Type aliases with return type inference works bad

[scabug]

Example below shows the problem:

trait SomeTag
type @@[+A, T] = A with T
def withSomeTag[A](a: A): A @@ SomeTag  = a.asInstanceOf[A @@ SomeTag]
// def need(a: String @@ SomeTag) = ??? // <- it works
// but this not:
type TypeAlias = String @@ SomeTag
def need(a: TypeAlias) = ???
need(withSomeTag("foo"))

Causes:

<console>:13: error: type mismatch;
 found   : String("foo")
 required: SomeTag with String
              need(withSomeTag("foo"))

Changing parameter's type to unaliased version works fine.
Because of this bug, using of scalaz.Tag / shapeless.tag.Tagged is difficult.

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Imported From: https://issues.scala-lang.org/browse/SI-8740?orig=1
Reporter: Arek Burdach (ark_adius)
Affected Versions: 2.11.1

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@paulp said:
Similar to #8709, but with reduced opportunity to deny that it's a bug.

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@retronym said:
This is an interesting one. I was the same corners of subtyping recently to investigate a problem with refinement types with an existential as one of the parents. My patch for that didn't solve this problem.

But I pushed on to analyze why this fails. See the comments in the test case here: https://github.com/retronym/scala/compare/scala:2.11.x...ticket/8740?expand=1

That branch also sketches out a fix, in which I reorder the descent into a refinement's parent types on the LHS to avoid overconstraining type variables.

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@milessabin said:
Any chance of this going in for 2.12.0-M4?

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@retronym said:
M4 is already a bit late, so we're planning to ship in the next few days once we have SAM support merged. M5 would be possible, though.

I was never all that happy with my solution, it seemed very arbitrary.

Maybe this version would be a fraction less ugly.

  def check = parents exists (retry(_, tp2))
  if (tp.isGround)  check
  else suspendingTypeVars(typeVarsInType(tp))(check) else check

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@benhutchison said:
It'd be great to see this fixed. I was wondering why I hadn't hit it myself, which lead me to a partial workaround:

In the failing example, TypeAlias is an alias that refers to another alias @@. Ive had success using aliases for tagged types, but Ive always defined directly in terms of unaliased types, eg

type Distance = Int with TagAs.Tagged[DistanceObj.type]

object TagAs {
  trait Tagged[+U]
}

That is, can't stack 2 aliases over a tagged type, but 1 seems OK.

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@cvogt said:
Jason, did the expected fix end up not being a full solution?

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B. Tommy Jensen (tommy-at-grindvoll.com) said:
I just bumped into this one today. Same behavior with 2.12.0. Any chance that a fix will be included in the near future?

[joroKr21]

Tweak the example a bit and I suspect the outlined fix breaks down again:

trait SomeTag
type @@[+A, T] = A with T
def withSomeTag[A, B](a: A): A @@ B  = a.asInstanceOf[A @@ B]
// def need(a: String @@ SomeTag) = ??? // <- it works
// but this not:
type TypeAlias = String @@ SomeTag
def need(a: TypeAlias) = ???
need(withSomeTag("foo"))

Now we have type variables in both parents of the refined type (not uncommon if we're defining an implicit typeclass instance for any tag) and isGround changes nothing.

[joroKr21]

This works though:

trait Tagged[T]
trait SomeTag
type @@[+A, T] = Tagged[T] with A
def withSomeTag[A, B](a: A): A @@ B  = a.asInstanceOf[A @@ B]
// def need(a: String @@ SomeTag) = ??? // <- it works
// but this not:
type TypeAlias = String @@ SomeTag
def need(a: TypeAlias) = ???
need(withSomeTag("foo"))

[joroKr21]

^ Sorry, now it infers the correct types, but the erasure is wrong 😩

[joroKr21]

A workaround that works:

object Main {
  trait SomeTag
  type @@[+A, T] <: A with T
  def withSomeTag[A, B](a: A): A @@ B  = a.asInstanceOf[A @@ B]
  // def need(a: String @@ SomeTag) = ??? // <- it works
  // but this not:
  type TypeAlias = String @@ SomeTag
  def need(a: TypeAlias) = ???
  need(withSomeTag("foo"))
}

The only downside is that primitives will be boxed.