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I've recently got into functional programming and Java 8 lambdas. I have an array of ints and I want to sort it in an ascending order.

The way I am trying to do this with lambda is as follows:

Arrays.stream(intArray).sorted((x, y) -> Integer.compare(x, y) == -1);

The issue with this is that my compiler says:

Error:(12, 32) java: method sorted in interface 
java.util.stream.IntStream cannot be applied to given types;
required: no arguments
found: (x,y)->Int[...]== -1
reason: actual and formal argument lists differ in length

What am I missing here?

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    Streams do not offer a way to sort an array, they allow to process the elements in a sorted order. The original array is not modified and in your case, since there is no actual operation, nothing will happens at all. To sort an array, use Arrays.sort(array); that’ll work for int[] and Integer[]. Commented Apr 2, 2016 at 8:12

3 Answers 3

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A Comparator takes two Object and returns an int.

Here, we're entering with two Object's but are returning a boolean expression (Integer.compare(x, y) == -1)

You actually just need the first part of it

Arrays.stream(intArray)
      .sorted((x, y) -> Integer.compare(x, y))
      .toArray(Integer[]::new);

Since you seem to be using an array of Integer objects (because an IntStream doesn't have a sorted(Comparator) method) you better be using a simple IntStream, it will simplify your life because the int will than be sorted in their natural order.

IntStream#sorted()

Returns a stream consisting of the elements of this stream in sorted order.

Arrays.stream(intArray)
      .mapToInt(x->x)
      .sorted()
      .toArray();
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7 Comments

That was my original approach. But even doing it this way, it still gives me the "Integer.Compare(x, y)" cannot be applied to <lambda expression> , <lambda expression>
@Kobek May you please post the full code? With the intArray, it'll help reproduce the problem.
Both of these are incomplete; because there is no terminal operation, neither does anything. You want a toArray() at the end (or similar.)
@BrianGoetz You were totally right. Is it better now?
Since the result is the sorted array returned by toArray() you should assign the result to a variable (in the first case, assigning it to the existing intArray would be fine, in the second, the result is int[] where the source is Integer[])…
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You can do

Arrays.stream(intArray).sorted().forEach(System.out::println);

The error in your case is because IntStream#sorted() does not take parameters.

DEMO

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Arrays.stream(intArray).sorted() Should be used for processing the array in a specific order. It can be used to sort arrays as explained in other answers but it will generate a new sorted array.

Arrays.stream(intArray)
      .sorted((x, y) -> Integer.compare(x, y))
      .toArray(Integer[]::new);

In case of very big sized data this will not be efficient.

If you want to do in-place sorting, you can use Collections.sort()

Arrays.sort(arr, (x,y) -> Integer.compare(x,y))

Or in the case of objects of type A and sorting based on field P

Arrays.sort(arr, Comparator.comparing(A::getP))

where getP() is the method that gets the field P value.

In case of reversed sorting 

 Arrays.sort(arr, Comparator.comparing(A::getP).reversed())

In case of sorting on 2 fields P and P2

Arrays.sort(arr, Comparator.comparing(A::getP).thenComparing(A::getP2))
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1 Comment

How to sort a raw array like int[] arr with a lambda that has multiple lines (various if elses). Something like this: Arrays.sort(arr, <multi-line lambda>);

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