Universal reference in a class template

Question

I'm writing a templatized container and wanted to use an universal reference for the Add(T&& object) function. Check out isocpp if you need an update on your universal/rvalue references. A quick example from that page states that, since the class is templatized, T&& is fully specified and therefore a rvalue, not an universal reference.

template <class T, class Allocator = allocator<T> >
class vector {
public:
 ...
 void push_back(T&& x); // fully specified parameter type ⇒ no type deduction;
 ... // && ≡ rvalue reference
};

This means that I cannot pass lvalues to the push_back function above.

vector<T> list;
list.Add(T()); // rvalue, compiles fine.
T data;
list.Add(data); // lvalue, does not compile.

In the case of std::vector this is solved by having a push_back(const T& object) overload. However, since the code in my Add function is quite complex, I didn't want to copy it over.

However, if I am to templatize my add function, it does become universal reference. So I came up with the following code.

#include <iostream>
#include <type_traits>
#include <vector>
using namespace std;
template <class T>
class Container
{
 public:
 template <class UR = T>
 void Add(UR&& object)
 {
 static_assert(std::is_same<std::remove_cv_t<std::remove_reference_t<UR>>, T>::value, "UR and T should be the same!");
 m_Vector.push_back(std::forward<UR>(object));
 }
 std::vector<T> m_Vector;
};
class Data
{
 public:
 int x;
 int y;
};
int main() {
 Container<Data> list;
 Data d;
 list.Add(d);
 return 0;
}

I've added the static assert to make sure that people do not actually templatize the Add function. Does anyone know a cleaner solution, since this feels a bit of template magic? Should I follow the std method and just copy my add function (which is a bit more complex than this minimal example)? Can anyone come up with a case where my solution will not work?

Here is a small ideone version for you to play around with.

Answer 1

Your solution is principally correct. If you want to play nice with potential generic traits (at the cost of a potentially uglier error message), you'd replace the static_assert with SFINAE:

template <
 class UR = T,
 class TypeMustBeT = std::enable_if_t<std::is_same<std::remove_reference_t<UR>, T>::value>
>
void Add(UR&& object)
{
 m_Vector.push_back(std::forward<UR>(object));
}

This way, the function Add will not exist at all when UR is not a forwarding reference for T¹.

If you don't like the "template magic" involved in this and/or the compiler errors generated from misuse, you can solve this by providing the two strongly-typed overloads as std does and forward to a common private implementation:

void Add(const T& object)
{
 AddImpl(object);
}
void Add(T&& object)
{
 AddImpl(std::move(object));
}
private:
template <class UR>
void AddImpl(UR&& object)
{
 static_assert(std::is_same<std::remove_cv_t<std::remove_reference_t<UR>>, T>::value, "UR and T should be the same!");
 m_Vector.push_back(std::forward<UR>(object));
}

This way, you can keep the static_assert-using version and keep the static assertion just to check your code's correctness, not the client code's.

Note that the static_assert needs to do std::remove_cv after std::remove_reference, otherwise it will fail for UR = const T&. The same applies to the original code in the question too.

Unrelated to the question at hand, but I assume using namespace std; is for this reduced example only and is not present in your real code. It's very bad practice to put that into header files in real code.

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¹ Note that "forwarding reference" is the preferred term instead of "universal reference," since it captures the semantics better.