Question

There are claims that Scala’s type system is Turing complete. My questions are:

1. Is there a formal proof for this?

2. How would a simple computation look like in the Scala type system?

3. Is this of any benefit to Scala – the language? Is this making Scala more “powerful” in some way compared languages without a Turing complete type system?

I guess this applies to languages and type systems in general.

Answer 1

There is a blog post somewhere with a type-level implementation of the SKI combinator calculus, which is known to be Turing-complete.

Turing-complete type systems have basically the same benefits and drawbacks that Turing-complete languages have: you can do anything, but you can prove very little. In particular, you cannot prove that you will actually eventually do something.

One example of type-level computation are the new type-preserving collection transformers in Scala 2.8. In Scala 2.8, methods like map, filter and so on are guaranteed to return a collection of the same type that they were called on. So, if you filter a Set[Int], you get back a Set[Int] and if you map a List[String] you get back a List[Whatever the return type of the anonymous function is].

Now, as you can see, map can actually transform the element type. So, what happens if the new element type cannot be represented with the original collection type? Example: a BitSet can only contain fixed-width integers. So, what happens if you have a BitSet[Short] and you map each number to its string representation?

someBitSet map { _.toString() }

The result would be a BitSet[String], but that’s impossible. So, Scala chooses the most derived supertype of BitSet, which can hold a String, which in this case is a Set[String].

All of this computation is going on during compile time, or more precisely during type checking time, using type-level functions. Thus, it is statically guaranteed to be type-safe, even though the types are actually computed and thus not known at design time.