Hey,

a few notes:

1. The (==) function in the second equation of the Maybe instance of (==) is not complete yet, since you need to match on instances of the Maybe type and "Just :: a -> Maybe a". Your idea "Just a == Just a" goes in the right direction, but please note that you can't bind two variable names ("a") in the same equation. You need to give each "boxed value" a different name. I'm sure you can work it out from there.

2. The right hand side of "(x:xs) == (y:ys)" is not implicitly recursive, but rather explicitly! Since we apply the function (==) to the smaller lists "xs" and "ys" until we arrive at a base case.

Greetings,

Tobias

----- Nachricht von trent shipley <[hidden email]> ---------

Datum: Sat, 15 Sep 2018 01:23:47 -0700

Von: trent shipley <[hidden email]>

Antwort an: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell <[hidden email]>

Betreff: [Haskell-beginners] Hutton Ex 8.9.7

An: Haskell Beginners <[hidden email]>

I couldn't get close on my own.

From:

https://github.com/pankajgodbole/hutton/blob/master/exercises.hs

{-

7. Complete the following instance declarations:

instance Eq a => Eq (Maybe a) where

...

instance Eq a => Eq [a] where

...

-}

-- suggested answer

instance Eq a => Eq (Maybe a) where

-- Defines the (==) operation.

Nothing == Nothing = True

Just == Just = True

-- why isn't this Just a == Just a ?

-- My guess is that a and Just a are different types and can't be == in Haskell

_ == _ = False

instance Eq a => Eq [a] where

-- Defines the (==) operation.

[] == [] = True

[x] == [y] = x == y

(x:xs) == (y:ys) = x==y && xs==ys -- I assume this is implicitly recursive.

_ == _ = False

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