letrec in Scala? (Immutable way to “Tie the knot?”)

问题

Suppose I have a stupid little case class like so:

case class Foo(name: String, other: Foo)

How can I define a and b immutably such that a.other is b, and b.other is a? Does scala provide some way to "tie the knot"? I'd like to do something like this:

val (a, b): (Foo, Foo) = (Foo("a", b), Foo("b", a)) // Doesn't work.

---

Possibilities

In Haskell I would do this:

data Foo = Foo { name :: String, other :: Foo }

a = Foo "a" b
b = Foo "b" a

Where the bindings to a and b are contained in the same let expression, or at the top level.

Or, without abusing Haskell's automagical letrec capabilities:

(a, b) = fix (\ ~(a', b') -> Foo "a" b', Foo "b" a')

Note the lazy pattern, ~(a', b'), that's important.

回答1:

You want Foo to be unchanged, but laziness in Scala is on the declaration site. It is impossible for Foo to be non-strict without changing it, and the pattern indicated in Haskell only works because Foo, there, is non-strict (that is, Foo "a" b doesn't evaluate b immediately).

Otherwise the solution is pretty much the same, allowing for the hoops necessary to get everything non-strict:

class Foo(name: String, other0: => Foo) { // Cannot be case class, because that mandates strictness
  lazy val other = other0 // otherwise Scala will always reevaluate
}
object Foo {
  def apply(name: String, other: => Foo) = new Foo(name, other)
}

val (a: Foo, b: Foo) = (Foo("a", b), Foo("b", a))

来源：https://stackoverflow.com/questions/10906797/letrec-in-scala-immutable-way-to-tie-the-knot