在 Scala 中将函数分配给变量的语法

Question

我正在使用 Scala 2.9.2。我想将一个未命名的函数分配给一个变量，但我似乎无法直接理解语法。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，还不错。此时我想我可以定义一个未命名的函数并将其分配给一个变量，但我似乎无法弄清楚语法。我尝试了几种变体，其中 none 有效。

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>

很抱歉这个基本问题，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() }

或者只是

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，换个方向...

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

Answer 2

我相信以下是您想要的：

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 ()}

在 Scala 中将函数分配给变量的语法

Syntax for assigning a function to a variable in Scala

syntax

scala