Use std::tuple for template parameter list instead of list of types

Question

I\'m trying to make a call to a templated function like this :

typedef std::tuple Instrument         


        

Answer 1

Your first solution is failing because the second overload to get is not visible at the point of its own return type declaration; to get around this you would need to separate out the return type computation into its own subprogram.

The second solution is closer; the problem is that you're only inferring the template std::tuple, not its arguments. An easy way to infer variadic arguments (e.g. type arguments to tuple) is through an empty tag structure, requiring one extra level of indirection:

template struct type_tag {};

class Cache {
    // ... (as above)

    template std::tuple get(type_tag>) {
        return get<0, Ts...>();
    }

public:
    template T get() {
        return get(type_tag{});
    }
};

You should check to see whether you can write the solution using pack expansion instead of recursion, for example:

template struct type_tag {};

class Cache {
    template std::tuple get(type_tag>) {
        return std::tuple{Ts{}...};
    }

public:
    template T get() {
        return get(type_tag{});
    }
};