Why does this.getClass give it's own class name rather than Anonymous class name?

Question

I have created anonymous class by implementing interface I inside public static void main() method. So, by java 8 for the abstract method test(), the implementation is provided from imple() method of class C.

So, inside public static void main() method, printing _interface.getClass(), I got

package_path.Main$$Lambda$1/310656974 which is absolutely fine. Bacause it print's the anonymous class name.

Also, _interface is pointing to an anonymous object in heap and hence I'm doing _interface.test();

So, the first statement that test() method has now is to print the class name,

But eventually what it print was, package_path.C (telling me C is the class name). How is that possible? Shouldn't package_path.Main$$Lambda$1/310656974 be printed again? Because 'this' means anonymous inside the test method right?

@java.lang.FunctionalInterface
interface I {
    void test();
}

class C {
    void imple() {
        System.out.println(this.getClass());
        System.out.println("Inside Implementation");
    }
}

class Main {
    public static void main(String[] args) {
        I _interface = new C()::imple;
        System.out.println(_interface.getClass());
        _interface.test();
    }
}

Naman · Accepted Answer · 2019-05-29T03:03:57.607

4

Hopefully, this might help you understand, that when you declare

I _interface = new C()::imple;

you've actually implemented the interface somewhat similar to (though not same as):

I _interface = new I() {
    @Override
    public void test() {
        new C().imple(); // creating an instance of class `C` and calling its 'imple' method
    }
};

Hence when the test method is called, it first creates an instance of C which prints

class x.y.z.C

as the class.

Because 'this' means anonymous inside the test method right?

~~Now as you can see above, there is no more anonymous class from which imple is being called from, hence this is not representing the anonymous class anymore.~~

As Holger clarified in comments further, despite the representation as lambda or anonymous class at the calling site, the this.getClass() inside a method of class C will evaluate to C.class, regardless of how the caller looks like.

Recommend: Continue to read and follow on Is there any runtime benefit of using lambda expression in Java?

edited May 29 '19 at 03:03

answered May 26 '19 at 05:05

Naman

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1

Got it, Thanks :) Probably with method reference, I was not able to understand this. With J7 and below implementation of anonymous class, things got clear :) – Manikandan Kbk DIP May 26 '19 at 05:13
1

@Naman your example implies that a new instance of `C` is created on each call of `test`, this is not correct; it is cached... – Eugene May 26 '19 at 05:49
1

@Eugene True, I guess it would be more or similar to `C c = new C(); I _interface = () -> c.imple();`. Just that the point I was deriving for OP was the new instance created for the class `C`. – Naman May 26 '19 at 06:03
1

This answer is misleading. The expression `this.getClass()` inside a method of class `C` will evaluate to `C.class`, regardless of how the caller looks like. But an occurrence of `getClass()` within a lambda expression would not evaluate to an anonymous class either, as lambda expressions and anonymous classes are *not* equivalent. – Holger May 27 '19 at 07:49
@Holger *The expression this.getClass() inside a method of class C will evaluate to C.class, regardless of how the caller looks like*... Agreed and that's one point I wanted to illustrate with the expansion for the OP to understand that `this` is not referred from within an anonymous class but class `C` itself. Related to *But an occurrence of `getClass()` within a lambda expression would not evaluate to an anonymous class either* ... I couldn't catch up with you here, did you mean to say that the class representation here is of a Lambda expression and not an Anonymous class? – Naman May 27 '19 at 07:58
2

Say `Supplier> s = () -> getClass();`. What would `s.get()` give you? It’s *not* an anonymous class implementing `Supplier` or, in other words, `s.get() != s.getClass()`. So rewriting the lambda expressions to an inner class the way you did in your answer will lead to wrong assumptions. – Holger May 27 '19 at 08:08
@Holger Thanks for sharing the link. Would update the answer accordingly. Over to you to close as duplicate if you believe the question is already answered as well. – Naman May 27 '19 at 14:37

Why does this.getClass give it's own class name rather than Anonymous class name?

1 Answers1