Got it to work this way but it got the wrong one. Still looking. instance Indexable (Tuple2 a) where first (Tuple2 b a) = a On Sun, Jan 24, 2016 at 11:39 PM, Daniel Hinojosa < dhinojosa@evolutionnext.com> wrote:
Quick adjustment, playing around too much with it, that should be:
class Indexable idx where first :: idx a -> a
Problem still exists.
On Sun, Jan 24, 2016 at 11:25 PM, Daniel Hinojosa < dhinojosa@evolutionnext.com> wrote:
I am pretty sure I have a good handle on type classes, but this one is perplexing me in Haskell. I created custom Tuples called Tuple2 and Tuple3 (I know Haskell already has Tuples, just thought I would create my own for this exercise). Then I wanted to create a type class that would have a method called first that would get the first element of the tuple regardless of what kind of Tuple it is. Tuple2, Tuple3, etc.
Here is what I have:
data Tuple3 a b c = Tuple3 a b c deriving (Show)
data Tuple2 a b = Tuple2 a b deriving (Show)
class Indexable idx where first :: idx c -> a
instance Indexable (Tuple2 a) where first (Tuple2 a b) = a
In my main, I try to get call putStrLn $ show $ first $ Tuple2 1 "One"
I was greeted with the following trace: Couldn't match expected type ‘a1’ with actual type ‘a’ ‘a’ is a rigid type variable bound by the instance declaration at TypeClasses.hs:35:10 ‘a1’ is a rigid type variable bound by the type signature for first :: Tuple2 a c -> a1 at TypeClasses.hs:36:4 Relevant bindings include a :: a (bound at TypeClasses.hs:36:18) first :: Tuple2 a c -> a1 (bound at TypeClasses.hs:36:4) In the expression: a In an equation for ‘first’: first (Tuple2 a b) = a
Help is appreciated.