Scala中的顺序是否影响循环答案

【问题标题】：Does the order in Scala for loop influenceScala中的顺序是否影响循环
【发布时间】：2016-05-24 08:22:24
【问题描述】：

我是 Scala 的新手。我对 foo 循环中的顺序有疑问。

type Occurrences = List[(Char, Int)]

lazy val dictionaryByOccurrences: Map[Occurrences, List[Word]] = dictionary.groupBy(x => wordOccurrences(x))

def wordAnagrams(word: Word): List[Word] = dictionaryByOccurrences.getOrElse(wordOccurrences(word), List())

def combinations(occurrences: Occurrences): List[Occurrences] = occurrences match {
    case List() => List(List())
    case head::tail => {
    for (o <- combinations(tail); x <- 1 to head._2)
    yield (head._1, x) :: o
}

如果我在for循环中改变顺序，那就错了

def combinations(occurrences: Occurrences): List[Occurrences] = occurrences match {
    case List() => List(List())
    case head::tail => {
    for (x <- 1 to head._2; o <- combinations(tail))
    yield (head._1, x) :: o
}

找不到原因

【问题讨论】：

标签： scala

【解决方案1】：

for(x <- xs; y <- ys; ...) yield f(x, y, ...) 的类型构造函数默认与xs 相同。现在combinations的返回类型是List[Occurrences]，那么期望的类型构造函数是List[_]，而1 to n的类型构造函数是Seq[_]。

以下代码有效：

def combinations(occurrences: Occurrences): List[Occurrences] = occurrences match {
    case List() => List(List())
    case head::tail => {
    for (x <- (1 to head._2).toList; o <- combinations(tail))
    yield (head._1, x) :: o
}

这也可以：

import collection.breakOut
def combinations(occurrences: Occurrences): List[Occurrences] = occurrences match {
    case List() => List(List())
    case head::tail => {
      (for (x <- 1 to head._2; o <- combinations(tail))
        yield (head._1, x) :: o)(breakOut)
    }
}

在深度上，for(x <- xs; y <- ys; ...) yield f(...) 等价于xs.flatMap(...)，List#flatMap 的完整签名如下：

def flatMap[B, That](f: (A) ⇒ GenTraversableOnce[B])
                    (implicit bf: CanBuildFrom[List[A], B, That]): That

你可以看到flatMap的返回类型是一个magicThat，默认是List[B]，你可以查看Scala 2.8 CanBuildFrom了解更多信息。

【讨论】：