function types as default arguments

Good catch!

The reason is: when generating default getters (the methods computing
defaults), the return type is set to be empty, i.e. the typer will infer it.

This is needed for:

  def foo[T](a: T = 1)
  def foo$default$1[T] = 1  // compiler infers Int return type

We can't set the parameter type "T" as return type, the default getter
would not type check.

In your case this gives:
  def transform(s: String, f: String => String = identity _)
  def transform$default$2 = identity _  // compiler infers Nothing => Nothing


What we *should* do: after type checking the parameter ValDef,
use the type computed for it's rhs as return type of the default getter.

It's not obvious how to do that, because when generating the default
getter (Namer), this this information is not yet available. I'll have a look.


Lukas


On Thu, Jun 4, 2009 at 19:38, Paul Phillips <paulp@improving.org> wrote:

Nothing strikes again:

scala> def transform(s: String, f: (String) => String = identity _) = f(s)
transform: (s: String,f: (String) => String)String

scala> transform("abc")
<console>:6: error: type mismatch;
 found   : (Nothing) => Nothing
 required: (String) => String
      transform("abc")
               ^

Oh, but I just noticed it works this way, which lessens the pain.  I
think these two versions ought to compile identically.

scala> def transform(s: String, f: (String) => String =
               identity[String] _) = f(s)
transform: (s: String,f: (String) => String)String

scala> transform("abc")
res0: String = abc

--
Paul Phillips      | Eschew mastication.
Caged Spirit       |
Empiricist         |
i'll ship a pulp   |----------* http://www.improving.org/paulp/ *----------

This is fixed now in trunk (r18069). If the parameter type does not dependon any type parameter of the method, it is re-used as return type of thedefault getter. Otherwise, the return type is inferred.

Lukas

On Fri, Jun 5, 2009 at 10:23, Lukas Rytz <lukas.rytz@epfl.ch> wrote:

Good catch!

The reason is: when generating default getters (the methods computing
defaults), the return type is set to be empty, i.e. the typer will infer it.

This is needed for:

  def foo[T](a: T = 1)
  def foo$default$1[T] = 1  // compiler infers Int return type

We can't set the parameter type "T" as return type, the default getter
would not type check.

In your case this gives:
  def transform(s: String, f: String => String = identity _)
  def transform$default$2 = identity _  // compiler infers Nothing => Nothing


What we *should* do: after type checking the parameter ValDef,
use the type computed for it's rhs as return type of the default getter.

It's not obvious how to do that, because when generating the default
getter (Namer), this this information is not yet available. I'll have a look.


Lukas


On Thu, Jun 4, 2009 at 19:38, Paul Phillips <paulp@improving.org> wrote:

Nothing strikes again:

scala> def transform(s: String, f: (String) => String = identity _) = f(s)
transform: (s: String,f: (String) => String)String

scala> transform("abc")
<console>:6: error: type mismatch;
 found   : (Nothing) => Nothing
 required: (String) => String
      transform("abc")
               ^

Oh, but I just noticed it works this way, which lessens the pain.  I
think these two versions ought to compile identically.

scala> def transform(s: String, f: (String) => String =
               identity[String] _) = f(s)
transform: (s: String,f: (String) => String)String

scala> transform("abc")
res0: String = abc

--
Paul Phillips      | Eschew mastication.
Caged Spirit       |
Empiricist         |
i'll ship a pulp   |----------* http://www.improving.org/paulp/ *----------