Perfect forwarding and std::tuple

Question

Consider the following code:

#include <iostream>
#include <tuple>
#include <utility>

// A.
template <typename... Args>
void f (const char* msg, Args&&... args)
{
    std::cout << "A. " << msg << "\n";
}

// B.
template <typename... Args>
void f (const char* msg, std::tuple<Args...>&& t)
{
    std::cout << "B. " << msg << "\n";
}

struct boo
{
    const std::tuple<int, int, long> g () const
    {
        return std::make_tuple(2, 4, 12345);
    }
};

int main ()
{
    f("First", 2, 5, 12345);
    f("Second", std::make_tuple(2, 5, 12345));

    boo the_boo;
    f("Third", the_boo.g());
    f("Fourth", std::forward<decltype(std::declval<boo>().g())>(the_boo.g()));

    return 0;
}

The output of that will be:

A. First
B. Second
A. Third
A. Fourth

From the output it's evident that it does not do what I would like it to do, that is I would like Third and Fourth to go through the B. version of the function. The std::forward from the Fourth call is superfluous as perfect forwarding does not happen there. In order to have perfect forwarding I know:

• I must have an rvalue reference in a type deducing context
• the type of the parameter must be a template type for the function

I understand it does not work. But I do not fully grasp:

• why the context is changed by using std::tuple in such a way that it fails to work as desired ? Why the template parameter cannot be the type for another templated type?

• how can I(elegantly) fix it ?

Answer 1

Your issue is that in Third and Fourth you are passing a const std::tuple where B. expects a non-const version.

When the compiler attempts to generate code for the call to f, it sees that you are calling with a const std::tuple and so deduces the type of Args... to be const std::tuple. Calling B. is not valid because the variable has a different const-qualification than expected.

To solve this, just make g() return a non-const tuple.

---

Edit:

In order for perfect forwarding to occur, you need a deduced context, as you say in your question. When you say std::tuple<Args...>&& in the function argument list, Args... is deduced, but std::tuple<Args...>&& is not; it can only by an rvalue reference. In order to fix this, that argument needs to be of the form T&& where T is deduced.

We can accomplish this using a custom type trait:

template <typename T>
struct is_tuple : std::false_type {};

template <typename... Args>
struct is_tuple <std::tuple<Args...>> : std::true_type {};

Then we use this trait to enable a single-argument template for tuples only:

// B.
template <typename T, typename = typename std::enable_if<
                          is_tuple<typename std::decay<T>::type>::value
                          >::type>
void f (const char* msg, T&& t)
{
    std::cout << "B. " << msg << "\n";
    std::cout << "B. is lval == " << std::is_lvalue_reference<T>() << "\n";
}

Or alternatively:

//! Tests if T is a specialization of Template
template <typename T, template <typename...> class Template>
struct is_specialization_of : std::false_type {};

template <template <typename...> class Template, typename... Args>
struct is_specialization_of<Template<Args...>, Template> : std::true_type {};

template <typename T>
using is_tuple = is_specialization_of<T, std::tuple>;

is_specialization_of taken from here and suggested by this question.

Now we have perfect forwarding!

int main ()
{
    f("First", 2, 5, 12345);
    f("Second", std::make_tuple(2, 5, 12345));

    boo the_boo;
    f("Third", the_boo.g());
    f("Fourth", std::forward<decltype(std::declval<boo>().g())>(the_boo.g()));

    auto the_g = the_boo.g();
    f("Fifth", the_g);

    return 0;
}

Outputs:

A. First
B. Second
B. is lval == 0
B. Third
B. is lval == 0
B. Fourth
B. is lval == 0
B. Fifth
B. is lval == 1