How does Java pick which overloaded function to call?

Question

This is a purely theoretical question.

Given three simple classes:

class Base {
}

class Sub extends Base {
}    

class SubSub extends Sub {
}
         


        

Answer 1

The formal specification can be found in part 15.12.2.5 of the Java Language Specification (JLS). Thanks to generics this is pretty complicated, so you might want to look at same section of the first edition of the JLS.

It basically says that the compiler tries to find a version of the method where all parameters including the object the method is called upon are most specific. If no such method exists (e.g. since you have method(Base, Sub) and method(Sub, Base) but not method(Sub, Sub)), then the compilation fails.

Note that the actual choice of method depends on the dynamic type of the target object for instance methods, but not for the parameters. Your example would still work the same on the instance level.

You should be able to give the compiler a helping hand by casting or redeclaring the type of ss. If the declared type of the variable matches a signature exactly then everything is clear for the compiler and maintenance programmers as well. It doesn't matter if you then assign a more specific type as long as the declared type matches.