template base class typedef members invisible
c++, dependent-name, templates, typedef
Solution
You probably want to do:
using MemberType = typename TBase<T>::MemberType; // new type alias syntax
or
typedef typename TBase<T>::MemberType MemberType; // old type alias syntax
The syntax `using Base::member;` can only be used to bring the declarations of non-type members into scope.
Also note that none of these are actually required, you can qualify each use (for types with the base, for non-types with either `this->` or the base) and that will make the symbol dependent.
Problem
I'm aware of the fact that the 'dependent names' are not visible to the compiler by default. But I was told in answers to other SO questions (here, here, and ultimately on the C++ faq) that a `using` declaration may help. So I tried. A template base class: ``` // regardless of the fact that members are exposed... template<typename T> struct TBase { typedef T MemberType; MemberType baseMember; MemberType baseFunction() { return MemberType(); } }; ``` And a derived class, using the base's members: ``` template<typename T> struct TDerived : public TBase<T> { // http://www.parashift.com/c++-faq-lite/nondependent-name-lookup-members.html // tells us to use a `using` declaration. using typename TBase<T>::MemberType; using TBase<T>::baseFunction; using TBase<T>::baseMember; void useBaseFunction() { // this goes allright. baseFunction(); ++baseMember; // but here, the compiler doesn't want to help... MemberType t; //error: expected `;' before ‘t’ } }; ``` I tried this out on ideone. It has gcc-4.3.3 and gcc-4.5.1 Is this expected behavior? How are we supposed to work around the 'dependent name' law for accessing parent template class' member typedefs?