向下或减少for循环?

In Scala, you often use an iterator to do a for loop in an increasing order like:

在Scala中，您经常使用迭代器在递增的顺序中执行for循环:

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

How would you do it so it goes from 10 to 1? I guess 10 to 1 gives an empty iterator (like usual range mathematics)?

怎么做呢，从10到1?我猜10比1给出一个空的迭代器(像通常的范围数学一样)?

I made a Scala script which solves it by calling reverse on the iterator, but it's not nice in my opinion, is the following the way to go?

我做了一个Scala脚本，通过在迭代器上调用反向来解决它，但是在我看来，它不是很好，是下面的方法吗?

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))

5 个解决方案

#1

186

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

#2

The answer from @Randall is good as gold, but for sake of completion I wanted to add a couple of variations:

@Randall的答案很好，但是为了完成我想添加一些变化:

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.

#3

Having programmed in Pascal, I find this definition nice to use:

用Pascal编程，我发现这个定义很好使用:

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
}

Used this way:

使用这种方式:

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

#4

Scala provides many ways to work on downwards in loop.

Scala提供了许多向下循环的方法。

1st Solution: with "to" and "by"

第一个解决方案:用“to”和“by”

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

2nd Solution: With "to" and "reverse"

第二种解决方案:用“to”和“reverse”

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

3rd Solution: with "to" only

第三个解决方案:只“to”。

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

#5

You can use Range class:

你可以使用Range类:

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

#1

186

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

#2