Java 8 Lambda Not working?

Question

Hi I have this (what i assume to be) really trivial code:

List<Integer> f = new LinkedList<Integer>();
    Collections.sort(f, (Integer f1, Integer f2) -> {
        Integer.compare(f1,f2);
    });

However, I get the following compile-error:

Cannot convert from Comparator<Integer> to Comparator<? super T>

This isn't very helpful - what is going wrong?

This is the place for the almost mandatory reference to “When to use LinkedList over ArrayList?” as everyone using LinkedList most likely haven’t read it… — Holger
– Holger, Commented Jan 23, 2017 at 14:26
You should always use Lists, where appropriate. But prefer ArrayList. There is almost never a reason for LinkedList nor Vector. — Holger
– Holger, Commented Jan 23, 2017 at 14:58
@bharal if you prefer/use Vectors, then you need to read this (I did too and learned !) : stackoverflow.com/questions/1386275/… — Exception_al
– Exception_al, Commented Jan 23, 2017 at 15:00

Community · Accepted Answer · 2017-05-23 12:32:05Z

12

You can use method reference in this case:

 List<Integer> f = new LinkedList<>();
 Collections.sort(f, Integer::compare);

In the original code there is missing return statement:

 Collections.sort(f, (f1 ,  f2) -> {
        return Integer.compare(f1,f2);
 });

return must be used if lambda contains {}

Same thing without return and brackets:

Collections.sort(f, (f1 ,  f2) -> 
         Integer.compare(f1,f2)
);

A few useful notes from comments section below:

It is possible to just use Collections.sort(f) and rely on natural ordering. by Jean-François Savard

Since Java 8 List interface has sort method which can also be used f.sort(null); f.sort(Comparator.naturalOrder()); or, Collections.sort(f, Comparator.naturalOrder()); by Holger

edited May 23, 2017 at 12:32

CommunityBot

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answered Jan 23, 2017 at 11:41

Anton Balaniuc

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6 Comments

bharal Over a year ago

Um, but what if i just wanted to do it how i did it - is that impossible? What am i doing wrong? Thanks for the answer, but it doesn't answer my question!

bharal Over a year ago

thanks so much, i didn't realise i'd messed up the return {} stuff, good to know.

Anton Balaniuc Over a year ago

@bharal I updated my original answer, but the shortest way is to use method reference.

Jean-François Savard Over a year ago

You could simply do Collections.sort(f) and rely on natural ordering.

Holger Over a year ago

Or f.sort(null);. Alternatively, f.sort(Comparator.naturalOrder()); or, Collections.sort(f, Comparator.naturalOrder()); will do as well.

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