On Sun, Apr 14, 2013 at 5:10 PM, Christopher Howard
christopher.how...@frigidcode.com wrote:
type Adjustment a = SaleVariables - a
[...]
instance Monad Adjustment where
(=) = ...
return = ...
Essentially, you can't partially apply type synonyms. I don't recall
the exact reasoning, but
It does not really make sense to make a type synonym an instance of some
class, because a type synonym is just just what is says -- another name for
some type. So making a type synonym for some type T an instance of a class
would be the same as making T itself an instance of the class.
Typically
On Sun, Apr 14, 2013 at 9:28 AM, Chris Wong
chrisyco+haskell-c...@gmail.com wrote:
On Sun, Apr 14, 2013 at 5:10 PM, Christopher Howard
christopher.how...@frigidcode.com wrote:
type Adjustment a = SaleVariables - a
[...]
instance Monad Adjustment where
(=) = ...
return = ...
Maybe you can try curried definition:
type Adjustment = (-) SaleVariables
I had a similar problem awhile ago.
Hth
—
On Sun, Apr 14, 2013 at 9:11 AM, Christopher Howard
christopher.how...@frigidcode.com=mailto:christopher.how...@frigidcode.com;
wrote:
I asked this question in
Oh, I see that I'm late to the party, sorry, wasn't able to push my mail for
some time
—
On Sun, Apr 14, 2013 at 3:09 PM, Daniil Frumin difru...@gmail.com wrote:
Maybe you can try curried definition:
type Adjustment = (-) SaleVariables
I had a similar problem awhile ago.
Hth
—
On Sun,
The point in not allowing partially applied type synonym instances is
that it'd make deciding whether a type is an instance of a class much
harder.
Cf. here[1] for a similar question with the Category class.
-- Steffen
[1] Attached message. Couldn't find it on the archives..
On 04/14/2013
I asked this question in Haskell-beginners, but I haven't heard anything
yet, so I'm forwarding to Cafe.
Original Message
Subject: [Haskell-beginners] Monad instances and type synonyms
Date: Sat, 13 Apr 2013 17:03:57 -0800
From: Christopher Howard
