CODEWITHSUNDEEP
scalafunctional-programming
opensas
opensas

Reputation: 63525

scala: define default parameters in a function (val) vs using a method (def)

I have the following method:

scala> def method_with_default(x: String = "default") = {x + "!"}
method_with_default: (x: String)java.lang.String

scala> method_with_default()
res5: java.lang.String = default!

scala> method_with_default("value")
res6: java.lang.String = value!

I'm trying to achieve the same with a val, but I get a syntax error, like this:

(with no default value, this one compiles ok)

scala> val function_with_default = (x: String) => {x + "!"}
function_with_default: String => java.lang.String = <function1>

(but I couldn't get this one to compile...)

scala> val function_with_default = (x: String = "default") => {x + "!"}
<console>:1: error: ')' expected but '=' found.
       val function_with_default = (x: String = "default") => {x + "!"}
                                              ^

any idea?

Upvotes: 13

Views: 1266

Answers (1)

Kim Stebel
Kim Stebel

Reputation: 42047

There is no way to do this. The best you can get is an object that extends both Function1 and Function0 where the apply method of Function0 calls the other apply method with the default parameter.

val functionWithDefault = new Function1[String,String] with Function0[String] {
  override def apply = apply("default")
  override def apply(x:String) = x + "!"
}

If you need such functions more often, you can factor out the default apply method into an abstract class DefaultFunction1 like this:

val functionWithDefault = new DefaultFunction1[String,String]("default") {
  override def apply(x:String) = x + "!"
}

abstract class DefaultFunction1[-A,+B](default:A)
               extends Function1[A,B] with Function0[B] {
  override def apply = apply(default)
}

Upvotes: 7

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