爲什麼List（_ *）在Scala匹配表達式中無法匹配List（4,18,52）？

Question

我通過奧賴利編程斯卡拉書的工作，並已運行到與此代碼示例中的絆腳石：

/* 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!"); 
} 
}  

根據文本，列表（_ *）應匹配任何列表中有零個或多個元素，但是當我執行此操作時，List（4,18,52）不匹配，並落入案例_部分（或者，如果刪除了，則拋出MatchError）。

任何想法，爲什麼這不匹配？自從這本書出版以來，這些語言有沒有變化？或者我剛剛有一個「你看不見自己的錯別字」的事情呢？

Answer 1

您使用的是什麼版本的Scala？

在斯卡拉2.8.1.final，它會抱怨最後case是無法訪問。

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> 

而且它適用於匹配空列表。

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>