accessing protected members of superclass in C++ with templates [duplicate]

Question

Why can't a C++ compiler recognize that g() and b are inherited members of Superclass as seen in this code:

template<typename T> struct Superclass {
 protected:
  int b;
  void g() {}
};

template<typename T> struct Subclass : public Superclass<T> {
  void f() {
    g(); // compiler error: uncategorized
    b = 3; // compiler error: unrecognized
  }
};

If I simplify Subclass and just inherit from Subclass<int> then it compiles. It also compiles when fully qualifying g() as Superclass<T>::g() and Superclass<T>::b. I'm using LLVM GCC 4.2.

Note: If I make g() and b public in the superclass it still fails with same error.

This happens because of the two-phase name lookup (which not all compilers use by default). There are 4 solutions to this problem: 1) Use the prefixes Superclass<T>::b and Superclass<T>::g(), 2) Use the prefixes this->a and this->g(), 3) Add statements using Superclass<T>::a and using Superclass<T>::g, 4) Use a global compiler switch that enables the permissive mode. The pros & cons of these solutions are described in stackoverflow.com/questions/50321788/… — George Robinson

Konrad Rudolph Konrad Rudolph · Accepted Answer · 2010-10-24T20:28:44

This can be amended by pulling the names into the current scope using using:

template<typename T> struct Subclass : public Superclass<T> {
  using Superclass<T>::b;
  using Superclass<T>::g;

  void f() {
    g();
    b = 3;
  }
};

Or by qualifying the name via the this pointer access:

template<typename T> struct Subclass : public Superclass<T> {
  void f() {
    this->g();
    this->b = 3;
  }
};

Or, as you’ve already noticed, by qualifying the full name.

The reason why this is necessary is that C++ doesn’t consider superclass templates for name resolution (because then they are dependent names and dependent names are not considered). It works when you use Superclass<int> because that’s not a template (it’s an instantiation of a template) and thus its nested names are not dependent names.

accessing protected members of superclass in C++ with templates [duplicate]

2 Answers