Java 8流操作执行顺序

Question

List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8);
List<Integer> twoEvenSquares = numbers.stream().filter(n -> {
    System.out.println("filtering " + n);
    return n % 2 == 0;
}).map(n -> {
    System.out.println("mapping " + n);
    return n * n;
}).limit(2).collect(Collectors.toList());


for(Integer i : twoEvenSquares)
{
    System.out.println(i);
}

filtering 1
filtering 2
mapping 2
filtering 3
filtering 4
mapping 4
4
16

filtering 1
filtering 2
filtering 3
filtering 4
mapping 2
mapping 4
4
16

因为在过滤2之后，我们还得再找到一个元素来分层极限(2)，操作，那么为什么输出不像我解释的那样呢？

Answer 1

流是基于拉的。只有终端操作（如collect)才会导致项目被消耗。

概念上，这意味着collect将从limit、map、map、filter和filter从流中询问项。

您的问题中的代码示意性地导致

collect
  limit (0)
    map
      filter
        stream (returns 1)
      /filter (false)
      filter
        stream (returns 2)
      /filter (true)
    /map (returns 4)
  /limit (1)
  limit (1)
    map
      filter
        stream (returns 3)
      /filter (false)
      filter
        stream (returns 4)
      /filter (true)
    /map (returns 16)
  /limit (2)
  limit (2)
  /limit (no more items; limit reached)
/collect