Syntax for assigning a function to a variable in Scala

Question

I'm working with Scala 2.9.2. I'd like to assign a unnamed function to a variable, but I can't seem to get the syntax straight. If I define a named function, and then assign the named function to the variable, it seems to work OK, as shown by the session here.

$ 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

So far, so good. At this point I am thinking that I can define an unnamed function and assign it to a variable, but it appears I can't figure out the syntax. I tried several variations and none of them worked.

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>

Sorry for the elementary question, but can someone point out the correct syntax?

Chris Martin · Accepted Answer · 2015-01-28 00:11:40Z

3

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

or just

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

or better yet:

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

And, to take it in a different direction...

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

answered Jan 28, 2015 at 0:11

Chris Martin

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Nami · Accepted Answer · 2015-01-28 00:13:14Z

2

I believe the following is what you want:

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

or more concisely:

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

answered Jan 28, 2015 at 0:13

Nami

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