blocks|key|1348403|text|List是一个不可变的single+linked+list。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1348404|使用List[Int](i)，您可以创建一个包含单个元素(i)的列表：|1348405|List(5)
//res0:+List[Int]+=+List(5)|code-block|syntax|javascript|1348406|实际上，您正在调用一个工厂方法List.apply。参见object+List+documentation。|1348407|您应该将Nil用作空列表：|1348408|var+getNumber:+List[Int]+=+Nil
//+or
var+getNumber+=+List.empty[Int]|1348409|使用getNumber.apply(n)+(或getNumber(n))，您可以尝试通过元素在列表中的索引来选择元素。参见List+documentation。所以你让IndexOutOfBoundsException和n+>+getNumber.size在这里。|1348410|您可以使用::方法将元素添加到列表中，如下所示：|1348411|getNumber+=+storeElemets(i)+::+getNumber
//+with+syntax+sugar:
getNumber+::=+storeElemets(i)|1348412|请注意，List是不可变的--您只能创建一个新列表。|1348413|您可以使用from+to+to而不是from+until+to%2B1。方法to+(和until)创建一个Range。您可以在任何scala集合(包括Range)上使用方法map，如下所示：|1348414|val+getNumber+=+(from+until+to).map{i+=>+storeElemets(i)}
//+with+for-comprehensions:
val+getNumber+=+
++for+{
++++i+<-+from+until+to
++}+yield+storeElemets(i)|1348415|请注意，Range上的map创建的是IndexedSeq，而不是List。在本例中不需要List，但是如果需要List，可以使用toList方法将集合转换为List|1348416|val+myList+=+getNumber.toList|1348417|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Linked_list#Singly_linked_list|1|http://www.scala-lang.org/api/2.10.3/index.html#scala.collection.immutable.List%2524|2|http://www.scala-lang.org/api/2.10.3/index.html#scala.collection.immutable.List|3|http://www.scala-lang.org/api/2.10.3/index.html#scala.collection.immutable.Range^0|0|4|B|I|0|0|2|C|T|1|0|0|F|A|Z|4|S|P|1|0|4|3|0|0|2|I|N|C|1P|4|2C|P|32|I|1P|I|2|0|5|2|0|0|4|4|0|5|A|I|F|10|2|15|5|1F|5|21|5|2C|3|1F|5|3|0|0|4|5|B|3|I|A|W|4|18|4|1J|4|1S|6|26|4|0|0^^$0|@$1|2|3|4|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@$9|1P|A|1Q|1|1R]]|E|$]]|$1|F|3|G|5|6|7|1S|8|@$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1X|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1Y|8|@$9|1Z|A|20|B|C]|$9|21|A|22|B|C]]|D|@$9|23|A|24|1|25]]|E|$]]|$1|O|3|P|5|6|7|26|8|@$9|27|A|28|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|J|7|29|8|@]|D|@]|E|$K|L]]|$1|S|3|T|5|6|7|2A|8|@$9|2B|A|2C|B|C]|$9|2D|A|2E|B|C]|$9|2F|A|2G|B|C]|$9|2H|A|2I|B|C]|$9|2J|A|2K|B|C]]|D|@$9|2L|A|2M|1|2N]]|E|$]]|$1|U|3|V|5|6|7|2O|8|@$9|2P|A|2Q|B|C]]|D|@]|E|$]]|$1|W|3|X|5|J|7|2R|8|@]|D|@]|E|$K|L]]|$1|Y|3|Z|5|6|7|2S|8|@$9|2T|A|2U|B|C]]|D|@]|E|$]]|$1|10|3|11|5|6|7|2V|8|@$9|2W|A|2X|B|C]|$9|2Y|A|2Z|B|C]|$9|30|A|31|B|C]|$9|32|A|33|B|C]|$9|34|A|35|B|C]|$9|36|A|37|B|C]|$9|38|A|39|B|C]]|D|@$9|3A|A|3B|1|3C]]|E|$]]|$1|12|3|13|5|J|7|3D|8|@]|D|@]|E|$K|L]]|$1|14|3|15|5|6|7|3E|8|@$9|3F|A|3G|B|C]|$9|3H|A|3I|B|C]|$9|3J|A|3K|B|C]|$9|3L|A|3M|B|C]|$9|3N|A|3O|B|C]|$9|3P|A|3Q|B|C]|$9|3R|A|3S|B|C]|$9|3T|A|3U|B|C]]|D|@]|E|$]]|$1|16|3|17|5|J|7|3V|8|@]|D|@]|E|$K|L]]|$1|18|3|-4|5|6|7|3W|8|@]|D|@]|E|$]]]|19|$1A|$5|1B|1C|1D|E|$1E|1F]]|1G|$5|1B|1C|1D|E|$1E|1H]]|1I|$5|1B|1C|1D|E|$1E|1J]]|1K|$5|1B|1C|1D|E|$1E|1L]]]]

