Create custom list data structure with "toList" function

Question

I want to create my own list data structure called Nodes. Then I will use my ListConverter class, which contains the 'toList' function, and create an instance of it.

data Nodes a = Empty
             | Node a (Nodes a)

class ListConverter a where
    toList :: a -> [Integer]

instance (Integral a) => ListConverter (Nodes a) where
    toList Empty = []
    toList (Node x Empty) = [x]
    toList (Node x y) = x : toList y

GHCi tells me, that the expected type is 'Integer' but is currently 'a'. Im very confused, because in the instance I give a type for a (Integral). Here's the error message:

error:
    * Couldn't match expected type `Integer' with actual type `a'
      `a' is a rigid type variable bound by
        the instance declaration
        at main.hs:7:10-48
    * In the expression: x
      In the expression: [x]
      In an equation for `toList': toList (Node x Empty) = [x]
    * Relevant bindings include
        x :: a (bound at main.hs:9:18)
        toList :: Nodes a -> [Integer] (bound at main.hs:8:5)
  |
9 |     toList (Node x Empty) = [x]
  |                              ^

x is bounded to Integrals but not all Integrals are Integers. I think that's the problem. — Brady Dean
– Brady Dean, Commented May 29, 2019 at 23:30
You said that you want to use ListConverter to "create an instance of it", where by "it" I believe you mean your Nodes type. If this means that you expect toList to give you a value of type Nodes a (for some a) then you will be disappointed, because it would take a value of type Nodes a and give a [Integer]. — Rein Henrichs
– Rein Henrichs, Commented May 29, 2019 at 23:52

typedfern · Accepted Answer · 2019-05-29 23:35:37Z

3

Your instance of ListConverter is expected to accept any value of class Integral for "a", but Integer is a specific type, not a class; you'd have to do this:

instance ListConverter (Nodes Integer) where

That or, conversely, make your ListConverter class capable of producing a list of whatever type your Nodes value contains:

class ListConverter f where
    toList :: f a -> [a]

instance ListConverter Nodes where
    toList Empty = []
    toList (Node x y) = x : toList y

(The second equation for toList - (Node x Empty) - is unnecessary)

answered May 29, 2019 at 23:35

typedfern

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Comments

Robin Zigmond · Accepted Answer · 2019-05-29 23:39:55Z

The problem with this instance is very simply stated. You have given the signature:

toList :: a -> [Integer]

But your attempted instance actually has type Nodes a -> [a]. This doesn't work unless a is the Integer type - yet you've claimed it works for all Integral a. This includes other types such as Int.

One solution is just to restrict your instance:

instance ListConverter (Nodes Integer) where...

This will work - but doesn't in my opinion really respect the spirit of what you probably intended the class for.

The best solution I think is to recognise that both lists and your Nodes type are paramaterised by another type, and define the class in such a way as to do the conversion over a common base class. That sounds more complicated than it is, I just mean:

class ListConverter l where
    toList :: l a -> [a]

Then you can write an instance ListConverter Nodes where..., and just copy your existing toList definition. (Whose middle line, I will point out in passing, is redundant.)

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