Re: How to make templated operator= more specific?
On Feb 16, 12:09 am, enjoy.c...@gmail.com wrote:
Greetings.
it'll be wonderful if somebody could help me.
I have next code:
// templated class Bar is kind of BarBase.
class Foo {
public:
template<typename T>
Foo(T f) { bb = new Bar<T>(f); }
template<typename T> operator Bar<T>() {
return static_cast<Bar<T>&>(*bb);
}
private:
BarBase *bb;
};
I have BarBase and family of inherited Bar<type> classes. When some
Bar<type> converts to BarBase (in constructor of Foo) it needs to be
converted then back to Bar<type> (via conversion operator in Foo).
Certainly, BarBase's pointer can be converted to any of it inherited
classes, including Bar<type> (which is right) and Bar<other_type>. The
latter i'd like to exclude. That means i want compile error when Foo
used like this:
Foo<int> fi;
This is inconsistent. Foo is not a template in your example. I assume
you mean, e.g.
Foo fi (42);
Bar<int> bi = fi; // Ok.
Bar<double> bd = fi; // Compile error.
How can i implement it?
You cannot, unless you make Foo a template; because the specific type
of the object pointed to by Foo::bb is lost after the constructor has
run. To recover the type you can use a dynamic cast, but that will
only detect a type mismatch at run-time:
template <typename T> operator Bar <T> ()
{return dynamic_cast <Bar <T>&> (*bb);}
To detect a type mismatch at compile-time you will have to make Foo a
template that carries along type information about the actual type of
the object pointed to by Foo::bb.
Regards,
Vidar Hasfjord
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