<code>List</code> is an immutable <a href="http://en.wikipedia.org/wiki/Linked_list#Singly_linked_list" rel="nofollow">single linked list</a>.

With <code>List[Int](i)</code> you are creating a list with a single element (<code>i</code>):

<pre><code>List(5)
//res0: List[Int] = List(5)
</code></pre>

Actually you are calling a factory method <code>List.apply</code>. See <a href="http://www.scala-lang.org/api/2.10.3/index.html#scala.collection.immutable.List%24" rel="nofollow">object <code>List</code> documentation</a>.

You should use <code>Nil</code> as an empty list:

<pre><code>var getNumber: List[Int] = Nil
// or
var getNumber = List.empty[Int]
</code></pre>

With <code>getNumber.apply(n)</code> (or <code>getNumber(n)</code>) you are trying to selects an element by its index in the list. See <a href="http://www.scala-lang.org/api/2.10.3/index.html#scala.collection.immutable.List" rel="nofollow"><code>List</code> documentation</a>. So you are getting <code>IndexOutOfBoundsException</code> here with <code>n &gt; getNumber.size</code>.

You could prepend element to list using <code>::</code> method like this:

<pre><code>getNumber = storeElemets(i) :: getNumber
// with syntax sugar:
getNumber ::= storeElemets(i)
</code></pre>

Note that <code>List</code> is immutable - you can only create a new list.

You could use <code>from to to</code> instead of <code>from until to+1</code>. Method <code>to</code> (and <code>until</code>) creates a <a href="http://www.scala-lang.org/api/2.10.3/index.html#scala.collection.immutable.Range" rel="nofollow"><code>Range</code></a>. You could use method <code>map</code> on any scala collection (including <code>Range</code>) like this:

<pre><code>val getNumber = (from until to).map{i =&gt; storeElemets(i)}
// with for-comprehensions:
val getNumber = 
 for {
 i &lt;- from until to
 } yield storeElemets(i)
</code></pre>

Note that <code>map</code> on <code>Range</code> creates <code>IndexedSeq</code> instead of <code>List</code>. You don't need <code>List</code> in this example, but if you need <code>List</code> you can convert collection to <code>List</code> using method <code>toList</code>:

<pre><code>val myList = getNumber.toList
</code></pre>

blocks|key|1354429|text|考虑一下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1354430|for(j<-0+until+numberOfQuery){
++val+lineSum+=+readLine.split("+").map+{+_.toInt+}.sum
++println(s"lineSum+=+$lineSum")
}|code-block|syntax|javascript|1354431|对于读取的每个字符串行，将字符串按空格拆分，将每个拆分项解析为整数，最后将所有整数值求和并打印结果。|1354432|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Consider 

<pre><code>for(j&lt;-0 until numberOfQuery){
 val lineSum = readLine.split(" ").map { _.toInt }.sum
 println(s"lineSum = $lineSum")
}
</code></pre>

which for each string line read, splits the string by the space, parses each split item into an integer, and finally sums up all the integer values and prints the result.

