2;} 
};

template<typename Type>
class Derived;

template<typename Type>
struct value_getter<Derived<Type> > {
    typedef Type value;
};

template<typename Type>
class Derived : public Base<Type, Derived>, public value_getter<Derived<Type> >
{
    public:
};

它不是很漂亮，但至少它是有效的。