Java lambda expressions - explicit target type

Question

Consider functional interfaces A and B extends A. Is it possible to create an instance implementing not only A but also B with a lambda expression when only A is required?

Example:

interface A {
    void foo();
}

interface B extends A {}

void bar(A arg) {
    System.out.println(arg instanceof B);
}

/* somewhere in a block */
    bar( () -> {} ); // Prints "false"

How could I call bar(A) so that it prints true?

I would find this useful in event listeners, where subinterfaces could specify additional data about the listener, e.g. if a Listener also implements ConcurrentListener, it is invoked concurrently.

Sweeper · Accepted Answer · 2018-06-04T13:59:09.847

8

Simply do a cast to tell the compiler the type you want:

A a = (B)() -> someAction();
System.out.println(a instanceof B); // true

I would find this useful in event listeners, where subinterfaces could specify additional data about the listener, e.g. if a Listener also implements ConcurrentListener, it is invoked concurrently.

That sounds like bad design (though I can't explain why...). Maybe you could have a isConcurrent method in the interface that implementers need to implement?

edited Jun 04 '18 at 13:59

answered Jun 04 '18 at 13:53

Sweeper

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1

Nice! For those like me who were wondering how a lambda can be automagically cast to an interface [see here](https://stackoverflow.com/q/28509596/314291). – StuartLC Jun 04 '18 at 14:01
Thanks for the reply! As for your workaround using new methods, that is exactly what I would have done if I wasn't trying to make listeners `@FunctionalInterface`s. – OLEGSHA Jun 04 '18 at 17:50

score 2 · Answer 2 · answered Jun 04 '18 at 14:07

2

You can explicitly cast lambda expression to B:

bar((B)() -> {});

or assign lambda expression to a variable:

B lambda = () -> {};
bar(lambda);

answered Jun 04 '18 at 14:07

frenchu

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Java lambda expressions - explicit target type

2 Answers2