Question

我正在使用Scala 2.9.2。我想为变量分配一个未命名的函数，但我似乎无法直接获得语法。如果我定义一个命名函数，然后将命名函数分配给变量，它似乎工作正常，如此处的会话所示。

$ scala
Welcome to Scala version 2.9.2 (OpenJDK Server VM, Java 1.7.0_65).
Type in expressions to have them evaluated.
Type :help for more information.

scala> type Foo = String => Int
defined type alias Foo

scala> def myFoo (s : String) : Int = s match {case "a" => 123 case _ => s.length ()}
myFoo: (s: String)Int

scala> val foo : Foo = myFoo
foo: String => Int = <function1>

scala> foo ("b")
res0: Int = 1

scala> foo ("a")
res1: Int = 123

到目前为止，这么好。此时我认为我可以定义一个未命名的函数并将其分配给变量，但似乎我无法弄清楚语法。我尝试了几种变化，但没有一种变化。

scala> val bar : Foo = (s : String) : Int = s match {case "a" => 123 case _ => s.length ()}
<console>:1: error: ';' expected but '=' found.
       val bar : Foo = (s : String) : Int = s match {case "a" => 123 case _ => s.length ()}
                                          ^

scala> val bar : Foo = (s : String) => Int = s match {case "a" => 123 case _ => s.length ()}
<console>:8: error: reassignment to val
       val bar : Foo = (s : String) => Int = s match {case "a" => 123 case _ => s.length ()}
                                           ^

scala> val bar : Foo = ((s : String) => Int) = s match {case "a" => 123 case _ => s.length ()}
<console>:1: error: ';' expected but '=' found.
       val bar : Foo = ((s : String) => Int) = s match {case "a" => 123 case _ => s.length ()}
                                             ^

scala> val bar : Foo = (s : String) : Int => s match {case "a" => 123 case _ => s.length ()}
<console>:1: error: ';' expected but 'match' found.
       val bar : Foo = (s : String) : Int => s match {case "a" => 123 case _ => s.length ()}
                                               ^

scala>

对于基本问题感到抱歉，但有人可以指出正确的语法吗？

Answer 1

val bar: String => Int = s => s match { case "a" => 123 case _ => s.length() }

或只是

val bar = (s: String) => s match { case "a" => 123 case _ => s.length() }

或更好：

val bar: String => Int = { case "a" => 123 case s => s.length() }

而且，从另一个方向来看......

val bar = { case "a" => 123 case s => s.length() }: String => Int

Answer 2

我相信以下是你想要的：

val bar: String => Int = s => s match {case "a" => 123 case _ => s.length ()}

或更简洁：

val bar: String => Int = { case "a" => 123 case s => s.length ()}

用于将函数分配给Scala中的变量的语法

2 个答案: