Question

I want to templatize the casting operator with a specialization for bool, but it’s not working.

template<typename T> //Don't know if the fact that C is templated itself is relevant here
class C
{ 
...
        template<typename U> operator U() const { return another_namespace::MyCast<U>(*this); }
        template<> operator bool() const { return IsValid(); } 
};

This gives me (g++ 4.6)

explicit specialization in non-namespace scope ‘class C< T>’

Now just

operator bool() const { return IsValid(); }

by itself works, as does MyCast (it’s a friend function declared in an external namespace). Is there any way I can get the intended behHoavior here?

Edit: I have subsequently found this, looks like the same basic question, however the answer (which gives a very complicated solution) looks designed specifically for strings. Also, the issue there turned out to be ambiguity, which I think is not the problem here–I get a very different compiler error message.

Answer 1

You can overload conversion operators, so simply using a non-templated version of the operator bool() should work:

template<typename T>
class C
{ 
...
        template<typename U> operator U() const { ... }
        operator bool() const { ... } 
};