传递给方法的无界通配符

Question

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);
    }
}

Really interested why this snippet is legal, because the signature of wildSub takes only ArrayList of T or derived from T , and arg of type T . But <?> means - some specific type, not known, and how it can satisfy the compiler? After all type <?> doesn't mean <? extends Long> <? extends Long> ...

Answer 1

The compiler is free to infer anything that is compatible with the types of the arguments and return type. In your case it can always infer T as Object . Which turns the signature into

static Object wildSub(ArrayList<?> holder, Object arg)

Which means it can take any ArrayList as first argument and anything as second. Since you don't do anything with the return value, Object will be okay.

Answer 2

If you think about it as the compiler using Object where ? is used, it makes sense why it would compile. That is all there is to it.

If you are doing any operations dependent on ? being a certain class, you will get a cast exception at run time if the wrong class is passed in.

Answer 3

As an addition to existing (correct) answers to make it more clear:

    ...
        Object result1 = ColTest.wildSub(list, lng); //compiles fine with Sun's javac
//      Long result2 = ColTest.wildSub(list, lng);   //does not compile without explicit casting
        Long result2 = (Long) ColTest.wildSub(list, lng);   //compiles fine 
    ...

Answer 4

This is due to capture conversion . Internally, compiler converts the type of an expression Foo<?> to Foo<X> , where X is a specific albeit unknown type.