template paramters: example from Modern C++ Design

Question

I have read "Modern C++ Design"
and I have a question in its sample code

in p278 p279
or refer to Loki's source if you don't have the book
BasicDipatcher::Add and BasicDispatcher::Go in MutilMethods.h

in page p278 bottom to p279 up
it has a piece of sample code

typedef BasicDispatcher<Shape> Dispatcher;
void HatchRectanglePoly(Shape& lhs, Shape& rhs) {...}
Dispatcher disp;
disp.Add<Rectangle, Poly>(HatchRectanglePoly);

I found in function Go, its arguments are BaseLhs&, BaseRhs&
which in this case, should be Shape&, Shape&

and in function Add, its arguments are SomeLhs&, SomeLhs&,
which in this case, should be Rectangle&, Poly&

so the key won't match anyway because they are different

therefore the callback(HatchRectanglePoly) won't be called
(If I add disp.Go.... in the samele code),
and instead, a std:runtime_error will be thrown

Am i correct??

thanks

Answer 1

The BaseLhs and BaseRhs are template parameters. Just like function parameter, the actual value will be provided when you use it (instantiate the template), not when you define it.

template
<
  class BaseLhs,
  class BaseRhs = BaseLhs,
  typename ResultType = void,
  typename CallbackType = ResultType (*)(BaseLhs&, BaseRhs&)
>
class BasicDispatcher

By default, BaseRhs is the same as BaseLhs.

typedef BasicDispatcher<Shape> Dispatcher;
Dispatcher disp;

Here, we instantiate a version of BasicDispatcher, the BaseLhs is Shape, the BaseRhs is Shape too (since we provide only 1 template argument). In this instantiation, the Go method is somewhat like this:

void Go(Shape& lhs, Shape& rhs);

The same goes for Add.

In short: The type name written in template<...> is just a placeholder, it will be substituted by actual type when use.

Hope you find this helpful.

P.S: About the class and typename inside the angle bracket, they have the same meaning, I guess it's just a hint to the reader that BaseLhs and BaseRhs will always be classes.