它可以引用`Data.Monoid。<>

Question

I have following declaration:

data Two a b = Two a b deriving (Eq, Show)

instance (Semigroup a, Semigroup b) => Semigroup (Two a b) where
  (Two a b) <> (Two c d) = Two (a <> c) (b <> d)

And tried it in the prelude:

*Main First Lib MonoidLaws Semi>   (Two a b) <> (Two c d) = Two (a <> c) (b <> d)

<interactive>:10:3: error:
    * Occurs check: cannot construct the infinite type: t1 ~ Two t1 t1
      Expected type: t1 -> t -> b
        Actual type: Two t1 t1 -> Two t t -> Two b b
    * Relevant bindings include
        (<>) :: t1 -> t -> b (bound at <interactive>:10:3)

How can I use mappend function from Semigroup for Two datatype in prelude?

Answer 1

Try it with something that's an instance of Semigroup, like List .

For example:

> (Two "1" "2") <> (Two "3" "4")

Two "13" "24"

In your attempt / example, a and b , c and d are not defined, so Haskell sees them as variables. Because you're using = between them, it's assuming you want to do pattern matching, so it's trying to match them to themselves, which is causing an infinite loop (as that is perfectly valid Haskell — to define values in terms of themselves). This is causing an error, though, because it would imply an infinite type , which it definitely seems is not what you want.

It might be worth starting with some simpler things, possibly. A good basic book which I helped author is http://happylearnhaskelltutorial.com but that doesn't deal with instantiating your own typeclasses. Having said that it'll give you a reasonably good understanding of pattern matching, variables, types and values which you'll need before you move on to understanding typeclasses enough to build your own instances.