Question

Consider an abstract class defining two properties

abstract class A {
  def a: Int
  def b: Int
  // real A has additional members
}

which is the base class for various case classes such as

case class Foo(a: Int, b: Int) extends A
case class Bar(a: Int, b: Int) extends A
// and many more

Goal: I would finally like to be able to create instances of the aforementioned case classes in two ways, namely

val b1 = Bar(1, 2)
val b2 = Bar(1) has 2
assert(b1 == b2) // must hold

Approach: It therefore seems reasonable to define a helper class that defines has and that allows me to partially construct As

case class PartialA(f: Int => A) {
  def has(b: Int) = f(b)
}

Problem: The current machinery doesn’t allow for calls like Bar(1) because this is actually an invocation of Bar.apply(1), that is, of the method apply as defined by the compiler-generated object Bar.

It would be great if I could force the compiler to generate the Bar object as object Bar extends PartialAConstructor, where

abstract class PartialAConstructor{
  def apply(a: Int, b: Int): A // abstract, created when the compiler creates
                               // object Bar
  def apply(a: Int) = PartialA((b: Int) => apply(a, b))
}

However, it doesn’t seem to be possible to influence the generation of companion objects of case classes.

Desired properties:

• Case classes: Foo, Bar etc. should remain case classes because I would like to use the compiler-generated goodies such as structural equality, copy and automatically generated extractors.

• “Full” structural equality: Defining the case classes as

  case class Bar(a: Int)(val b: Int)
  

  is not an option, because the compiler-generated equals method only considers the first list of arguments, and thus the following would hold erroneously:

  assert(Foo(1)(0) == Foo(1)(10))
  

• As little code repetition as possible: For example, it is of course possible to define a

  def Bar(a: Int) = PartialA((b: Int) => Bar(a, b))
  

  but that would have to be done for every case class extending A, that, is Foo, Bar etc.

Answer 1

You could heavily rely on currrying (and on the fact that Foo.apply, as any method, will automatically get promoted to a function) and on a little helper to enhance syntax:

object partially {
  def apply[A1,A2,R]( f: (A1, A2) => R ) = f.curried
  def apply[A1,A2,R]( f: (A1, A2) => R, a1: A1 ) = f.curried( a1 )

  def apply[A1,A2,A3,R]( f: (A1, A2, A3) => R ) = f.curried
  def apply[A1,A2,A3,R]( f: (A1, A2, A3) => R, a1: A1 ) = f.curried( a1 )
  def apply[A1,A2,A3,R]( f: (A1, A2, A3) => R, a1: A1, a2: A2 ) = f.curried( a1 )( a2 )


  def apply[A1,A2,A3,A4,R]( f: (A1, A2, A3, A4) => R ) = f.curried
  def apply[A1,A2,A3,A4,R]( f: (A1, A2, A3, A4) => R, a1: A1 ) = f.curried( a1 )
  def apply[A1,A2,A3,A4,R]( f: (A1, A2, A3, A4) => R, a1: A1, a2: A2 ) = f.curried( a1 )( a2 )
  def apply[A1,A2,A3,A4,R]( f: (A1, A2, A3, A4) => R, a1: A1, a2: A2, a3: A3 ) = f.curried( a1 )( a2 )( a3 )
  // ... and so on, potentially up to 22 args
}

Then you can do:

scala> val x = partially(Foo)(1)
x: Int => Foo = <function1>
scala> x(2)
res37: Foo = Foo(1,2)

If you really want to use your has method (instead of just directly applying the function), throw in an implicit classes on top of that:

implicit class Func1Ops[-A,+R]( val f: A => R ) extends AnyVal { 
  def has( arg: A ): R = f( arg ) 
}

and now you can do:

scala> val x = partially(Foo)(1)
x: Int => Foo = <function1>

scala> x has 2
res38: Foo = Foo(1,2)