Is C# a single dispatch or multiple dispatch language?
OK, I understood the subtle difference where function overloading is different from multiple-dispatch.
Basically, the difference is whether which method to call is chosen at run-time or compile-time. Now, I know everybody's said this, but without a clear example this sounds VERY obvious, given that C# is statically typed and multiple-dispatch languages (apparently to me, at least) seem to be dynamically typed. Up to now, with just that definition multiple-dispatch and function overloading sounded exactly the same to me.
The case where this makes a real difference is when you have 2 overloads of a method which differ on the type of a parameter, but the 2 types are polymorphic, and you call with a reference declared as the higher type, which has an object of the lower type... (If someone can think of a better way to express this, please feel free to edit this answer)
Example:
int CaptureSpaceShip(IRebelAllianceShip ship) {}
int CaptureSpaceShip(XWing ship) {}
void Main() {
IRebelAllianceShip theShip = new XWing();
CaptureSpaceShip(theShip);
}
XWing obviously implements IRebelAllianceShip. In this case, the first method will be called, whereas if C# implemented multiple-dispatch, the second method would be called.
Sorry about the doc rehash... This seems to me the clearest way to explain this difference, rather than just reading the definitions for each dispatch method.
For a more formal explanation: http://en.wikipedia.org/wiki/Double_dispatch#Double_dispatch_is_more_than_function_overloading
For those that find this article using a search engine, C# 4.0 introduces the dynamic keyword. The code would look like the following.
int CaptureSpaceShip(IRebelAllianceShip ship) {}
int CaptureSpaceShip(XWing ship) {}
void Main() {
IRebelAllianceShip theShip = new XWing();
CaptureSpaceShip((dynamic)theShip);
}
C# is single dispatch but there are some blog posts which by their title looks like they are trying to emulate multimethods. If I can get one of the articles to load I will update my answer here.