从匿名内部类中调用重写的默认方法

Question

Consider this code:

interface A {
    default void doA() {
        System.out.println("a");
    } 
}

interface B {
    void doB(); 
}

class Test implements A {

    @Override
    public void doA() {        
        // Works
        B b = () -> A.super.doA();
        b.doB();

        // Does not compile
        /*
        new B() {      
            public void doB() {  
                A.super.doA();
            }       
        }.doB();
        */
    }

    public static void main(String[] args) {
        new Test().doA();
    }

}

This is contrived, but basically Test::doA() tries to wrap this as a B and have B::doB() call its super function A.super.doA() .

I can call A.super.doA() in a lambda of type B just fine. But I cannot figure out the syntax of calling A.super.doA() inside an anonymous B . See the commented out code.

Any ideas?

Answer 1

Code in lambdas and anonymous classes is treated differently

Unlike code appearing in anonymous class declarations , the meaning of names and the this and super keywords appearing in a lambda body, along with the accessibility of referenced declarations, are the same as in the surrounding context (except that lambda parameters introduce new names).

^{The transparency of this (both explicit and implicit) in the body of a lambda expression - that is, treating it the same as in the surrounding context - allows more flexibility for implementations, and prevents the meaning of unqualified names in the body from being dependent on overload resolution.
Practically speaking, it is unusual for a lambda expression to need to talk about itself (either to call itself recursively or to invoke its other methods), while it is more common to want to use names to refer to things in the enclosing class that would otherwise be shadowed (this, toString()). If it is necessary for a lambda expression to refer to itself (as if via this ), a method reference or an anonymous inner class should be used instead.}

^{JLS 10 - 15.27.2. Lambda Body}

Code in lambdas

The keyword this may be used in a lambda expression only if it is allowed in the context in which the lambda expression appears. Otherwise, a compile-time error occurs.

^{JLS 10 - 15.8.3. this}

I think it also can be applied to the keyword super .

The statement A.super.doA(); would be working in the enclosing context (the body of the method Test#doA ), so it is allowed in lambdas as well.

class Test implements A {

    @Override
    public void doA() {
        B b = () -> {
            System.out.println(super.getClass());
            System.out.println(Arrays.toString(super.getClass().getInterfaces()));
        };
        b.doB();

        // ...
    }

}

This snippet prints

class Test
[interface A]

We will compare it with the anonymous class result.

Code in anonymous classes

class Test implements A {

    @Override
    public void doA() {
        // ...

        new B() {
            public void doB() {
                System.out.println(super.getClass());
                System.out.println(Arrays.toString(super.getClass().getInterfaces()));
            }
        }.doB();
    }

}

The snippet outputs

class Test$1
[interface B]

Keeping in mind that an anonymous class has own this and super and it does not inherit A (and can't do it), it becomes clear that A.super.doA(); can't be compiled in its context.

Workarounds

A workaround could be remembering the enclosing context by a lambda, and invoking that lambda in the method of an anonymous class:

class Test implements A {

    @Override
    public void doA() { 
        Runnable doA = () -> A.super.doA();

        new B() {
            public void doB() {
                doA.run();
            }
        }.doB();
    }

}

If B inherited A , it would be possible to call doA() or B.super.doA() referring to the default method:

class Test implements A {

    @Override
    public void doA() {
        new B() {
            public void doB() {
                doA(); // or B.super.doA();
            }
        }.doB();
    }

}

Answer 2

As said in this answer , this is possible in lambda expressions due to the different meaning of this and super (compared to inner classes).

The impossibility to do the same with inner classes has been addressed in The Java® Language Specification, §15.12.1 explicitly:

The TypeName . super syntax is overloaded: traditionally, the TypeName refers to a lexically enclosing type declaration which is a class, and the target is the superclass of this class, as if the invocation were an unqualified super in the lexically enclosing type declaration.

…

To support invocation of default methods in superinterfaces, the TypeName may also refer to a direct superinterface of the current class or interface, and the target is that superinterface.

…

No syntax supports a combination of these forms, that is, invoking a superinterface method of a lexically enclosing type declaration which is a class, as if the invocation were of the form InterfaceName . super in the lexically enclosing type declaration.

 class Subclass3 implements Superinterface { void foo() { throw new UnsupportedOperationException(); } Runnable tweak = new Runnable() { void run() { Subclass3.Superinterface.super.foo(); // Illegal } }; } 

A workaround is to introduce a private method in the lexically enclosing type declaration, that performs the interface super call.

Answer 3

I don't think that it is possible.

This one :

B b = () -> A.super.doA();

or this one :

A.super.doA();

are valid as these statements use the A instance method as context.

From the anonymous class, things are different as you don't have access to the A instance context. So A cannot be referenced.
The method in the anonymous class can reference final variables of the enclosing method or the instance of the enclosing method (by prefixing classname.this ) but the method cannot behave as if it was executed in the context of an A instance method : what A.super.doA() means.

I think that a section of the JLS has to specified this point.