CODEWITHSUNDEEP
javajavafx
GOXR3PLUS
GOXR3PLUS

Reputation: 7255

JavaFX remove InvalidatorListener with lambda expression

I am using this code:

    renameWindow.showingProperty().addListener(new InvalidationListener() {
                @Override
                public void invalidated(Observable observable) {

                    //Remove the Invalidator Listener
                    renameWindow.showingProperty().removeListener(this);

               }
            });

And i want to do in lambda way and i use:

renameWindow.showingProperty().addListener(listener->{
                renameWindow.showingProperty().removeListener(this);
            });

and i am getting error cause maybe the listener is the Observable interface or what?.I want to do this using lambda expression.How can this be done?How to remove the InvalidationListener using lambda.

Upvotes: 1

Views: 1382

Answers (1)

fabian
fabian

Reputation: 82461

this refers to a instance of the class containing the code, not to the lambda expression itself.

jls-15.27.2: Lambda Body

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).

Therefore you're trying to call removeListener with a object as parameter that doesn't implement InvalidationListener or ChangeListener, which means none of the removeListener methods is applicable.

The only way to get a reference to the lambda expression is by using some expression access a reference to it from the lambda body that is evaluated at the time the lambda body is executed.

This could be done e.g. by assigning it to a field.

Example

private InvalidationListener listener = observable -> renameWindow.showingProperty().removeListener(this.listener);

Upvotes: 2

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