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Any reason why scala does not explicitly support dependent types?

There are path dependent types and I think it is possible to express almost all the features of such languages as Epigram or Agda in Scala, but I'm wondering why Scala does not support this more explicitly like it does very nicely in other areas (say, DSLs) ? Anything I'm missing like "it is not necessary" ?

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Ashkan Kh. Nazary Avatar asked Oct 17 '12 13:10

Ashkan Kh. Nazary


1 Answers

Syntactic convenience aside, the combination of singleton types, path-dependent types and implicit values means that Scala has surprisingly good support for dependent typing, as I've tried to demonstrate in shapeless.

Scala's intrinsic support for dependent types is via path-dependent types. These allow a type to depend on a selector path through an object- (ie. value-) graph like so,

scala> class Foo { class Bar } defined class Foo  scala> val foo1 = new Foo foo1: Foo = Foo@24bc0658  scala> val foo2 = new Foo foo2: Foo = Foo@6f7f757  scala> implicitly[foo1.Bar =:= foo1.Bar] // OK: equal types res0: =:=[foo1.Bar,foo1.Bar] = <function1>  scala> implicitly[foo1.Bar =:= foo2.Bar] // Not OK: unequal types <console>:11: error: Cannot prove that foo1.Bar =:= foo2.Bar.               implicitly[foo1.Bar =:= foo2.Bar] 

In my view, the above should be enough to answer the question "Is Scala a dependently typed language?" in the positive: it's clear that here we have types which are distinguished by the values which are their prefixes.

However, it's often objected that Scala isn't a "fully" dependently type language because it doesn't have dependent sum and product types as found in Agda or Coq or Idris as intrinsics. I think this reflects a fixation on form over fundamentals to some extent, nevertheless, I'll try and show that Scala is a lot closer to these other languages than is typically acknowledged.

Despite the terminology, dependent sum types (also known as Sigma types) are simply a pair of values where the type of the second value is dependent on the first value. This is directly representable in Scala,

scala> trait Sigma {      |   val foo: Foo      |   val bar: foo.Bar      | } defined trait Sigma  scala> val sigma = new Sigma {      |   val foo = foo1      |   val bar = new foo.Bar      | } sigma: java.lang.Object with Sigma{val bar: this.foo.Bar} = $anon$1@e3fabd8 

and in fact, this is a crucial part of the encoding of dependent method types which is needed to escape from the 'Bakery of Doom' in Scala prior to 2.10 (or earlier via the experimental -Ydependent-method types Scala compiler option).

Dependent product types (aka Pi types) are essentially functions from values to types. They are key to the representation of statically sized vectors and the other poster children for dependently typed programming languages. We can encode Pi types in Scala using a combination of path dependent types, singleton types and implicit parameters. First we define a trait which is going to represent a function from a value of type T to a type U,

scala> trait Pi[T] { type U } defined trait Pi 

We can than define a polymorphic method which uses this type,

scala> def depList[T](t: T)(implicit pi: Pi[T]): List[pi.U] = Nil depList: [T](t: T)(implicit pi: Pi[T])List[pi.U] 

(note the use of the path-dependent type pi.U in the result type List[pi.U]). Given a value of type T, this function will return a(n empty) list of values of the type corresponding to that particular T value.

Now let's define some suitable values and implicit witnesses for the functional relationships we want to hold,

scala> object Foo defined module Foo  scala> object Bar defined module Bar  scala> implicit val fooInt = new Pi[Foo.type] { type U = Int } fooInt: java.lang.Object with Pi[Foo.type]{type U = Int} = $anon$1@60681a11  scala> implicit val barString = new Pi[Bar.type] { type U = String } barString: java.lang.Object with Pi[Bar.type]{type U = String} = $anon$1@187602ae 

And now here is our Pi-type-using function in action,

scala> depList(Foo) res2: List[fooInt.U] = List()  scala> depList(Bar) res3: List[barString.U] = List()  scala> implicitly[res2.type <:< List[Int]] res4: <:<[res2.type,List[Int]] = <function1>  scala> implicitly[res2.type <:< List[String]] <console>:19: error: Cannot prove that res2.type <:< List[String].               implicitly[res2.type <:< List[String]]                     ^  scala> implicitly[res3.type <:< List[String]] res6: <:<[res3.type,List[String]] = <function1>  scala> implicitly[res3.type <:< List[Int]] <console>:19: error: Cannot prove that res3.type <:< List[Int].               implicitly[res3.type <:< List[Int]] 

(note that here we use Scala's <:< subtype-witnessing operator rather than =:= because res2.type and res3.type are singleton types and hence more precise than the types we are verifying on the RHS).

In practice, however, in Scala we wouldn't start by encoding Sigma and Pi types and then proceeding from there as we would in Agda or Idris. Instead we would use path-dependent types, singleton types and implicits directly. You can find numerous examples of how this plays out in shapeless: sized types, extensible records, comprehensive HLists, scrap your boilerplate, generic Zippers etc. etc.

The only remaining objection I can see is that in the above encoding of Pi types we require the singleton types of the depended-on values to be expressible. Unfortunately in Scala this is only possible for values of reference types and not for values of non-reference types (esp. eg. Int). This is a shame, but not an intrinsic difficulty: Scala's type checker represents the singleton types of non-reference values internally, and there have been a couple of experiments in making them directly expressible. In practice we can work around the problem with a fairly standard type-level encoding of the natural numbers.

In any case, I don't think this slight domain restriction can be used as an objection to Scala's status as a dependently typed language. If it is, then the same could be said for Dependent ML (which only allows dependencies on natural number values) which would be a bizarre conclusion.

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Miles Sabin Avatar answered Sep 29 '22 18:09

Miles Sabin