Existential types and type members

Question

What works (Part A)

Suppose I have a trait with a type parameter:

trait A[T]

I can use an existential type to write a method that will take a collection of As that all have the same T:

def foo(as: Seq[A[X]] forSome { type X }) = true

Note that this is different from the following:

def otherFoo(as: Seq[A[X] forSome { type X }]) = true

Or the equivalent:

def otherFoo(as: Seq[A[_]]) = true

In these cases the scope of the existential is inside the Seq, so the As can have different Ts. With my original foo (with the existential scoping over the Seq), the following is fine:

foo(Seq(new A[Int] {}, new A[Int] {}))

But make the type parameters different and it doesn't compile:

scala> foo(Seq(new A[Int] {}, new A[String] {}))
<console>:10: error: type mismatch;
 found   : Seq[A[_ >: java.lang.String with Int]]
 required: Seq[A[X]] forSome { type X }

              foo(Seq(new A[Int] {}, new A[String] {}))
                     ^

This is all pretty straightforward.

What works (Part B)

Now suppose I have a similar trait with a type member instead of a type parameter:

trait B { type T }

I can write a method that will only take a B with some specified T:

scala> def bar[X](b: B { type T = X }) = true
bar: [X](b: B{type T = X})Boolean

scala> bar[Int](new B { type T = Int })
res5: Boolean = true

scala> bar[String](new B { type T = Int })
<console>:10: error: type mismatch;
 found   : java.lang.Object with B
 required: B{type T = String}
              bar[String](new B { type T = Int })
                          ^

Again, this works exactly the way you'd expect it to.

What doesn't work

When we try to write the equivalent of our foo above, but for type members, things get weird.

scala> def baz(bs: Seq[B { type T = X }] forSome { type X }) = true
baz: (as: Seq[B{type T = X}] forSome { type X })Boolean

scala> baz(Seq(new B { type T = Int }, new B { type T = String }))
res7: Boolean = true

That the last line compiles makes no sense to me. I've told it that I want all the type members to be the same. My foo shows that I can do this for type parameters, and bar shows that I can constrain a type based on its type members. But I can't combine the two.

I've tried this on 2.9.2 and 2.10.0-M5.

Motivation

This question is inspired by this one, where my first thought was, oh, just use an existential type (setting aside for a second the issue that it seems to be impossible to get an existential type to scope of the type of a repeated parameter, which would be handy here):

def accept(rs: Seq[RList[Int] { type S = X }] forSome { type X }) = true

But this doesn't actually work—you get the same weird result as in the simplified example above.

Nicolas Nicolas · Accepted Answer · 2012-08-21T13:31:36

I've finally sort it out (at least I hope so). Let's do it the other way. We build our trait:

scala> trait B {type T}
defined trait B

We try and build a sequence of B:

scala> Seq(new B {type T = Int}, new B {type T = String})
res0: Seq[B{type T >: String with Int}] = List($anon$1@592b12d, $anon$2@61ae0436)

Damn, it works! Ok, we don't have an equality for type T but let's play with it:

scala> res0 : (Seq[B {type T = X}] forSome {type X >: String with Int})
res1: Seq[B{type T = X}] forSome { type X >: String with Int } = List($anon$1@592b12d, $anon$2@61ae0436)

It's closer. No wait, it's not closer, It's better than what you've proposed as a parameter of baz, we do not only provide a raw type, we also have a upper bound! Thus, we can clearly pass it to baz. This is why it doesn't work as you expected.