On 04/14/2013 04:09 AM, Brent Yorgey wrote:
On Sat, Apr 13, 2013 at 05:03:57PM -0800, Christopher Howard wrote:
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
Thank you everyone for your patience. I believe what you say about it already being an instance of Monad, but I don't seem to have convinced the compiler: code: -------- import Control.Monad -- ...snip... type Adjustment = (->) SaleVariables addTax :: Cash -> Adjustment Cash addTax cash = \v -> cash * (1 + salesTax v) -- obviously silly to add taxes twice, but work with me here testrun = addTax 10.00 >>= \c -> addTax c -------- gives me code: -------- No instance for (Monad ((->) SaleVariables)) arising from a use of `>>=' Possible fix: add an instance declaration for (Monad ((->) SaleVariables)) In the expression: addTax 10.00 >>= \ c -> addTax c In an equation for `testrun': testrun = addTax 10.00 >>= \ c -> addTax c -------- -- frigidcode.com