[Haskell-cafe] Restrictions on associated types for classes
martin.sulzmann.haskell at googlemail.com
Thu Dec 17 15:15:27 EST 2009
class Cl [a] => Cl a
instance Cl a => Cl [a]
(I omit the type family constructor in the head for simplicyt)
state the same (logical) property:
For each Cl t there must exist Cl [t].
Their operational meaning is different under the dictionary-passing
The instance declaration says we build dictionary Cl [a] given the
dictionary Cl [a]. The super class declaration says that the dictionary for
must be derivable (extractable) from Cl a's dictionary. So, here
we run into a cycle (on the level of terms as well as type inference).
However, if we'd adopt a type-passing translation  (similar to
dynamic method lookup in oo languages) then there isn't
necessarily a cycle (for terms and type inference). Of course,
we still have to verify the 'cyclic' property which smells like
we run into non-termination if we assume some inductive reason
(but we might be fine applying co-induction).
 Cordelia V. Hall, Kevin
Simon L. Peyton
Type Classes in Haskell. ACM Trans. Program. Lang. Syst.
 Satish R. Thatte: Semantics of Type Classes Revisited. LISP and
On Thu, Dec 17, 2009 at 6:40 PM, Simon Peyton-Jones
<simonpj at microsoft.com>wrote:
> | > Hmm. If you have
> | > class (Diff (D f)) => Diff f where
> | >
> | > then if I have
> | > f :: Diff f => ...
> | > f = e
> | > then the constraints available for discharging constraints arising
> | > from e are
> | > Diff f
> | > Diff (D f)
> | > Diff (D (D f))
> | > Diff (D (D (D f)))
> | > ...
> | >
> | > That's a lot of constraints.
> | But isn't it a bit like having an instance
> | Diff f => Diff (D f)
> A little bit. And indeed, could you not provide such instances? That is,
> every time you write an equation for D, such as
> type D (K a) = K Void
> make sure that Diff (K Void) also holds.
> The way you it, when you call f :: Diff f => <blah>, you are obliged to
> pass runtime evidence that (Diff f) holds. And that runtime evidence
> includes as a sub-component runtime evidence that (Diff (D f)) holds. If
> you like the, the evidence for Diff f looks like this:
> data Diff f = MkDiff (Diff (D f)) (D f x -> x -> f x)
> So you are going to have to build an infinite data structure. You can do
> that fine in Haskell, but type inference looks jolly hard.
> For example, suppose we are seeking evidence for
> Diff (K ())
> We might get such evidence from either
> a) using the instance decl
> instance Diff (K a) where ...
> b) using the fact that (D I) ~ K (), we need Diff I, so
> we could use the instance
> instance Diff I
> Having two ways to get the evidence seems quite dodgy to me, even apart
> from the fact that I have no clue how to do type inference for it.
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