Visitor and templated virtual methods - c++

Question

In a typical implementation of the Visitor pattern, the class must account for all variations (descendants) of the base class. There are many instances where the same method content in the visitor is applied to the different methods. A templated virtual method would be ideal in this case, but for now, this is not allowed.
So, can templated methods be used to resolve virtual methods of the parent class?
Given (the foundation):
struct Visitor_Base; // Forward declaration.
struct Base
{
virtual accept_visitor(Visitor_Base& visitor) = 0;
};
// More forward declarations
struct Base_Int;
struct Base_Long;
struct Base_Short;
struct Base_UInt;
struct Base_ULong;
struct Base_UShort;
struct Visitor_Base
{
virtual void operator()(Base_Int& b) = 0;
virtual void operator()(Base_Long& b) = 0;
virtual void operator()(Base_Short& b) = 0;
virtual void operator()(Base_UInt& b) = 0;
virtual void operator()(Base_ULong& b) = 0;
virtual void operator()(Base_UShort& b) = 0;
};
struct Base_Int : public Base
{
void accept_visitor(Visitor_Base& visitor)
{
visitor(*this);
}
};
struct Base_Long : public Base
{
void accept_visitor(Visitor_Base& visitor)
{
visitor(*this);
}
};
struct Base_Short : public Base
{
void accept_visitor(Visitor_Base& visitor)
{
visitor(*this);
}
};
struct Base_UInt : public Base
{
void accept_visitor(Visitor_Base& visitor)
{
visitor(*this);
}
};
struct Base_ULong : public Base
{
void accept_visitor(Visitor_Base& visitor)
{
visitor(*this);
}
};
struct Base_UShort : public Base
{
void accept_visitor(Visitor_Base& visitor)
{
visitor(*this);
}
};
Now that the foundation is laid, here is where the kicker comes in (templated methods):
struct Visitor_Cout : public Visitor_Base
{
template <class Receiver>
void operator() (Receiver& r)
{
std::cout << "Visitor_Cout method not implemented.\n";
}
};
Intentionally, Visitor_Cout does not contain the keyword virtual in the method declaration. All the other attributes of the method signatures match the parent declaration (or perhaps specification).
In the big picture, this design allows developers to implement common visitation functionality that differs only by the type of the target object (the object receiving the visit). The implementation above is my suggestion for alerts when the derived visitor implementation hasn't implement an optional method.
Is this legal by the C++ specification?
(I don't trust when some says that it works with compiler XXX. This is a question against the general language.)

Answer 1

oh, I see what you're after. Try something like this:
template < typename Impl >
struct Funky_Visitor_Base : Visitor_Base
{
// err...
virtual void operator()(Base_Int& b) { Impl::apply(b) }
virtual void operator()(Base_Long& b) { Impl::apply(b) }
virtual void operator()(Base_Short& b) { Impl::apply(b) }
virtual void operator()(Base_UInt& b) { Impl::apply(b) }
virtual void operator()(Base_ULong& b) { Impl::apply(b) }
// this actually needs to be like so:
virtual void operator()(Base_UShort& b)
{
static_cast<impl *const>(this)->apply(b)
}
};
struct weird_visitor : Funky_Visitor_Base<weird_visitor>
{
// Omit this if you want the compiler to throw a fit instead of runtime error.
template < typename T >
void apply(T & t)
{
std::cout << "not implemented.";
}
void apply(Base_UInt & b) { std::cout << "Look what I can do!"; }
};
That said, you should look into the acyclic visitor pattern. It has misunderstood visitors built into the framework so you don't have to implement functions for stuff you'll never call.
Funny thing is that I actually used something very similar to this to build an acyclic visitor for a list of types. I applied a metafunction that basically builds Funky_Visitor_Base and turns an operator (something with an apply() like I show) into a visitor for that complete list. The objects are reflective so the apply() itself is actually a metafunction that builds based on whatever type it's hitting. Pretty cool and weird actually.

Answer 2

In your derived visitor class, Visitor_Cout, the operator() template does not override the operator() in the Visitor_Base. Per the C++03 standard (14.5.2/4):
A specialization of a member function template does not override a virtual function from a base class. [Example:
class B {
virtual void f(int);
};
class D : public B {
template <class T> void f(T); // does not override B::f(int)
void f(int i) { f<>(i); } // overriding function that calls
// the template instantiation
};
—end example]