Re: How to make templated operator= more specific?
On Feb 16, 2:09 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;
Bar<int> bi = fi; // Ok.
Bar<double> bd = fi; // Compile error.
What's Foo<int> ? In your code Foo is not a template. If you want Foo
to be a template simply do this:
template<class T>
class Foo {
public:
Foo() { }
operator Bar<T>() {
return bb;
}
private:
Bar<T> bb;
};
But somehow I suspect that's not what you wanted.
Yechezkel Mett
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