使用并行流列出设置空值

Question

我有以下代码

final String arr[] = {"A", "B", "C", "D"};

final List<String> list1 = new ArrayList<>();

final List<String> list2 = Arrays.asList(arr).parallelStream()

                          .peek(s -> list1.add(s + s))

                          .map(s -> s + s)

                          .collect(Collectors.toList());

System.out.println(list1);

System.out.println(list2);

我得到以下结果：

[null, AA, DD, null]

[AA, BB, CC, DD]

为什么某些元素显示null在 list1 中？我知道我应该使用SynchronizedListwith parallelStream，但null值让我感到惊讶。一旦您向列表中添加了一些内容，就应该设置该插槽，并且下一次执行应该为其添加另一个值。

Answer 1

ArrayList是不是线程安全的集合，并尝试从多个线程更新它parallelStream，你看到的是结果resizing的，内部阵列ArrayList-该操作不是线程安全的要么，你先增加数组并再移动元素它：可能是元素尚未移动而您尝试访问它。为了使它更加明确，调整大小时发生一个新的阵列与增大的尺寸创建-如果充满了这种新的阵列null小号只复制发生之后。真的无法预测会发生什么，只是不要依赖任何这些。编辑我不能将此作为评论发布，但这里有一些代码证明null可以看到：List<Integer> result = IntStream.iterate(0, i -> i + 1)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .takeWhile(i -> {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; String arr[] = { "A", "B", "C", "D", "E", "F", "G", "H", "R", "T" };&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; List<String> list = new ArrayList<>();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Arrays.stream(arr)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .parallel()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .peek(list::add)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .forEach(x -> {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; });&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (list.contains(null)) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.println("It took " + i + " steps to see a null");&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return false;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return true;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; })&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .boxed()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .collect(Collectors.toList());