Replacing lambda expression with method reference

Question

I have a static method which takes the parameter Stream<Double> stream. Coming from either a arraylist.stream() or Arrays.stream(array).

The method's job is to return the sum of all integers that are divisible by three.

return stream.filter(i -> i.intValue() % 3 == 0).mapToInt(i -> i.intValue()).sum()

This method works, however IntelliJ is suggesting the following:

This inspection reports lambdas which can be replaced with method references.

I'm not overly familiar with method references, especially the referencing instance methods using class names scenario.

I have tried the following which gives an error.

stream.filter(i -> i.intValue() % 3 == 0).mapToInt(Integer::intValue).sum()

Any suggestions?

Since i is a Double, you should use Double::intValue. IntelliJ not only inspects, but also proposes the change. Just hit alt-enter. — mrjink
– mrjink, Commented Oct 7, 2016 at 11:24
If IntelliJ suggest such a replacement, you can direct it to do that replacement. By the way, it would be more efficient to unbox only once, i.e. .mapToInt(Double::intValue).filter(i -> i % 3 == 0) — Holger
– Holger, Commented Oct 7, 2016 at 11:24
@Holger does reversing the order not run the possibility of rounding doubles to ints, which are then divisible by three and change the result? — Liam Ferris
– Liam Ferris, Commented Oct 7, 2016 at 12:08
Your original code, i.intValue() % 3 == 0, already does exactly that. — Holger
– Holger, Commented Oct 7, 2016 at 12:28

Utkan Ozyurek · Accepted Answer · 2016-10-07 11:30:10Z

7

As you said the parameter stream is type of Double, so you should do

stream.mapToInt(Double::intValue).filter(i -> i % 3 == 0).sum()

because you are calling Double class' intValue function.

answered Oct 7, 2016 at 11:30

Utkan Ozyurek

6387 silver badges20 bronze badges

Sign up to request clarification or add additional context in comments.

1 Answer 1