Question

I’m working (mainly for learning purposes) on own implementation of tuple and I’ve just encountered a problem. I have following code:

namespace Rose
{
    template<typename T>
    struct RemoveReference
    {
        typedef T Type;
    };

    template<typename T>
    struct RemoveReference<T &>
    {
        typedef T Type;
    };

    template<typename... Elems>
    class Tuple;

    template<typename First, typename... Elems>
    class Tuple<First, Elems...>
    {
    public:
        Tuple(First a, Elems... more)
            : More(more...), Element(a)
        {
        }

        Tuple<First, Elems...> & operator=(const Tuple<RemoveReference<First>::Type,
                                           RemoveReference<Elems>::Type...> & rhs)
        {
            this->Element = rhs.Element;
            this->More = rhs.More;

            return *this;
        }

    private:
        Tuple<Elems...> More;
        First Element;
    };

    template<typename Only>
    class Tuple<Only>
    {
    public:
        Tuple(Only a) : Element(a)
        {
        }

        Tuple<Only> & operator=(const Tuple<RemoveReference<Only>::Type> & rhs)
        {
            this->Element = rhs.Element;

            return *this;
        }

    private:
        Only Element;
    };
}

int main()
{
    Rose::Tuple<int, float, int> t(1, 1.f, 2);
}

Which causes following error (there are more of them, but this one is essential):

error: type/value mismatch at argument 1 in template parameter list for ‘template struct Rose::Tuple’
error: expected a type, got ‘Rose::RemoveReference::Type’

I don’t really understand what’s this about. The RemoveReference trait works, when used alone.

Here are two testcases:

• not working code I pasted above
• proof that RemoveReference works when used alone

I have tried this code with G++ 4.6.1, 4.5.1 and Clang++ 2.9.

What’s the reason for those errors to appear?

Answer 1

RemoveReference<T>::Type is a dependent type, so you need to add typename here:

        Tuple<First, Elems...> & operator=(const Tuple<typename RemoveReference<First>::Type,
                                                       typename RemoveReference<Elems>::Type...> & rhs)

and probably other places.