I'm really inexperienced at this :
---
{-# OPTIONS_GHC -fglasgow-exts -funbox-strict-fields
-fallow-undecidable-instances -O2 #-}
class Gadget g where
fInit :: g - a - g
data FString = FString !Int !String deriving Show
instance Gadget FString where
fInit (FString n _) s = FString
david48 wrote:
| I'm really inexperienced at this :
class Gadget g where
fInit :: g - a - g
data FString = FString !Int !String deriving Show
instance Gadget FString where
fInit (FString n _) s = FString n (take n s)
The types of:
fInit :: g - a - g
and:
take :: Int - [a] - [a]
On Dec 20, 2007 5:03 PM, Claude Heiland-Allen
[EMAIL PROTECTED] wrote:
You're trying to apply 'take n' to a value of type 'a' ('take n'
requires [a]), moreover putting the value of 'take n s' into the FString
further constrains its type to be [Char] == String.
First of all, thanks a lot for
david48 wrote:
class Gadget g where
fInit :: g - a - g
data FString = FString !Int !String deriving Show
instance Gadget FString where
at this point fInit has this type:
FString - a - FString
fInit (FString n _) s = FString n (take n s)
but your implementation has this type
Tillmann Rendel wrote:
david48 wrote:
class Gadget g where
fInit :: g - a - g
Tillman's two suggestions (below) are probably your answer.
Just to say what everyone else has said in a bunch of different ways:
your class says that for ANY Gadget, fInit will work with ANY OTHER type a.
On Dec 20, 2007 5:26 PM, Tillmann Rendel
[EMAIL PROTECTED] wrote:
at this point fInit has this type:
FString - a - FString
fInit (FString n _) s = FString n (take n s)
but your implementation has this type
FString - String - FString
These types are incompatible, your fInit
On Dec 20, 2007 5:44 PM, david48 [EMAIL PROTECTED] wrote:
fString :: Int - FString
fString n = FString n
Oo do I feel dumb for writing this !
Problem solved :)
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On Dec 20, 2007 5:36 PM, Jules Bean [EMAIL PROTECTED] wrote:
2. Maybe you want lots of possible different as for each g. Then you
make a a parameter of the class too.
3. Maybe you want just one particular a for each g. I.e. g
determines a. Then you can proceed as for (2), but add the
