Re: Implicit conversion and method call

From:

Victor Bazarov <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Wed, 29 Apr 2009 10:43:44 -0400

Message-ID:

<gt9p2h$nr$1@news.datemas.de>

GeeRay wrote:

Hi all,
how can I create a template class to decorate a type and use it as the
type itself?

For example:

I want to do this:

#include <iostream>
class A
{
public:
    A(){};
    virtual ~A(){};
    void foo(){ std::cout << "foo" << std::endl;};
};

template<class T>
class B
{
public:
    B(){};
    virtual ~B(){};

    operator T(){return instance;};
private:
    T instance;
};

int main(int argn, char* argv[])
{
    B<A> b();

Drop the parentheses, otherwise you're declaring a function. My answer
assumes that the definition of 'b' is like this:

B<A> b;

b.foo();
}

Is it possible?

No. For the member function calls (like the . you use to access the
'foo' member) the conversions are not considered. You can overload the
member access operator for pointers (pretending that your 'B' class is a
pointer), like so:

    template<class T> class B { ...

       T* operator->() { return &instance; }
    };

, then you could write something like

    B<A> b;
    b->foo();

which is not necessarily the best syntax, of course...

V
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