blocks|key|5288375|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5288376|你可以用List(10,+20,+30,+40,+11,+22,+33,+44,+15,+5)声明你的列表。您可以通过追加方法.toList+append从任何集合中构建列表。|ordered-list-item|offset|length|style|CODE|5288377|您的第一个错误是您没有创建大小为to%2B1的列表，它只创建了一个具有单个元素to%2B1的列表。尝试使用List.fill(to%2B1)(0)用to%2B1+0填充列表。|5288378|5288379|此外，忘记分号，使用val而不是var，这样可以确保不变性，这对于大型程序来说非常方便。|5288380|storeElemets是未定义的，但是如果它返回i，那么在第二次迭代中，索引对于单例列表getNumber来说太高了|5288381|如果输入如下，则为|5288382|：|5288383|5288384|0+5+10+20+30+40+11+22+33+44+155+a+b|5288385|其中a和b是所考虑列表的下级索引和上级索引，则应执行以下操作：|5288386|++val+z+=+readLine.split("+")
++val+from+=+z(0).toInt
++val+to+=+z(1).toInt
++val+getNumber+=+z.toList.drop(2).drop(from).take(to-from%2B1)
++val+sum+=+getNumber.sum+//+Equivalent+to+getNumber.foldLeft(0){+case+(res,+n)+=>+res+%2B+n+}
++println(sum)
++println(getNumber.min)|code-block|syntax|javascript|5288387|entityMap|0|LINK|mutability|MUTABLE|url|http://www.scala-lang.org/api/2.10.3/index.html#scala.collection.immutable.List%2524^0|0|4|17|1R|7|0|1D|I|1D|I|0|0|0|A|3|G|3|0|0|C|P|1|19|9|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|1A|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1B|7|@$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]]|8|@]|9|$]]|$1|H|3|I|4|C|6|1G|7|@$D|1H|E|1I|F|G]]|8|@$D|1J|E|1K|1|1L]]|9|$]]|$1|J|3|-4|4|5|6|1M|7|@]|8|@]|9|$]]|$1|K|3|L|4|5|6|1N|7|@$D|1O|E|1P|F|G]|$D|1Q|E|1R|F|G]]|8|@]|9|$]]|$1|M|3|N|4|5|6|1S|7|@$D|1T|E|1U|F|G]|$D|1V|E|1W|F|G]|$D|1X|E|1Y|F|G]]|8|@]|9|$]]|$1|O|3|P|4|5|6|1Z|7|@]|8|@]|9|$]]|$1|Q|3|R|4|C|6|20|7|@]|8|@]|9|$]]|$1|S|3|-4|4|5|6|21|7|@]|8|@]|9|$]]|$1|T|3|U|4|5|6|22|7|@]|8|@]|9|$]]|$1|V|3|W|4|5|6|23|7|@]|8|@]|9|$]]|$1|X|3|Y|4|Z|6|24|7|@]|8|@]|9|$10|11]]|$1|12|3|-4|4|5|6|25|7|@]|8|@]|9|$]]]|13|$14|$4|15|16|17|9|$18|19]]]]

<ol>
<li>You can declare your list using <code>List(10, 20, 30, 40, 11, 22, 33, 44, 15, 5)</code>. You can build a list from any collection by the method <code>.toList</code> append to it.</li>
<li>Your first mistake is that you do not create a list of size to+1, it just creates a list with a single element which is to+1. Try <a href="http://www.scala-lang.org/api/2.10.3/index.html#scala.collection.immutable.List%24" rel="nofollow"><code>List.fill(to+1)(0)</code></a> to fill a list with to+1 zeros.</li>
</ol>

Besides, forget about semicolons and use <code>val</code> instead of <code>var</code> so that you ensure immutability, which is very convenient for high-scale programs.

<code>storeElemets</code> is undefined, but if it returns <code>i</code>, then at the second iteration the index is too high for the singleton list <code>getNumber</code>

<ol>
<li>If the input is the following:

0 5 10 20 30 40 11 22 33 44 15 5
a b</li>
</ol>

where a and b are the lower and upper index of the considered list, then you should do the following:

<pre><code> val z = readLine.split(" ")
 val from = z(0).toInt
 val to = z(1).toInt
 val getNumber = z.toList.drop(2).drop(from).take(to-from+1)
 val sum = getNumber.sum // Equivalent to getNumber.foldLeft(0){ case (res, n) =&gt; res + n }
 println(sum)
 println(getNumber.min)
</code></pre>

blocks|key|5288419|text|如果你真的想要，你可以使用构建器：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5288420|val+b+=+List.newBuilder[Int]+//+creates+a+list+builder

for(i+<-from+until+to%2B1){
++b+%2B=+storeElemets(i)+//+stores+an+element
++println(storeElemets(i))
}

b.result()+//+you+need+to+call+it+to+retrieve+a+collection+from+builder|code-block|syntax|javascript|5288421|附注:没有必要(甚至在scala风格指南中也不推荐)在行尾使用;。此外，您可以在没有类定义的情况下调用println。|offset|length|style|CODE|5288422|entityMap^0|0|0|V|1|1F|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

