Re: Double Dispatch Obsolete?

Gianni Mariani <>
Tue, 04 Sep 2007 14:28:19 -0700
DeMarcus wrote:

Since I started with OO I've been told switching on typeid is a big
no-no. E.g.

void Washer::wash( Vehicle myVehicle )
 if( typeid(myVehicle) == typeid(Car) )
  Washer::washCar( myVehicle );
 else if( typeid(myVehicle) == typeid(Bike)
  Washer::washBike( myVehicle );
 else if( typeid(myVehicle) == typeid(Boat)
  Washer::washBoat( myVehicle );

Yes - no-no - not extensible. Code like this tends to proliferate and
is prone to error.

The alternative is the more correct Double Dispatch. E.g.

void Washer::wash( Vehicle myVehicle )
 myVehicle.washer( this )

void Car::washer( Washer w )
 w.washCar( this );

Kind of.

It would be more like:

myVehicle.DoSomthing( Wash ).

void Car::DoSomthing( Dispatcher & i_dispatch )
     i_dispatch.WashCar( * this );

Now, consider we change Washer to XMLConverter and wash() to write().
This will still work, but when we want to go backwards and read XML and
write a Vehicle we need to switch on some kind of type id label anyway.

Vehicle XMLConverter::readVehicle( XMLdoc doc )
 Vehicle v;
 string s = doc.readAttr();
 if( s == "Car" )
  v = new Car();
 else if( s == "Bike" )
  v = new Bike();
 else if( s == "Boat" )
  v = new Boat();

 return v;

This is usually solved by a generic factory system like Austria C++'s
factory thing.

v = at::FactoryRegister< Interface, std::string >::Get().Create( s )();

So why not just give every MyObject a typeName() method and switch or
std::map<char*, fncPtr> on that throughout all dispatchers?

That can be one way. If done correctly, the double dispatch technique
is able to pick up when you miss a case by using pure virtual methods.
In this example if a new type of vehicle is make and the method is not
implemented, it can cause a compile time error which can flag missing

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