Why does the Scala compiler disallow overloaded methods with default arguments?

Question

While there might be valid cases where such method overloadings could become ambiguous, why does the compiler disallow code which is neither ambiguous at compile time nor at

Answer 1

Here is a generalization of @Landei answer:

What you really want:

def pretty(tree: Tree, showFields: Boolean = false): String = // ...
def pretty(tree: List[Tree], showFields: Boolean = false): String = // ...
def pretty(tree: Option[Tree], showFields: Boolean = false): String = // ...

Workarround

def pretty(input: CanPretty, showFields: Boolean = false): String = {
  input match {
    case TreeCanPretty(tree)       => prettyTree(tree, showFields)
    case ListTreeCanPretty(tree)   => prettyList(tree, showFields)
    case OptionTreeCanPretty(tree) => prettyOption(tree, showFields)
  }
}

sealed trait CanPretty
case class TreeCanPretty(tree: Tree) extends CanPretty
case class ListTreeCanPretty(tree: List[Tree]) extends CanPretty
case class OptionTreeCanPretty(tree: Option[Tree]) extends CanPretty

import scala.language.implicitConversions
implicit def treeCanPretty(tree: Tree): CanPretty = TreeCanPretty(tree)
implicit def listTreeCanPretty(tree: List[Tree]): CanPretty = ListTreeCanPretty(tree)
implicit def optionTreeCanPretty(tree: Option[Tree]): CanPretty = OptionTreeCanPretty(tree)

private def prettyTree(tree: Tree, showFields: Boolean): String = "fun ..."
private def prettyList(tree: List[Tree], showFields: Boolean): String = "fun ..."
private def prettyOption(tree: Option[Tree], showFields: Boolean): String = "fun ..."

Answer 2

My understanding is that there can be name collisions in the compiled classes with default argument values. I've seen something along these lines mentioned in several threads.

The named argument spec is here: http://www.scala-lang.org/sites/default/files/sids/rytz/Mon,%202009-11-09,%2017:29/named-args.pdf

It states:

 Overloading If there are multiple overloaded alternatives of a method, at most one is
 allowed to specify default arguments.

So, for the time being at any rate, it's not going to work.

You could do something like what you might do in Java, eg:

def foo(a: String)(b: Int) =  a + (if (b > 0) b else 42)

Answer 3

I'd like to cite Lukas Rytz (from here):

The reason is that we wanted a deterministic naming-scheme for the generated methods which return default arguments. If you write

def f(a: Int = 1)

the compiler generates

def f$default$1 = 1

If you have two overloads with defaults on the same parameter position, we would need a different naming scheme. But we want to keep the generated byte-code stable over multiple compiler runs.

A solution for future Scala version could be to incorporate type names of the non-default arguments (those at the beginning of a method, which disambiguate overloaded versions) into the naming schema, e.g. in this case:

def foo(a: String)(b: Int = 42) = a + b
def foo(a: Int)   (b: Int = 42) = a + b

it would be something like:

def foo$String$default$2 = 42
def foo$Int$default$2 = 42

Someone willing to write a SIP proposal?

Answer 4

I can't answer your question, but here is a workaround:

implicit def left2Either[A,B](a:A):Either[A,B] = Left(a)
implicit def right2Either[A,B](b:B):Either[A,B] = Right(b)

def foo(a: Either[Int, String], b: Int = 42) = a match {
  case Left(i) => i + b
  case Right(s) => s + b
}

If you have two very long arg lists which differ in only one arg, it might be worth the trouble...

Answer 5

What worked for me is to redefine (Java-style) the overloading methods.

def foo(a: Int, b: Int) = a + b
def foo(a: Int, b: String) = a + b
def foo(a: Int) = a + "42"
def foo(a: String) = a + "42"

This ensures the compiler what resolution you want according to the present parameters.

Answer 6

One of the possible scenario is

  def foo(a: Int)(b: Int = 10)(c: String = "10") = a + b + c
  def foo(a: Int)(b: String = "10")(c: Int = 10) = a + b + c

The compiler will be confused about which one to call. In prevention of other possible dangers, the compiler would allow at most one overloaded method has default arguments.

Just my guess:-)