You can use a builder if you really want:

<pre><code>val b = List.newBuilder[Int] // creates a list builder

for(i &lt;-from until to+1){
 b += storeElemets(i) // stores an element
 println(storeElemets(i))
}

b.result() // you need to call it to retrieve a collection from builder
</code></pre>

P.S. There is no need (and even not recommended in scala style guide) to use <code>;</code> at line ends. Also, you can call <code>println</code> without class definition.

blocks|key|5288462|text|在人们注意到的其他问题之上(可变性与不可变性，分号，不知道storeElemets做了什么，等等)您还选择了一些模棱两可且难以理解的值名称，这些名称与常用术语冲突，从而使代码难以阅读。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5288463|从字里行间看起来，storeElemets就是你所谓的“输入”。|5288464|我的观点是这样的：|5288465|val+storeElemets+=+IndexedSeq(10,20,30,40,11,22,33,44,15,5)+//immutable

for(qry+<-+0+until+numberOfQuery)+{
++val+elems+=+readLne+split+"+"+//split+into+an+array
++val+Array(start,+end)+=+elems+take+2+map+(_.toInt)+//pattern+matching

++val+stored+=+(start+to+end)+map+storeElemets+//also+immutable
++//alternatively,+this+is+more+efficient+in+*most*+cases...
++val+stored+=+storeElemets+drop+start+take+(end-start%2B1)

++//println+is+a+*side+effect*,+so+we+want+to+isolate+it+from+the+other+logic
++stored+foreach+println
++println(stored.min)
}|code-block|syntax|javascript|5288466|entityMap^0|T|C|0|9|C|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

On top of other issues people have noted (mutability vs. immutability, semicolons, not knowing what <code>storeElemets</code> does, etc.) You've also chosen some ambiguous and difficult value names that clash with common terms and help make this code difficult to read.

Reading between the lines it looks as though <code>storeElemets</code> is what you are calling your "input"

My take would be something like this:

<pre><code>val storeElemets = IndexedSeq(10,20,30,40,11,22,33,44,15,5) //immutable

for(qry &lt;- 0 until numberOfQuery) {
 val elems = readLne split " " //split into an array
 val Array(start, end) = elems take 2 map (_.toInt) //pattern matching

 val stored = (start to end) map storeElemets //also immutable
 //alternatively, this is more efficient in *most* cases...
 val stored = storeElemets drop start take (end-start+1)

 //println is a *side effect*, so we want to isolate it from the other logic
 stored foreach println
 println(stored.min)
}
</code></pre>

blocks|key|5288490|text|作为一个略高于初学者的人，我将尝试回答你们的问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5288491|Scala+1)在中声明列表的正确方式是什么？|offset|length|style|BOLD|5288492|有一些正确的方法，例如：|5288493|val+numbersList+=列表(10，20，30，40，11，22，...)|5288494|val+numbersList+=+10+:：20+:：30+:：40+:：...+:：Nil|5288495|2)我的错误是什么?|5288496|您应该注意到，scala在某些方面与其他编程语言非常不同。一方面是你不应该使用var(变量)，只能使用val的，这是因为在scala中你可以有并行工作的操作，而这不能用变量来完成(或者至少是非常困难的)。因此，使用scala的一个非常重要的要点是：|5288497|不要使用变量(除非万不得已)！|5288498|事实证明，变量在大多数情况下并不是必需的。但是，要在没有变量的情况下编程，必须从头开始学习一些东西(至少这是我的经验)。所以val+(Value)是一个不可变的对象，一旦你定义了它就不能改变它。对于列表对象也是如此。这意味着，一旦您使用以下命令定义了列表对象numbersList：|5288499|val+numbersList+=列表(10，20，30，40，11，22，...)|5288500|您不能更改它，这是您在scala中应该采用的工作方式。您可以在numbersList中更改一项，但随后您会得到一个完整的新列表，其中只有一项会因此而更改。因此，您也不应该使用循环(因为对于循环，您需要计数器变量)。与vars一起使用的for-construkt毕竟不是一个循环，而是一个被翻译成带有map、flatMap和filter+(并且可以并行运行)的表达式。|5288501|3)添加数字的正确方法是什么？|5288502|在上面给出的列表对象numbersList的情况下，它只是|5288503|val结果=+numbersList.sum|5288504|我认为，当你知道一种或某些计算机语言，并试图将这种语言的知识转移到scala上时，学习scala是非常困难的，因为概念上的差异。|5288505|举个例子，我推荐scala的发明人Martin+Odersky教授在coursera上的免费在线课程：“scala中的函数式编程原则”。|5288506|当然，任何人都可以随心所欲地使用一种计算机语言，但有时知道一种语言的使用方式是很好的，这样才能最大限度地利用它。|5288507|entityMap^0|0|9|E|0|0|0|0|0|A|0|0|0|0|0|0|0|F|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1E|8|@$D|1F|E|1G|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1H|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1I|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1J|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1K|8|@$D|1L|E|1M|F|G]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1N|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1O|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1P|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1R|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1S|8|@$D|1T|E|1U|F|G]]|9|@]|A|$]]|$1|11|3|12|5|6|7|1V|8|@]|9|@]|A|$]]|$1|13|3|14|5|6|7|1W|8|@]|9|@]|A|$]]|$1|15|3|16|5|6|7|1X|8|@]|9|@]|A|$]]|$1|17|3|18|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|19|3|1A|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|1B|3|-4|5|6|7|20|8|@]|9|@]|A|$]]]|1C|$]]

