سؤال

say I have the following code

public class A {
    int x;
    public boolean is() {return x%2==0;}
    public static boolean is (A a) {return !a.is();}
}

and in another class...

List<A> a = ...
a.stream().filter(b->b.isCool());
a.stream().filter(A::is); 
//would be equivalent if the static method is(A a) did not exist

the question is how do I refer to the instance method version using the A::is type notation? Thanks a lot

هل كانت مفيدة؟

المحلول

In your example, both the static and the non-static method are applicable for the target type of the filter method. In this case, you can't use a method reference, because the ambiguity can not be resolved. See §15.13.1 Compile-Time Declaration of a Method Reference for details, in particular the following quote and the examples below:

If the first search produces a static method, and no non-static method is applicable [..], then the compile-time declaration is the result of the first search. Otherwise, if no static method is applicable [..], and the second search produces a non-static method, then the compile-time declaration is the result of the second search. Otherwise, there is no compile-time declaration.

In this case, you can use a lambda expression instead of a method reference:

a.stream().filter(item -> A.is(item));

The above rule regarding the search for static and non-static methods is somewhat special, because it doesn't matter, which method is the better fit. Even if the static method would take an Object instead of A, it's still ambiguous. For that reason, I recommend as a general guideline: If there are several methods with the same name in a class (including methods inherited from base classes):

  • All methods should have the same access modifiers,
  • All methods should have the same final and abstract modifiers,
  • And all methods should have the same static modifier

نصائح أخرى

We can not use not static methods or non-global methods by using className::methodName notation. If you want to use methods of a particular class you have to have an instance of the class.

So if you want to access is() method then you can use : 
A a = new A();
a.is();
OR 
(new A()).is();

Thanks.

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