# type signatures with existentially quantified data constructors

**Avi Pfeffer
**
avi@eecs.harvard.edu

*Thu, 16 Aug 2001 01:11:25 -0400 (EDT)*

Hi all,
I'm trying to write an interpreter for a language embedded in Haskell,
that "inherits" Haskell types and values. So I create a type Expr a, to
represent expressions of type a, and an evaluator that takes an Expr a
and produces an a. (To keep things simple, let's assume there are no
variables.) I start with
data Expr a = Haskell a
| If (Expr Bool) (Expr a) (Expr a)
eval :: Expr a -> a
eval (Haskell x) = x
eval (If x y z) = if eval x then eval y else eval z
Next, applications. To correctly type an application expression, I need
to introduce an existentially quantified type variable:
data Expr a = ... | forall b . Appl (Expr (b->a)) (Expr a)
eval (Appl x y) = (eval x) (eval y)
Works fine. Things get tricky when I try to add pairs. In order to type
a pair expression, I find I need to create a Pair class and introduce a
class constraint:
class Pair a b c | a -> b c, b c -> a
instance Pair (b,c) b c
data Expr a = ... | forall b c . Pair a b c => MkPair (Expr b) (Expr c)
This seems to express the right type relationships. But now I find I
can't write eval for MkPair:
eval (MkPair x y) = (eval x, eval y)
GHC produces the error message
Cannot unify the type-signature variable `a'
with the type `(t, t1)'
Expected type: a
Inferred type: (t, t1)
in the definition of function `eval': (eval x, eval y)
I have tried various ways of adding type annotations, but I can't seem to
make it work. The problem seems to be the combination of class constraint
with existential variables. How can I tell the compiler that it should in
fact unify a with (t,t1) here? Or am I misunderstanding something, and
there is a reason it can't? Perhaps there's a way of typing MkPair
without introducing the Pair class? Or is there a completely different
approach to my problem? All suggestions welcome.
--Avi