Beeing just a bit above a beginner, i will try to answer your Questions:

1) What is the correct way to declare a List in Scala?

There are some correct ways, for example:

val numbersList = List(10, 20, 30, 40, 11, 22, ...)

val numbersList = 10 :: 20 :: 30 :: 40 :: ... :: Nil

2) What is my mistake ? 

You should notice that scala is in some aspects very different from other programming languages. One aspect is that you are not supposed to use var's (Variables), only val's.
This is because in scala you can have operations that work in parallel, and this can not be accomplished with variables (or it is at least very difficult). 
So one very importend point in using scala is: 

Don't use variables (unless you absolutely have to) !!!

As it turns out, variables are not necesseay most of the time. But to programm without variables one has to learn some things from the beginning again (This was at least my experience). 
So a val (Value) is an immutable object that can not be changed once you defined it. And this is also true for a List object.
That means, once you define the list-object numbersList with:

val numbersList = List(10, 20, 30, 40, 11, 22, ...)

you can not change it, and this is the way you are supposed to work in scala.
You can change one item in the numbersList, but then you get a complete new List with only that one item changed as a result.
As a result of this, you are not supposed to use loops either (because for a loop you need a counter variable).
And the for-construkt used with vars isn't a loop after all but an expression that is translated in to an expression with map, flatMap and filter (and that can run in parallel)

3) What is the right way to add numbers?

In the case of a given List object numbersList from above, it is just

val result = numbersList.sum

I think it is very hard to learn scala when you know one or some computer languages and try to transfer your knowledge of this languages to scala, because of the conceptional differences.

Insted i would recommend the free online-course on coursera: "functional programming principles in scala" given by Prof. Martin Odersky, the inventor of scala.

Of course anyone can use a computer-language the way he wants, but it is sometimes good to know the way a language is meant to be used, to get the best out of it.

I am just started learning Scala. I want to add numbers through loop and want to find out the minimum number. So, I did this

<pre><code>for(j&lt;-0 until numberOfQuery){
 val z = readLine.split(" ");
 var from = Integer.parseInt(z(0));
 var to = Integer.parseInt(z(1));
 var i=from;
 var getNumber = List[Int](to+1);
 var counter = 0;
 for(i &lt;-from until to+1){
 getNumber.apply(storeElemets(i));
 System.out.println(storeElemets(i));
 }
 System.out.println(getNumber.min);
 }
</code></pre>

for input <code>10 20 30 40 11 22 33 44 15 5</code> and <code>0 5</code> the minimum number will be <code>10</code>. But, it returning <code>java.lang.IndexOutOfBoundsException</code>. So,

1) What is the correct way to declare a <code>List</code> in Scala?
2) What is my mistake ? What is the right way to add numbers.

How to add a number on List in Scala(Beginner)

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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我刚刚开始学习Scala。我想要通过循环添加数字，并想找出最小数字。所以，我做了这个for(j<-0 until numberOfQuery){ val z = readLine.split(" "); var from = Integer.parseInt(z(0)); var to =...

问如何在Scala中添加列表中的数字(初学者)
EN

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问如何在Scala中添加列表中的数字(初学者)EN