typedef a shared pointer that contains a templated class
c++, class, forward-declaration, shared-ptr, templates
Solution
You also probably want template typedef. Read up on Sutter's article.
In C++03, you need a hack as:
template <typename Arg> struct ArgPtr {
typedef std::shared_ptr<Arg> ArgPtrType;
};
In C++11, you can use template aliasing directly with the `using` keyword:
template <typename T>
using ArgPtrType = std::shared_ptr<Arg<T>>;
Problem
Suppose I have some template class forward declared and I want to typedef a shared pointer to it. How would I do this? ``` template<typename T> class Arg; typedef std::tr1::shared_ptr<Arg> ArgPtr; // Compiler error ```