Re: Is this pointer required to be used in template inheritance?

From:

"Alf P. Steinbach" <alfps@start.no>

Newsgroups:

comp.lang.c++

Date:

Thu, 12 Feb 2009 23:58:59 +0100

Message-ID:

<gn29fs$ec3$1@news.motzarella.org>

* Alessandro [AkiRoss] Re:

Hello there,
I was trying this code:

#include <iostream>

template <typename T>
struct Base {
protected:
    T value;
};

template <typename T>
struct Derived: public Base<T> {
    void set(const T &d) {
        this->value = d; // Must be referred using this
    }
    void get() {
        //std::cout << "Value: " << value << std::endl; // ERROR
    }
};

int main() {
    Derived<int> d;
    d.set(5);
    d.get();
}

In the Derived::get() method, g++ tells me that value isn't in that
scope. While in set() it is, because I'm accessing it using the this
pointer.
Why is it necessary? It isn't required if the classes aren't
templates.

I don't think the language rule is /necessary/, but it provides some safety that
the code has only one possible meaning and that that is the indtended meaning.

I.e. for the general case it's practical.

Consider:

<code>
#include <iostream>

template< typename T > class Base;

static int const value = 666;

template< typename T >
class Derived: public Base<T>
{
public:
     void foo() const
     {
         std::cout << value << std::endl;
         std::cout << this->value << std::endl;
     }
};

template<>
class Base<double>
{
protected:
     int value;
public:
     Base(): value( 42 ) {}
};

int main()
{
     Derived<double>().foo();
}
</code>

<output>
666
42
</output>

Instead of "this"-qualification you might use class name qualification or a
"using" declaration.

Cheers & hth.,

- Alf