[Haskell-cafe] Fwd: [Haskell-beginners] Monad instances and type synonyms

[Daniil Frumin]

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.howard at frigidcode.com="mailto:christopher.howard at frigidcode.com">> wrote:

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 <christopher.howard at frigidcode.com> 
Reply-To: The Haskell-Beginners Mailing List - Discussion of primarily 
beginner-level topics related to Haskell <beginners at haskell.org> 
To: Haskell Beginners <beginners at haskell.org> 

I am playing around with some trivial code (for learning purposes) I 
wanted to take 

code: 
-------- 
-- SaleVariables a concrete type defined early 

-- `Adjustment' represents adjustment in a price calculation 
-- Allows functions of type (a -> Adjustment a) to be composed 
-- with an appropriate composition function 
type Adjustment a = SaleVariables -> a 
-------- 

And put it into 

code: 
-------- 
instance Monad Adjustment where 

 (>>=) = ... 
 return = ... 
-------- 

If I try this, I get 

code: 
-------- 
Type synonym `Adjustment' should have 1 argument, but has been given none 
In the instance declaration for `Monad Adjustment' 
-------- 

But if I give an argument, then it doesn't compile either (it becomes a 
"*" kind). And I didn't want to make the type with a regular "data" 
declaration either, because then I have to give it a constructor, which 
doesn't fit with what I want the type to do. 

-- 
frigidcode.com 




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