Question

我正在阅读O＆＃39; Reilly Programming Scala 一书，并且使用此代码示例遇到了绊脚石：

/* matching on sequences */
val willWork = List(1, 3, 23, 90 );
val willNotWork = List( 4, 18, 52 );
val empty = List();

for( l <- List(willWork, willNotWork, empty ))
{
    l match
{
  case List( _, 3, _, _ ) => println( "Four elements, with the second being '3'." );
  case List( _* ) => println( "Any other list with zero or more elements" );
  case _ => println( "Uh, oh!" );
}
}

根据文本，List（_ *）应该匹配任何具有零个或多个元素的List，但是当我执行它时，List（4,18,52）不匹配，并且属于case _ section （或者，如果删除了它，则抛出MatchError）。

知道为什么这不匹配？自本书出版以来，语言是否发生过变化，或者我是否只有其中一个“你自己无法看到自己”的错别字。事情还在继续？

Answer 1

您使用的是什么版本的Scala？

在Scala 2.8.1.final中，它会抱怨最后case无法访问。

scala> val willWork = List(1, 3, 23, 90 );
willWork: List[Int] = List(1, 3, 23, 90)

scala> val willNotWork = List( 4, 18, 52 );
willNotWork: List[Int] = List(4, 18, 52)

scala> val empty = List();
empty: List[Nothing] = List()

scala> 

scala> for( l <- List(willWork, willNotWork, empty ))
     | {
     |     l match
     | {
     |   case List( _, 3, _, _ ) => println( "Four elements, with the second being '3'." );
     |   case List( _* ) => println( "Any other list with zero or more elements" );
     |   case _ => println( "Uh, oh!" );
     | }
     | }     
<console>:15: error: unreachable code
         case _ => println( "Uh, oh!" );
                          ^

scala>

它可以很好地匹配空列表。

scala> val willWork = List(1, 3, 23, 90 );
willWork: List[Int] = List(1, 3, 23, 90)

scala> val willNotWork = List( 4, 18, 52 );
willNotWork: List[Int] = List(4, 18, 52)

scala> val empty = List();
empty: List[Nothing] = List()

scala> 

scala> for( l <- List(willWork, willNotWork, empty ))
     | {
     |     l match
     | {
     |   case List( _, 3, _, _ ) => println( "Four elements, with the second being '3'." );
     |   case List( _* ) => println( "Any other list with zero or more elements" );
     | }
     | }
Four elements, with the second being '3'.
Any other list with zero or more elements
Any other list with zero or more elements

scala>

为什么List（_ *）与Scala匹配表达式中的List（4,18,52）不匹配？

1 个答案: