Potential existential variable escape

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scala> val x : ({ val y : T } forSome { type T }) = new { val y = 3 }
x: AnyRef{def y: T} forSome { type T } = $$anon$$1@bb0f79

scala> x.y
res0: T = 3

scala> val z : T = x.y
<console>:5: error: not found: type T
       val z : T = x.y

[scabug]

Imported From: https://issues.scala-lang.org/browse/SI-481?orig=1
Reporter: Geoffrey Alan Washburn (washburn)

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@michelou said:
I don't any problem here: the type name "T" returned by the interpreter can't be used in user code, right !?

[scabug]

Geoffrey Alan Washburn (washburn) said:
I would be happier if it returned something of type

T forSome { type T }

I have not figured out a way to introduce an unsoundness or compiler crash yet, but it is possible to write a reasonable program that won't typecheck:

scala> { val x : ({ val y : T } forSome { type T }) = new { val y = 3 }
     |   new { val z = x.y }
     | }
<console>:5: error: type mismatch;
 found   : $$anon(in object $$iw) where type $$anon(in object $$iw) <: java.lang.Object{def z: T}
 required: $$anon(in value res1) forSome { type $$anon(in value res1) <: java.lang.Object{def z: T}; type T }
  val res1 =
      ^

versus

scala> { val x : ({ val y : T } forSome { type T }) = new { val y = 3 }
     | new { val z : T forSome { type T } = x.y }
     | }
res5: java.lang.Object{def z: T forSome { type T }} = $$anon$$2@5f396f