考虑以下代码:

#include <vector>

template<typename T> class Container;
template<typename T> Container<Container<T>> make_double_container(const std::vector<std::vector<T>>&);

template<typename T>
class Container {
    std::vector<T> v;
    friend Container<Container<T>> make_double_container<T>(const std::vector<std::vector<T>>&);

public:
    Container() {}
    explicit Container(std::vector<T> v) : v(v) {}
};

template<typename T>
Container<Container<T>> make_double_container(const std::vector<std::vector<T>>& v) {
    Container<Container<T>> c;
    for(const auto& x : v) {
        c.v.push_back(Container<T>(x));
    }
    return c;
}

int main() {
    std::vector<std::vector<int>> v{{1,2,3},{4,5,6}};
    auto c = make_double_container(v);
    return 0;
}

编译器告诉我:
main.cpp: In instantiation of 'Container<Container<T> > make_double_container(const std::vector<std::vector<T> >&) [with T = int]':
main.cpp:27:37:   required from here
main.cpp:8:20: error: 'std::vector<Container<int>, std::allocator<Container<int> > > Container<Container<int> >::v' is private
     std::vector<T> v;
                    ^
main.cpp:20:9: error: within this context
         c.v.push_back(Container<T>(x));

我相信是正确的,因为make_double_containerContainer<T>的 friend ,但不是Container<Container<T>>的 friend 。在这种情况下,如何使make_double_container工作?

最佳答案

显然,您可以将make_double_container的每一个专业知识作为 friend :

template <typename U>
friend Container<Container<U>> make_double_container(const std::vector<std::vector<U>>&);

如果您希望在不局部特化的情况下将友谊降至最低,请尝试
template <typename> struct extract {using type=void;};
template <typename U> struct extract<Container<U>> {using type=U;};
friend Container<Container<typename extract<T>::type>>
     make_double_container(const std::vector<std::vector<typename extract<T>::type>>&);

Demo

09-10 00:07
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