Scala 和 dropWhile

Question

我是 HS 的大四学生，也是函数式编程和 Scala 的新手。我在 Scala REPL 中尝试了一些结构，需要一些关于返回响应的指导

//Defined a tuple

scala> val x =(2.0, 3.0, 1)
x: (Double, Double, Int) = (2.0,3.0,1)

//This made sense to me.  Result is a list of values that are of type Ints
scala> x.productIterator.dropWhile(_.isInstanceOf[Double]).toList
res1: List[Any] = List(1)

**//This DID NOT make sense to me.  Why are Double values included?**
scala> x.productIterator.dropWhile(_.isInstanceOf[Int]).toList
res0: List[Any] = List(2.0, 3.0, 1)


//filter operator seems to work
scala> x.productIterator.toList.filter(x => x.isInstanceOf[Double])
res7: List[Any] = List(2.0, 3.0)

Answer 1

Iterator.dropWhile 只要与提供的谓词匹配，就会删除任何值，并返回迭代器的其余部分:

Skips longest sequence of elements of this iterator which satisfy given predicate p, and returns an iterator of the remaining elements.

您传递的提供的谓词对于第一个元素失败，该元素的类型为 Double ，因此它是您具体化为 List[A] 的整个迭代器.

例如，如果您选择在 isInstanceOf[Double] 时放下，您将收到一个包含单个元素的列表 1 :

scala> x.productIterator.dropWhile(_.isInstanceOf[Double]).toList
res13: List[Any] = List(1)