java - 了解 java Stream.limit() 操作

Question

我在学习Stream.limit()说:

Returns a stream consisting of the elements of this stream, truncated to be no longer than maxSize in length.

这可以从以下方面理解:

Stream.of(1,2,3,4,5,6,7,8)
    .limit(3)
    .forEach(i -> {
        System.out.print(i + ",");
    });  //prints: 1,2,3,

但是与其他流方法一起使用时，它具有批量处理元素的效果:

Stream.of(1,2,3,4,5,6,7)
    .filter(i -> {
        System.out.println("Filtering "  + i + ": " + (i%2 == 0));
        return i%2 == 0;
    })
    .map(i-> {
        System.out.println("Mapping " + i + " to " + i*i);
        return i*i;
    })
    .limit(2)
    .forEach(i -> System.out.println("---> " + i));

打印:

Filtering 1: false
Filtering 2: true
Mapping 2 to 4
---> 4
Filtering 3: false
Filtering 4: true
Mapping 4 to 16
---> 16

在这里，您可以看到元素以 2 个为一组进行处理。

我有以下疑问:

1. 为什么它不只处理前两个元素 1 和 2？也就是说，为什么输出不只是:

    Filtering 1: false
    Filtering 2: true
    Mapping 2 to 4
    ---> 4
   

2. 为什么它没有处理最后四个元素 5、6、7 和 8 并打印以下内容？:

    Filtering 5: false
    Filtering 6: true
    Mapping 6 to 36
    ---> 36
    Filtering 7: false
    Filtering 8: true
    Mapping 8 to 64
    ---> 64
   

Answer 1

1. 1 和 2 均已处理。 1 是奇数，因此 filter 将其从流管道中移除。 2 是偶数并通过整个流管道导致 4 在 forEach 中打印出来。

2. 流本质上是惰性的。由于您使用了 limit(2)，因此只有两个元素可以通过 limit 管道步骤。 2 和 4 遍历了整个流管道，导致 4 和 16 在 forEach< 中打印出来.

limit() 是短路状态中间操作，根据 docs :

Short-circuiting operations such as limit(n) or findFirst() can allow computations on infinite streams to complete in finite time.

...

Further, some operations are deemed short-circuiting operations. An intermediate operation is short-circuiting if, when presented with infinite input, it may produce a finite stream as a result. A terminal operation is short-circuiting if, when presented with infinite input, it may terminate in finite time. Having a short-circuiting operation in the pipeline is a necessary, but not sufficient, condition for the processing of an infinite stream to terminate normally in finite time.