Java Stream Collectors.toList() wont compile

Question

Can any one explain why the below code will not compile but the second one does?

Do not compile

private void doNotCompile() {

    List<Integer> out;
    out = IntStream
            .range(1, 10)
            .filter(e -> e % 2 == 0)
            .map(e -> Integer.valueOf(2 * e))
            .collect(Collectors.toList());

    System.out.println(out);
}

Compilation errors on the collect line

The method collect(Supplier, ObjIntConsumer, BiConsumer) in the type IntStream is not applicable for the arguments (Collector>)
Type mismatch: cannot convert from Collector> to Supplier

Compiles

private void compiles() {
    List<Integer> in;

    in = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9);
    List<Integer> out;
    out = in.stream()
            .filter(e -> e % 2 == 0)
            .map(e -> 2 * e)
            .collect(Collectors.toList());

    System.out.println(out);
}

Next time, add the error message instead of just stating it won't compile. — AxelH, Oct 19 '17 at 10:36

Eran · Accepted Answer · 2017-10-19T11:18:00.430

IntStream doesn't have a collect method that accepts a Collector. If you want a List<Integer>, you have to box the IntStream into a Stream<Integer>:

out = IntStream
        .range(1, 10)
        .filter(e -> e % 2 == 0)
        .map(e -> 2 * e)
        .boxed()
        .collect(Collectors.toList());

An alternative to .map().boxed() is mapToObj():

out = IntStream
        .range(1, 10)
        .filter(e -> e % 2 == 0)
        .mapToObj(e -> 2 * e)
        .collect(Collectors.toList ());

or you can use the IntStream collect method:

out = IntStream
        .range(1, 10)
        .filter(e -> e % 2 == 0)
        .map(e -> 2 * e)
        .collect(ArrayList<Integer>::new, ArrayList::add, ArrayList::addAll);

"IntStream doesn't have a collect method that accepts a Collector" ... specifically because there is no such thing as a `List` — Michael, Oct 19 '17 at 10:41

Michael · Answer 2 · 2017-10-19T11:02:45.230

2

In the first example, you are operating over a stream of primitive integers. Primitive integers can't go into a List, basically because generics in Java are less than ideal. Java language designers are working on potentially fixing this.

In the meantime, to solve this you need to box these primitive ints into an Integer wrapper first. See Eran's answer for a code sample.

In the second example, you're already iterating over Integers so It Just Works™.

I thought I was boxing those int to Integers when I did Integer.valueOf in the mapper

The map function of IntStream takes a IntUnaryOperator which is a functional interface that takes a primitive int and returns a primitive int.

The Integer you get from valueOf is unboxed to match the functional interface.

edited Oct 19 '17 at 11:02

answered Oct 19 '17 at 10:48

Michael

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Thanks, but I thought I was boxing those int to Integers when I did Integer.valueOf in the mapper – Vasco Oct 19 '17 at 10:54
Updated my answer :) – Michael Oct 19 '17 at 11:02
I get it now, it also works if I use maptoObj instead of map – Vasco Oct 19 '17 at 11:07

score 1 · Answer 3 · answered Oct 19 '17 at 10:38

The collect method of IntStream is different in the non-compilable case.

public <R> R collect(Supplier<R> supplier, ObjIntConsumer<R> accumulator, BiConsumer<R,R> combiner)

Therefore it does not accept the parameter you supplied to collect

You could get around this by casting List<Integer> to the IntStream

Java Stream Collectors.toList() wont compile

3 Answers3

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