Scala向下还是向下循环？

在Scala中，您经常使用迭代器以for递增顺序执行循环，例如：

for(i <- 1 to 10){ code }

您将如何处理，使其从10变为1？我猜10 to 1给定一个空的迭代器（如通常的范围数学）？

我制作了一个Scala脚本，可以通过在迭代器上调用reverse来解决该问题，但是我认为这不好，遵循以下方法吗？

def nBeers(n:Int) = n match {

    case 0 => ("No more bottles of beer on the wall, no more bottles of beer." +
               "\nGo to the store and buy some more, " +
               "99 bottles of beer on the wall.\n")

    case _ => (n + " bottles of beer on the wall, " + n +
               " bottles of beer.\n" +
               "Take one down and pass it around, " +
              (if((n-1)==0)
                   "no more"
               else
                   (n-1)) +
                   " bottles of beer on the wall.\n")
}

for(b <- (0 to 99).reverse)
    println(nBeers(b))

scala> 10 to 1 by -1
res1: scala.collection.immutable.Range = Range(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)

@Randall的答案和黄金一样好，但是为了完整起见，我想添加一些变化：

scala> for (i <- (1 to 10).reverse) {code} //Will count in reverse.

scala> for (i <- 10 to(1,-1)) {code} //Same as with "by", just uglier.

Scala提供了许多向下循环工作的方法。

第一种解决方案：使用“至”和“通过”

//It will print 10 to 0. Here by -1 means it will decremented by -1.     
for(i <- 10 to 0 by -1){
    println(i)
}

第二种解决方案：带有“至”和“反向”

for(i <- (0 to 10).reverse){
    println(i)
}

第三种解决方案：仅使用“至”

//Here (0,-1) means the loop will execute till value 0 and decremented by -1.
for(i <- 10 to (0,-1)){
    println(i)
}

用Pascal编程后，我发现此定义很好用：

implicit class RichInt(val value: Int) extends AnyVal {
  def downto (n: Int) = value to n by -1
  def downtil (n: Int) = value until n by -1
}

使用这种方式：

for (i <- 10 downto 0) println(i)

您可以使用Range类：

val r1 = new Range(10, 0, -1)
for {
  i <- r1
} println(i)

您可以使用 ： for (i <- 0 to 10 reverse) println(i)

for (i <- 10 to (0,-1))

循环将执行直到值== 0，每次减小-1。