[英]Java Stream stateful behavior example
The package summary of java.util.stream
states the following: java.util.stream
的软件包摘要声明如下:
An example of a stateful lambda is the parameter to
map()
in:有状态lambda的一个例子是
map()
的参数:Set<Integer> seen = Collections.synchronizedSet(new HashSet<>()); stream.parallel().map(e -> { if (seen.add(e)) return 0; else return e; })...
Here, if the mapping operation is performed in parallel, the results for the same input could vary from run to run, due to thread scheduling differences, whereas, with a stateless lambda expression the results would always be the same.
这里,如果映射操作是并行执行的,则由于线程调度差异,相同输入的结果可能因运行而不同,而对于无状态lambda表达式,结果将始终相同。
I don't understand why this wouldn't produce consistent results, given that the set is synchronized and can only process one element at a time. 我不明白为什么这不会产生一致的结果,因为该集是同步的,并且一次只能处理一个元素。 Can you complete the above example in a way that demonstrates how the result could vary due to parallelization?
您能否以一种演示结果如何因并行化而变化的方式完成上述示例?
List<Integer> numbers = // fill it in with [1, 3, 3, 5]
List<Integer> collected = numbers.stream().parallel().map(...).collect(Collectors.toList());
collected
could contain either [0, 0, 3, 0] or [0, 3, 0, 0], depending on which of the middle two elements was processed first. collected
可以包含[0,0,3,0]或[0,3,0,0],具体取决于首先处理中间两个元素中的哪一个。
I'm not sure if this counts as a stateful per-se (if it does not I'll just delete this), but relying on anything that has state inside map
and the like is bad. 我不确定这是否算作有状态本身(如果它不是我只会删除它),但依赖于任何具有状态内部
map
等的东西都是坏的。
int[] arr = new int[3];
Stream.of(1, 2, 3)
.map(i -> {
arr[i] = i + 5;
return i * 2;
})
.count();
System.out.println(Arrays.toString(arr));
In jdk-9 this will produce an array with zeroes only, since there are no operations that change the Stream size ( flatmap
or filter
), thus map
is never executed. 在jdk-9中,这将生成仅具有零的数组,因为没有操作可以更改Stream大小(
flatmap
或filter
),因此从不执行map
。
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