On Sat, Apr 13, 2013 at 05:03:57PM -0800, Christopher Howard wrote:
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.
Sorry, what you're trying to do is simply not possible. Type synonyms must always be fully applied. So if you want to make Adjustment an instance of Monad then you have to make it a newtype. However... Adjustment already *is* an instance of Monad! (In particular ((->) e) is an instance for any type e.) So there's no need for you to redeclare an instance yourself. These days I think you just have to import Control.Monad to bring the instance in scope. -Brent