Question

Does anyone know if something like this is possible in Scala:

case class Thing(property:String)

def f(thing:Thing, prop:String = thing.property) = println(prop)

The above code doesn’t compile; giving the error error: not found: value thing at thing.property

The following shows the expected behaviour:

f(Thing("abc"), "123") // prints "123"
f(Thing("abc"))        // prints "abc"

I realise I could make the prop argument an Option[String] and do the check in the function definition, but I was wondering if there was a way around it with the new named/default argument support in 2.8.0.

Answer 1

Yes, it’s possible in Scala 2.8. Here’s a quote from the “Named and Default Arguments in Scala 2.8” design document:

Since the scope of a parameter extends
over all subsequent parameter lists
(and the method body), default
expressions can depend on parameters
of preceding parameter lists (but not
on other parameters in the same
parameter list). Note that when using
a default value which depends on
earlier parameters, the actual
arguments are used, not the default
arguments.

def f(a: Int = 0)(b: Int = a + 1) = b // OK

And another example:

def f[T](a: Int = 1)(b: T = a + 1)(c: T = b)
// generates:
// def f$default$1[T]: Int = 1
// def f$default$2[T](a: Int): Int = a + 1
// def f$default$3[T](a: Int)(b: T): T = b

According to this, your code may look as follows:

scala> case class Thing(property:String)
defined class Thing

scala> def f(thing:Thing)(prop:String = thing.property) = println(prop)
f: (thing: Thing)(prop: String)Unit

scala> f(Thing("abc"))("123")
123

scala> f(Thing("abc"))()
abc