Java generics Eclipse compiler bug?

Question

My question is a follow up of an other guy's question: Unbounded wildcard passed to method

He was interested why the following code compiled:

public class ColTest {
    static<T> T wildSub(ArrayList<? extends T> holder, T arg){
        T t=holder.get(0);
        return t;
    }

    public static void main(String[] args) {
        ArrayList<?> list=new ArrayList<Long>(Arrays.asList(2L,3L,7L));
        Long lng=1L;
        ColTest.wildSub(list, lng);
    }
}

We came to a conclusion that the trick was that compiler inferred the ? as an Object and made the following Long argument pass thanks to the trivial inheritance of Object->Long.

The code does compile using Sun/Oracle javac (I use 1.6.0_26-b03), but does not compile in Eclipse (I use Helios) where it displays the following compilation error:

The method wildSub(ArrayList<? extends T>, T) in the type ColTest is not applicable for the arguments (ArrayList<capture#2-of ?>, Long)

My question is:

Is this a bug in the Java compiler implementation Eclipse uses or some kind of ambiguity in Java 'generics inference algorithm' specification that is valid and just implemented differently by Eclipse?

irreputable · Accepted Answer · 2011-10-20 00:29:24Z

4

It appears to be an Eclipse bug.

T should be inferred to as Object, per 15.12.2.7.

15.12.2.8 also has a catch-all clause: "Any remaining type variables that have not yet been inferred are then inferred to have type Object"

with T=Object, per 15.12.2.2, the method is applicable.

http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.12

answered Oct 20, 2011 at 0:29

irreputable

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1 Comment

MarianP Over a year ago

I need to analyze it a little more (what a beast!), but thanks for being the only person to pull the effort with such a nerdy question with not much reputation to be gained. Respect!

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