Type inference issue for generic method

Question

Is it possible to resolve the problem with weak inference without defining additional variables or type casts for code below?

public class GenericClass<T> {
    public <R> R m(Class<R> cl) {
        return null;
    }
}

GenericClass<SomeClass> v2 = new GenericClass<SomeClass>()
    .m(GenericClass.class)
    .m(GenericClass.class); // <- Object cannot be converted to GenericClass<SomeClass>

Darth Android · Answer 1 · 2016-04-08T21:07:10.723

3

Yes:

public class GenericClass<T> {
    public <R> R m(Class<? super R> cl) {
        return null;
    }
}

GenericClass<SomeClass> v2 = new GenericClass<SomeClass>()
    .<GenericClass<SomeClass>>m(GenericClass.class)
    .m(GenericClass.class);

We need to fix the fact that cl might be an erased type (i.e., a super type of the generic type, R), and then we need to tell the compiler what the real type R is since the method argument is only indicating the super type.

The second call to m doesn't not need to have the generic type specified because it is inferred from the assignment.

edited Apr 08 '16 at 21:07

answered Apr 08 '16 at 20:46

Darth Android

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19

The generic class is a third-party library class I cannot modify. But thanks anyway. – J. Smith Apr 10 '16 at 15:31
You will need to cast then. – Darth Android Apr 11 '16 at 03:54

score 0 · Answer 2 · answered Apr 08 '16 at 21:01

in order to call you method m in chain it needs to return its on class "this". In your case is GenericClass. This should work.

class GenericClass<T> {
    public <R> GenericClass<T> m(Class<R> cl) {
        // code to do something here
        return this;
    }
}

for every call to the "m" method it will return it's own class then you can call it again.

I hope that helps.

Type inference issue for generic method

2 Answers2