Deriving a class from a class template

Question

I am new to C++. During my learning phase I encountered with below issue. I am trying to derive a class stack from a class template Queue. Compiler throws below error in constructor of stack

..\src\TrainingDay2.cpp:44:3: error: 'b' was not declared in this scope
   b=a;

Please help to find out the root cause.

#include <iostream>
using std::cout;
using std::endl;

template<class T> class Queue //Base class
{
private:
    T ArrQueue[20];
protected:
    T* a;
    T* b;
public:

    Queue() { cout<<"QUEUE CONST "<<  endl; }
    void push(T x);
    void pop(void);
};


template <class T>
class stack :public Queue<T> // Derived class
{
public:
    stack():Queue<T>() {
        b=a;
    }
    void pop() {
        b--;
    }
};


int main()
{
    stack<int> S;
    return 0;
}

In visual studio 2013 it compiles fine. Did you post whole code? Are those classes split across several files? — Sly_TheKing, Mar 13 '15 at 11:33
@Sly_TheKing That's because VS is nonconformant in this regard, it doesn't do proper two-phase lookup. — Angew is no longer proud of SO, Mar 13 '15 at 11:36
You are using advanced C++ features but then why do you hardcode your queue to size 20? Either make the size another template parameter, or use a vector with dynamic size. — Neil Kirk, Mar 13 '15 at 11:44

score 2 · Answer 1 · answered Mar 13 '15 at 11:36

Because the base class is a template, whose instantiation depends on a template parameter of the derived class, and you're trying to name a member of the base class, two-phase lookup mandates that you write this->b, not b.

(And that default constructor call is not needed.)

stack()
{
    this->b = this->a;
}

void pop()
{
    this->b--;
}

(live demo)

Welcome to C++… :P

[C++11: 14.6.2/3]: In the definition of a class or class template, if a base class depends on a template-parameter, the base class scope is not examined during unqualified name lookup either at the point of definition of the class template or member or during an instantiation of the class template or member. [..]

score 2 · Answer 2 · answered Mar 13 '15 at 11:41

The reason is that inside a template, two-phase name lookup rules apply:

Names which do not depend on template parameters are looked up (=resolved) when the template is defined (=parsed).
Names which do depend on template parameters are looked up when the template is instantiated (when you provide the template arguments).
Base classes which depend on template parameters are only searched during name lookup at instantiation time.

The reason for this two-phase lookup is that until the template arguments are known, the compiler cannot know what the definition of the base class (or other dependent construct) will be. Remember that template specialisation exists.

You need to somehow tell the compiler that the name b is dependent. You basically have three ways to do that:

Prefix the name with this->; since you're in a class template, all your members depend on template parameters implicitly:
```
this->b = a;
```
Use full qualification for the name. This will make the dependency on template parameters explicit:
```
Queue<T>::b = a;
```

Put a using declaration into your class to inform the compiler that the name comes from a dependent base class:

template <class T>
class stack :public Queue<T> // Derived class
{
protected:
    using Queue<T>::b;

public:
    stack():Queue<T>()
    {
        b=a;
    }
     void pop()
    {
        b--;
    }
};

Excellent answer. Probably a little too complex for this OP but well worth having on the question. — Lightness Races in Orbit, Mar 13 '15 at 11:42
@LightnessRacesinOrbit Thanks. I tried to keep it understandable at non-expert level too; I don't know how well I did. — Angew is no longer proud of SO, Mar 13 '15 at 12:26

score 0 · Answer 3 · edited May 23 '17 at 11:43

0

In a template definition, unqualified names will no longer find members of a dependent base (as specified by [temp.dep]/3 in the C++ standard). For example, You must make the names dependent, e.g. by prefixing them with this->.

gcc-problem-using-a-member-of-a-base-class-that-depends-on-a-template-argument

And yes, it can be valid in VS.

edited May 23 '17 at 11:43

Community

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answered Mar 13 '15 at 11:37

senfen

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Deriving a class from a class template

3 Answers3

(live demo)