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 Cl [a]
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 Hammond
Simon L. Peyton Jones
Type Classes in Haskell. ACM Trans. Program. Lang. Syst. 18
 Satish R. Thatte: Semantics of Type Classes Revisited. LISP and
Functional Programming 1994
On Thu, Dec 17, 2009 at 6:40 PM, Simon Peyton-Jones
<[email protected] <mailto:[email protected]>> 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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