Method reference of an object in variable vs. returned by method

Question

I noticed that if I write something like:

View view = getView();
foo(error -> view.showError(error));

Android Studio (and probably IntelliJ too) shows the suggestion "Can be replaced with method reference".

Instead, if I write

foo(error -> getView().showError(error));

Android Studio doesn't say anything.

But in both cases I can use method references:

foo(view::showError)

and

foo(getView()::showError)

, respectively.

Are these two forms functionally different? They seem to be doing the same thing, but Android Studio seems to disagree.

you look at size of apk yet? – danny117 Dec 14 '17 at 14:41 — danny117, Dec 14 '17 at 14:41

score 3 · Accepted Answer · answered Dec 14 '17 at 16:20

Note that none of these alternatives is exactly equivalent.

View view = getView();
foo(error -> view.showError(error));

will evaluate getView() immediately, but invoke showError(error) only when the function is actually evaluated (but then, each time). If view is null, a NullPoinerException will be thrown when the function is evaluated.

View view = getView();
foo(view::showError);

will evaluate getView() immediately and throw a NullPoinerException immediately if view is null. showError(error) will be invoked when the function is actually evaluated; at this time, it’s guaranteed that view is not null.

foo(error -> getView().showError(error));

will evaluate getView() only when the function is actually evaluated; it could evaluate to a different result each time. Therefore, a NullPoinerException will be thrown in a particular function evaluation if getView() returns null in this specific evaluation.

Your IDE suggests converting the first variant to the second, because it’s semantically equivalent, as long as view is not null. In contrast, the third variant is significantly different even in the non-null case, as reevaluating getView() each time may lead to different results than the early bound receiver instance. See also “What is the equivalent lambda expression for System.out::println”.

Of course, if getView() is a trivial getter invariably returning the same instance each time, the transformation would be legit, but I suppose that the IDE didn’t look into the implementation of getView() to make such a decision. It’s up to you to decide whether this change is valid in your application.

score 0 · Answer 2 · answered Dec 14 '17 at 14:48

Both are exactly the same, but if you do not use an object more than once, then android studio gives the suggestion of using the method reference directly. Cause there is no point in saving the object reference if you are not going to use it again. Hope it helps.

Method reference of an object in variable vs. returned by method

2 Answers2