Java Stream Collectors.toList() wont compile

▼魔方 西西 提交于 2019-12-19 09:28:30

问题


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);
}

回答1:


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);



回答2:


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




回答3:


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.



来源:https://stackoverflow.com/questions/46827927/java-stream-collectors-tolist-wont-compile

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!