You want all instances of B1 to be instances of A and all instances of B2 to
be instances of A.
This means:
forall a. B1 a => A a
and
forall a. B2 a => A a
I think this is not possible with type-classes and the type class extensions
in Haskell, because the compiler can't construct the proofs
deterministically anymore. The compiler has two choices, use an instance of
B1 to construct a proof for A, or use an instance of B2 to construct a proof
for A.
However, if the compiler has to proof (A Int) and there is no instance (B1
Int), but there is an instance (B2 Int). I can imagine that a compiler is
smart enough to use (B2 Int) ....
Gerrit
Gerrit,
Thanks for your help. Yes, your suggestion below will allow me to give
only an instance declaration for B1 (and get the A declaration for free,
if you will).
However, I meant to suggest that there were other T's which are not
instances of B1 but of B2. I will be back in the same situation for
those T's as I was previously in with T. I can't add
instance B2 a => A a
in addition to
instance B1 a => A a
because of the duplicate instance situation. So, my problem remains.
--
Robin Bate Boerop
On 22-Apr-06, at 2:01 PM, Gerrit van den Geest wrote:
Robin,
The following does NOT work, because of a duplicate instance
declaration for A:
class A a
class B1 b1
class B2 b2
instance B1 x => A x
instance B2 x => A x -- duplicate instance, won't compile
data T = T
instance B1 T
Yes, this doesn't work and I think there is no GHC extension that
supports this.
The following DOES work, but it requires that I explicitly give
instance declarations of A for all instances of B1 and B2:
class A a
class A a => B1 a
class A a => B2 a
data T = T
instance A T -- I don't want to have to specify this!
instance B1 T
If you replace the instance you don't want to specify with:
instance B1 a => A a
and use the following flags to start ghc:
-fglasgow-exts -fallow-undecidable-instances
You can add other datatypes, and you only have to give an instance for
class B1.
Good luck!
Gerrit
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