路径依赖类型没有路径？

Question

Consider this trivial example:

class Outer {
  case class Inner()
  def test(i: Inner) = {}
}

As expected, this doesn't compile because of a type mismatch:

val o1 = new Outer()
val o2 = new Outer()

o1.test(o2.Inner())  // doesn't compile

What if we want to define a standalone function?

This is no good

def test[X <: Outer](a: X#Inner, b: X#Inner) = {}

because this compiles as if everything is OK

test(o1.Inner(), o2.Inner())

This works

def test(x: Outer)(a: x.Inner, b: x.Inner) = {}

because this compiles:

test(o1)(o1.Inner(), o1.Inner())

and this doesn't:

test(o1)(o1.Inner(), o2.Inner())

However we have to pass an extra Outer argument to test . Is it possible to avoid this? Ideally, the following should work:

test(o1.Inner(), o1.Inner()) // ok
test(o1.Inner(), o2.Inner()) // compilation error

Answer 1

This seems to work:

scala> def t[X <: Outer#Inner, Y <: Outer#Inner](a: X, b: Y)(implicit e: X =:= Y)=1
test: [X <: Outer#Inner, Y <: Outer#Inner](a: X, b: Y)(implicit ev: X =:= Y)Unit

scala> test(o1.Inner(), o1.Inner()) // ok

scala> test(o1.Inner(), o2.Inner())
<console>:15: error: Cannot prove that o1.Inner =:= o2.Inner.
       test(o1.Inner(), o2.Inner())
           ^

You might also want to experiment with two parameter lists as it sometimes affects type inference.

Answer 2

I don't think that, out of the box, it's possible to enforce it in a satisfying way. For instance, one possible solution might be:

scala> def test[X <: Outer#Inner](a: X)(b: X) = ()
test: [X <: Outer#Inner](a: X)(b: X)Unit

scala> test(o1.Inner())(o1.Inner())

scala> test(o1.Inner())(o2.Inner())
<console>:16: error: type mismatch;
 found   : o2.Inner
 required: o1.Inner
       test(o1.Inner())(o2.Inner())
                                ^

Looks good, but you can circumvent it by explicitly passing in the type arguments. (same goes for @OlivierBlanvillain's solution by the way)

scala> test[Outer#Inner](o1.Inner())(o2.Inner())

Now let's try the following:

scala> def test[X <: Outer](a: X#Inner)(b: X#Inner) = ()
test: [X <: Outer](a: X#Inner)(b: X#Inner)Unit

scala> test(o1.Inner())(o2.Inner())

Doesn't work, scalac infers X to be Outer , which isn't specific enough, and we could supply Outer as explicit type argument anyway. We need a way to force X to be a singleton type so that it can only represent the path o1 or o2 , and not some general type that can be inhabited by an infinite amount of values. There is a way. Scala has the marker trait Singleton for this purpose. Let's try it:

scala> def test[X <: Outer with Singleton](a: X#Inner)(b: X#Inner) = ()
test: [X <: Outer with Singleton](a: X#Inner)(b: X#Inner)Unit

scala> test(o1.Inner())(o1.Inner())
<console>:15: error: inferred type arguments [Outer] do not conform to method test's type parameter bounds [X <: Outer with Singleton]
       test(o1.Inner())(o1.Inner())
       ^
<console>:15: error: type mismatch;
 found   : o1.Inner
 required: X#Inner
       test(o1.Inner())(o1.Inner())
                    ^

Now our valid case doesn't work anymore! The problem is that scalac refuses to infer singleton types. We have to pass them in explicitly:

scala> test[o1.type](o1.Inner())(o1.Inner())

The invalid cases don't work anymore:

scala> test(o1.Inner())(o2.Inner())
<console>:16: error: inferred type arguments [Outer] do not conform to method test's type parameter bounds [X <: Outer with Singleton]
       test(o1.Inner())(o2.Inner())
       ^
<console>:16: error: type mismatch;
 found   : o1.Inner
 required: X#Inner
       test(o1.Inner())(o2.Inner())
                    ^

scala> test[o1.type](o1.Inner())(o2.Inner())
<console>:16: error: type mismatch;
 found   : o2.Inner
 required: o1.Inner
       test[o1.type](o1.Inner())(o2.Inner())
                                         ^

scala> test[Outer](o1.Inner())(o2.Inner())
<console>:17: error: type arguments [Outer] do not conform to method test's type parameter bounds [X <: Outer with Singleton]
       test[Outer](o1.Inner())(o2.Inner())
           ^

So this enforces the rules we want, but you have to pass in the types explicitly...

---

EDIT

Actually it turns out you can enforce this without losing type inference and without help from any external libraries, but you're probably not going to like it :-p

META EDIT as pointed out in the comments this can still be circumvented if you try hard enough, so I guess you're stuck with the above solution.

scala> import scala.language.existentials
import scala.language.existentials

scala> def test[X <: x.Inner forSome { val x: Outer }](a: X, b: X) = ()
test: [X <: x.Inner forSome { val x: Outer }](a: X, b: X)Unit

scala> test(o1.Inner(), o1.Inner())

scala> test(o1.Inner(), o2.Inner())
<console>:16: error: inferred type arguments [Outer#Inner] do not conform to method test's type parameter bounds [X <: x.Inner forSome { val x: Outer }]
       test(o1.Inner(), o2.Inner())
       ^
<console>:16: error: type mismatch;
 found   : o1.Inner
 required: X
       test(o1.Inner(), o2.Inner())
                    ^
<console>:16: error: type mismatch;
 found   : o2.Inner
 required: X
       test(o1.Inner(), o2.Inner())
                                ^

scala> test[o1.Inner](o1.Inner(), o2.Inner())
<console>:16: error: type mismatch;
 found   : o2.Inner
 required: o1.Inner
       test[o1.Inner](o1.Inner(), o2.Inner())
                                          ^