How to Access Overloaded Methods

A chocolate cake is a specific kind of cake. A cake is a specific kind of dessert. A buttered scone is a specific kind of scone. A scone is a specific kind of dessert.

Suppose we have three methods all with the same name and each with the same number of parameters. However, the parameters are of different object types:

  1. void arrange(Dessert d, Scone s);
  2. void arrange(Cake c, Dessert d);
  3. void arrange(ChocolateCake cc, Scone s);

Which version of the method will the following statements invoke?

arrange(dessertRef, sconeRef);
This invokes No 1 because the parameter types are an exact match.

arrange(chocolateCakeRef, dessertRef);
This invokes No 3 because chocolate cake is a more specific form of cake and scone is a more specific kind of dessert.

arrange(chocolateCakeRef, butteredSconeRef);
This invokes No 3 because chocolate is an exact match and buttered scone is a more specific type of scone.

arrange(cakeRef, sconeRef);
This would tend to invoke Nos 1 and 2 equally and therefore cannot invoke either. It is an invalid invocation.

The same rules apply to primitive types. Eg. an int is assignable to a float just as a buttered scone is assignable to a scone. But if you declare two methods which both compute the distance to a navigation station: 

int dist(int lat, int lng){   }
short dist(int lat, int lng){   }
and then you invoke one of them by the statement: 
double d = dist(lat, lng);
the compiler has no way of knowing which one to invoke here - even if the two methods throw different exceptions.

The moral is that when in doubt use the method's complete class reference.

This page's parent within this Web Site. About this Web Site. Its home page. Email its Author.