Why can't I use alias from a base class in a derived class with templates?

Question

Consider this C++ code :

template
class Step
{
public:
   using Session_ptr = boost::shared_ptr;
protected:
   Session_pt         


        

Answer 1

Here I give another example with the solutions already given in other answers presented.

std::iterator is a dependent base class of Myterator. In order to lookup there, you'll have to do qualified name lookup.

// std::iterator example  from http://www.cplusplus.com/reference/iterator/iterator/
//***************************************************************************************
#include      // std::cout
#include      // std::iterator, std::input_iterator_tag

template 
class MyIterator : public std::iterator
{
  // The working alternatives, one per row
  //typename std::iterator::pointer  p;
  //using pointer = typename std::iterator::pointer; pointer p;
  //using typename std::iterator::pointer; pointer p;
  //using Iter = typename std::iterator; typename Iter::pointer p;
  pointer p; // This does not work while any of alternatives in comments above do work
public:
  MyIterator(MyType* x) :p(x) {}
  MyType& operator*() {return *p;}

};

int main () {
  int numbers[]={10,20,30,40,50};
  MyIterator from(numbers);
  std::cout << *from << ' ';
  std::cout << '\n';

  return 0;
}