blocks|key|328062|text|scala>+10+to+1+by+-1
res1:+scala.collection.immutable.Range+=+Range(10,+9,+8,+7,+6,+5,+4,+3,+2,+1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|328063|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>scala&gt; 10 to 1 by -1
res1: scala.collection.immutable.Range = Range(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)
</code></pre>

blocks|key|1072567|text|来自@Randall的答案非常好，但为了完成，我想添加几个变体：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1072568|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.|code-block|syntax|javascript|1072569|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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

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

blocks|key|1616865|text|在用Pascal编写程序后，我发现这个定义很好用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1616866|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
}|code-block|syntax|javascript|1616867|这样使用：|1616868|for+(i+<-+10+downto+0)+println(i)|1616869|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

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

<pre><code>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
}
</code></pre>

Used this way:

<pre><code>for (i &lt;- 10 downto 0) println(i)
</code></pre>

blocks|key|1076032|text|Scala提供了许多在循环中向下工作的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1076033|第一个解决方案:使用"to“和"by”|1076034|//It+will+print+10+to+0.+Here+by+-1+means+it+will+decremented+by+-1.+++++
for(i+<-+10+to+0+by+-1){
++++println(i)
}|code-block|syntax|javascript|1076035|第二种解决方案:使用"to“和"reverse”|1076036|for(i+<-+(0+to+10).reverse){
++++println(i)
}|1076037|第三种解决方案:仅使用"to“|1076038|//Here+(0,-1)+means+the+loop+will+execute+till+value+0+and+decremented+by+-1.
for(i+<-+10+to+(0,-1)){
++++println(i)
}|1076039|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Scala provides many ways to work on downwards in loop. 

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

<pre><code>//It will print 10 to 0. Here by -1 means it will decremented by -1. 
for(i &lt;- 10 to 0 by -1){
 println(i)
}
</code></pre>

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

<pre><code>for(i &lt;- (0 to 10).reverse){
 println(i)
}
</code></pre>

3rd Solution: with "to" only

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

blocks|key|3025447|text|您可以使用Range类：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3025448|val+r1+=+new+Range(10,+0,+-1)
for+{
++i+<-+r1
}+println(i)|code-block|syntax|javascript|3025449|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can use Range class:

<pre><code>val r1 = new Range(10, 0, -1)
for {
 i &lt;- r1
} println(i)
</code></pre>

blocks|key|327097|text|您可以使用：for+(i+<-+0+to+10+reverse)+println(i)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|327098|entityMap^0|6|11|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

You can use : 
<code>for (i &lt;- 0 to 10 reverse) println(i)</code>

blocks|key|1076053|text|for+(i+<-+10+to+(0,-1))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1076054|循环将一直执行到值==+0为止，每次递减-1。|unstyled|1076055|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>for (i &lt;- 10 to (0,-1))
</code></pre>

The loop will execute till the value == 0, decremented each time by -1.

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

<pre><code>for(i &lt;- 1 to 10){ code }
</code></pre>

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

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?

<pre><code>def nBeers(n:Int) = n match {

 case 0 =&gt; ("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 _ =&gt; (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 &lt;- (0 to 99).reverse)
 println(nBeers(b))
</code></pre>

Scala downwards or decreasing for loop?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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