迭代器上的Scala图不产生副作用 [英] Scala map on iterator does not produce side effects

问题描述

为什么这样，

scala> List(1,2,3,4).iterator.map((x: Int) => println(x))

不打印出

1
2
3
4

而

List(1,2,3,4).map((x: Int) => println(x))
List(1,2,3,4).foreach((x: Int) => println(x))
List(1,2,3,4).iterator.foreach((x: Int) => println(x))

所有做？

换句话说，为什么是地图在一个将T型映射到Unit并具有副作用的迭代器上，不能显示这些副作用？

In other words, why is it that a map on a iterator that maps type T to Unit and has side effects unable to show those side effects?

编辑：

另外为什么下面的lazyMap的调用实际上会计算新的迭代器（提供完整的新的迭代器）从开始到结束，如果迭代器是懒惰的

Also why does the following invocation of lazyMap actually computes the new iterator (provide the complete new iterator) from beginning to end if iterator is lazy?

def lazyMap[T, U](coll: Iterable[T], f: T => U) = new Iterable[U] {
  def iterator = coll.iterator map f
}

scala> lazyMap(List(1,2,3,4), (x: Int) => x + 1)
res4: java.lang.Object with Iterable[Int] = (2, 3, 4, 5)

推荐答案

因为迭代器上的映射是懒惰的，你需要一些严格性：

Cause map on iterator is lazy and you need some strictness:

scala> List(1,2,3,4).iterator.map((x: Int) => println(x))
res0: Iterator[Unit] = non-empty iterator

// nothing actually happened yet, just remember to do this printing things

scala> res0.toList
1
2
3
4
res1: List[Unit] = List((), (), (), ())

当您在迭代器上进行foreach时，很明显您正在做副作用，所以懒惰将是不希望的。我不会这样说，关于地图。

When you doing foreach on iterator it is quite obvious that you're doing side effects, so lazyness will be undesired. I wouldn't said so about map.

UPD

编辑：这种行为的原因是存在toString语句结果的隐式调用，这反过来限制了迭代器 - 自己尝试这个代码：

As for your edit: the reason for such behaviour, is that there is implicit call of toString for statement result which in turn stricts the iterator -- try this code on your own:

scala> { lazyMap(List(1,2,3,4), {(x: Int) => println(x); x + 1}); 1 }

，你会看到这个函数 f 从不称为

and you'll see that function f is never called

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