CODEWITHSUNDEEP
haskell
RandomB
RandomB

Reputation: 3749

How to write map/fmap analogue (a->b) -> F m a -> F m b

Good day/night everyone! I have type:

data FixItem m a = KeepItem|SkipItem|FixItem (m (a -> a))
fixItem f = FixItem $ pure f

and I want to write function mapFix :: (a -> b) -> FixItem m a -> FixItem m b. When I try:

mapFix f SkipItem = SkipItem -- good
mapFix f KeepItem = fixItem f -- error "rigid type"!!!
mapFix f (FixItem mf) = FixItem $ pure (.) <*> (pure f) <*> mf -- too!

So, I get error:

    • Couldn't match type ‘b’ with ‘a’
      ‘b’ is a rigid type variable bound by
        the type signature for:
          mapFix :: forall (m :: * -> *) a b.
                    Applicative m =>
                    (a -> b) -> FixItem m a -> FixItem m b
        at src/test.hs:235:11
      ‘a’ is a rigid type variable bound by
        the type signature for:
          mapFix :: forall (m :: * -> *) a b.
                    Applicative m =>
                    (a -> b) -> FixItem m a -> FixItem m b
        at src/test.hs:235:11
      Expected type: b -> b
        Actual type: a -> b
    • In the first argument of ‘fixItem’, namely ‘f’
      In the expression: fixItem f
      In an equation for ‘mapFix’: mapFix f KeepItem = fixItem f
    • Relevant bindings include
        f :: a -> b (bound at src/test.hs:236:8)
        mapFix :: (a -> b) -> FixItem m a -> FixItem m b
          (bound at src/test.hs:236:1)

How to write mapFix or implement Functor instance for such type (FixItem fixes a to a, not to b, i.e. fix is a -> a, not a -> b)?

Upvotes: 1

Views: 82

Answers (1)

Shersh
Shersh

Reputation: 9179

You can't implement Functor type class for your data type. It's because of a -> a inside one of your constructors. When you have functions, you should be more careful. But in short, you have type variable a in contravariant position so you can't implement Functor over this type variable.

Though you can implement Invariant for your data type. Because a in both covariant and contravariant positions, your data type is invariant functor.

Can help you:

Example of Invariant Functor?

What is a contravariant functor?

Some blog post

Upvotes: 4

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