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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迭代器上的Scala图不产生副作用
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