Re: Anyone understand method invocation in Java?

From:
Joshua Cranmer <Pidgeot18@verizon.invalid>
Newsgroups:
comp.lang.java.programmer
Date:
Mon, 22 Sep 2008 08:47:12 -0400
Message-ID:
<gb844g$lnp$1@news-int2.gatech.edu>
Scott wrote:

Looking at method invocation in the language spec

"15.12.2.5 Choosing the Most Specific Method
....
It is possible that no method is the most specific, because there are
two or more methods that are maximally specific. In this case:

If all the maximally specific methods have override-equivalent (?8.4.2)
signatures, then:

If exactly one of the maximally specific methods is not declared
abstract, it is the most specific method.

Otherwise, if all the maximally specific methods are declared abstract,
and the signatures of all of the maximally specific methods have the
same erasure (?4.6), then the most specific method is chosen arbitrarily
among the subset of the maximally specific methods that have the most
specific return type. However, the most specific method is considered to
throw a checked exception if and only if that exception or its erasure
is declared in the throws clauses of each of the maximally specific
methods.

Otherwise, we say that the method invocation is ambiguous, and a
compile-time error occurs.
..."


?15.12.2.6 is used for disambiguating which overridden method to choose
if multiple potentially applicable methods are found.

The first thing to note is that the corner case is where generics comes
into play; override-equivalent basically means "the two methods erase to
the same signature."

Can you give an example of a method invocation giving rise to a set of
abstract methods all of which are maximally specific all having a most
specific return type (is that necessarily the same type?).


What this is specifically doing is giving precedence to the generics
methods. Example:

public abstract class GenericBase<T> {
   public T someMethod() { return null; }
}

class GenericSubclass extends GenericBase<String> {
   public String someMethod() { return "foo!"; }
   public GenericSubclass() {
     someMethod();
   }
}

The compiler sees the call and has two applicable applicable methods,
i.e., the two calls to someMethod. They are override-equivalent: only
the return types differ. String is more specific than Object (the
erasure of T), so the String method is called.

What is the meaning of the phrase
"However, the most specific method is *considered* to throw..." it
either throws an exception or not.


What this means is that we consider the list of possible exceptions to
throw based on the intersection of all of the methods that could be
arbitrarily chosen.

Since the most specific method is arbitrarily chosen is it possible to
give an example where a different arbitrary choice would would throw a
correct list of exceptions?


Constructing the example requires some generics stuff which I don't
have time the to post right now; I'll do it later.

--
Beware of bugs in the above code; I have only proved it correct, not
tried it. -- Donald E. Knuth

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