c++ - 如何向可变类声明模板友元函数

Question

如何在类声明的内部和外部将模板函数声明为可变类的友元？

例如，这就是我认为您会在类声明中编写它的方式，但我收到 undefined reference 错误。

#include <tuple>
#include <vector>

namespace ns
{
    template<class...>
    class Example;

    template<class A, class... Bs>
    std::vector<A>& get(Example<Bs...>& pExample);

    template<class... As>
    class Example
    {
    private:
        std::tuple<std::vector<As>...> mVectors;

    public:
        explicit Example(std::size_t pCapacity)
                : mVectors(std::vector<As>(pCapacity)...)
        {}

        template<class B> //Error: undefined reference to `std::vector<int,std::allocator<int> >& ns::get<int, int, float>(ns::Example<int, float>&)
        friend std::vector<B>& get(Example<As...>& pExample)
        {
            return std::get<std::vector<B>>(pExample.mVectors);
        }
    };
}

Answer 1

友元函数不继承模板参数。

#include <tuple>
#include <vector>

namespace ns
{
    template<class... As>
    class Example;

    template<class B, class... As>
    std::vector<B>& get(Example<As...>& pExample);

    template<class... As>
    class Example
    {
    private:
        std::tuple<std::vector<As>...> mVectors;

    public:
        explicit Example(std::size_t pCapacity)
            : mVectors(std::vector<As>(pCapacity)...)
        {}

        template<class B, class... Cs> // <----
        friend std::vector<B>& get(Example<Cs...>& pExample)
        {
            return std::get<std::vector<B>>(pExample.mVectors);
        }
    };
}

int main()
{
    ns::Example<int,short,char> e(10);
    auto v = ns::get<int>(e);
}

Live example