On Wed, Sep 29, 2010 at 9:13 PM, Alexander Solla a...@2piix.com wrote:
On 09/29/2010 02:15 PM, DavidA wrote:
instance Monad (\v - Vect k (Monomial v))
Yes, that is exactly what I am trying to say. And since I'm not allowed to
say
it like that, I was trying to say it using a type synonym
Ryan Ingram ryani.spam at gmail.com writes:
Haskell doesn't have true type functions; what you are really saying is
instance Monad (\v - Vect k (Monomial v))
Yes, that is exactly what I am trying to say. And since I'm not allowed to say
it like that, I was trying to say it using a type
On Wednesday 29 September 2010 23:15:14, DavidA wrote:
Ryan Ingram ryani.spam at gmail.com writes:
Haskell doesn't have true type functions; what you are really saying
is
instance Monad (\v - Vect k (Monomial v))
Yes, that is exactly what I am trying to say. And since I'm not allowed
It's hard. Here's a simple example:
type Foo f = f Int
class C (f :: (* - *) - *) where
thingy :: f [] - f IO
-- Should this ever typecheck? I would say no; there's no way to
unify f [] with [Int].
callThingy :: [Int] - IO Int
callThingy = thingy
-- but what if you say this?
instance C
On Wed, Sep 29, 2010 at 11:15 PM, DavidA polyom...@f2s.com wrote:
Ryan Ingram ryani.spam at gmail.com writes:
Haskell doesn't have true type functions; what you are really saying is
instance Monad (\v - Vect k (Monomial v))
Yes, that is exactly what I am trying to say. And since I'm not
On 09/29/2010 02:15 PM, DavidA wrote:
instance Monad (\v - Vect k (Monomial v))
Yes, that is exactly what I am trying to say. And since I'm not allowed to say
it like that, I was trying to say it using a type synonym parameterised over v
instead.
Why not:
instance Monad ((-) Vect k
On 09/29/2010 09:13 PM, Alexander Solla wrote:
On 09/29/2010 02:15 PM, DavidA wrote:
instance Monad (\v - Vect k (Monomial v))
Yes, that is exactly what I am trying to say. And since I'm not
allowed to say
it like that, I was trying to say it using a type synonym
parameterised over v
David,
Ryan Ingram wrote:
Haskell doesn't have true type functions; what you are really saying
is
instance Monad (\v - Vect k (Monomial v))
Daniel Fischer wrote:
I think there was a theoretical reason why that isn't allowed (making type
inference undecidable? I don't remember, I don't
