std :: get以元组为成员的自己的类

Question

I have a class like this:

 template<typename ... TTypes>
 class Composite {
 public:
     std::tuple<TTypes...> &getRefValues() { return values; }   
 private:
     std::tuple<TTypes...> values;
 };

Can I define std::get for my class Composite? It should basically call the already defined std::get for the private tuple values.

I was able to implement a customized get function when the return type is known (eg for an int array member) but I don't know how to realize get when the return type can be an arbitrary type (depending on the components' type of the tuple values)?

Answer 1

You may do:

template <std::size_t I, typename... Ts>
auto get(Composite<Ts...>& composite)
-> decltype(std::get<I>(composite.getRefValues()))
{
    return std::get<I>(composite.getRefValues());
}

Note: In C++14, you may omit the -> decltype(..) part.

Answer 2

For completeness, here is my solution. Thanks everyone:

template<typename ... TTypes>
class Composite {
public:
    Composite(TTypes... t) {
        std::tuple<TTypes...> tuple(t...);
        values = tuple;
    } 

    std::tuple<TTypes...> &getRefValues() { return values; }    
private:
    std::tuple<TTypes...> values;
};

namespace std {
template<size_t I, typename ... TTypes>
    auto get(Composite<TTypes ...> &t) -> typename std::tuple_element<I, std::tuple<TTypes...>>::type { 
        return std::get<I>(t.getRefValues());
    }
}

int main() {
    Composite<int, char, double> c(13, 'c', 13.5);
    std::cout << std::get<0>(c) << std::endl;
    std::cout << std::get<1>(c) << std::endl;
    std::cout << std::get<2>(c) << std::endl;

    return 0;
}