为什么 Scala 编译器不允许使用默认参数的重载方法？

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 run time?

Example:

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

// This fails, too. Even if there is no position in the argument list,
// where the types are the same.
def foo(a: Int)   (b: Int = 42) = a + b
def foo(a: String)(b: String = "Foo") = a + b

// This is OK:
def foo(a: String)(b: Int) = a + b
def foo(a: Int)   (b: Int = 42) = a + b    

// Even this is OK.
def foo(a: Int)(b: Int) = a + b
def foo(a: Int)(b: String = "Foo") = a + b

val bar = foo(42)_ // This complains obviously ...

Are there any reasons why these restrictions can't be loosened a bit?

Especially when converting heavily overloaded Java code to Scala default arguments are a very important and it isn't nice to find out after replacing plenty of Java methods by one Scala methods that the spec/compiler imposes arbitrary restrictions.

Answer 1

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, eg 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 2

It would be very hard to get a readable and precise spec for the interactions of overloading resolution with default arguments. Of course, for many individual cases, like the one presented here, it's easy to say what should happen. But that is not enough. We'd need a spec that decides all possible corner cases. Overloading resolution is already very hard to specify. Adding default arguments in the mix would make it harder still. That's why we have opted to separate the two.

Answer 3

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 4

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 5

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 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:-)

Answer 7

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)