How a lambda expression maps into a functional interface?

Question

Let's look at the below code.

    List<String> names = Arrays.asList("Adam", "Brian", "Supun");
    List<Integer> lengths = names.stream()
                                 .map(name -> name.length())
                                 .collect(Collectors.toList());

And simply then will look at the javadoc for streams.map. There the signature for map method appears like this.

<R> Stream<R> map(Function<? super T,? extends R> mapper)

Can somebody please explain how JVM maps the lambda expression we gave (name -> name.length()) in terms of Function<? super T,? extends R> mapper?

What are you trying to accomplish? You shouldn't ever need to create local variables of the Stream type. Are you looking for names.stream().map(...).collect(Collectors.toList()) ? Also, don't use the raw List type. — Michael
– Michael, Commented Oct 16, 2017 at 12:53
@Michael actually I needed the viewers to focus on the lambda expression part. SO I didn't add the latter additions you have mentioned. This is just to understand the mapping of lambda into a functional interface. Nothing else. :)) — Supun Wijerathne
– Supun Wijerathne, Commented Oct 16, 2017 at 12:57
@Michael got your point. Thank You. :)) Hope I have corrected the code. — Supun Wijerathne
– Supun Wijerathne, Commented Oct 16, 2017 at 13:05

Supun Wijerathne · Accepted Answer · 2017-10-16 13:20:59Z

9

A Function is something that takes X and returns Y.

 ? super T     == String
 ? extends R   == Integer

basically with name -> name.length() you are implementing the @FunctionlInterface Function<T,R> with overriding the single abstract method R apply(T t).

You can also shorten that with a method reference :

Stream<Integer> lengths = names.stream().map(String::length);

edited Oct 16, 2017 at 13:20

Supun Wijerathne

13.1k11 gold badges73 silver badges101 bronze badges

answered Oct 16, 2017 at 12:50

Eugene

122k17 gold badges219 silver badges335 bronze badges

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

Comments

ByeBye · Accepted Answer · 2017-10-16 12:51:49Z

2

Check apply method from Function:

R apply(T t);

? extends R is return type, ? super T is taken type

As Function class has only one non-default public method, it can map your lambda to Function instance

answered Oct 16, 2017 at 12:51

ByeBye

7,0168 gold badges36 silver badges71 bronze badges

Collectives™ on Stack Overflow

How a lambda expression maps into a functional interface?

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related