Is it possible to define a function on a subset of an existing type?

Question

I'm new to Haskell and would like to know whether it's possible to define a function that is only defined on a subset of an already existing type, without actually having to define a new type.

Example: I want to create a function that only accepts even integers (or even natural numbers, etc.) and returns, e.g. that number squared, like:

squared :: 2*Integer -> Integer
squared n = n*n

The above two lines do not work, of course.

I know I could write it like this:

squared' :: Integer -> Integer
squared' n 
  | (even n) = n*n
  | otherwise = error "n is not even!"

or something similar, but I want to know whether something like the non-working example is possible, as well.

I hope this question is not completely stupid (or was already answered) but I really don't know a lot of Haskell yet (so searching for an answer was kind of difficult as well)...

Answer 1

In general no. Such a thing is called a subset type, it's a hallmark of dependent types which Haskell doesn't have. Usually it's implemented by boxing a value with a proof that the value satisfies some property, but since we have no notion of proofs in Haskell, we're stuck.

Usually the way to fake it is with "smart constructors".

newtype Even = Even {unEven :: Integer} deriving (Eq, Show, Ord)

toEven :: Integer -> Maybe Even
toEven a | even a = Just $ Even a
         | otherwise = Nothing

And then hide the Even constructor.

If you really really want it, you can switch to a language that can interop with Haskell that has dependent types (Coq and Agda spring to mind).

Answer 2

No. The type system would need to support refinement types (or full dependent types, as suggested by @jozefg).

Here is a Haskell extension with refinement types.

http://goto.ucsd.edu/~rjhala/liquid/haskell/blog/blog/2013/01/01/refinement-types-101.lhs/