CODEWITHSUNDEEP
haskell
user3749167
user3749167

Reputation: 161

Function that behaves differently depending on which type classes the argument belong to

In Haskell, is there any way to declare function it behaves differently depending on whether the type of a argument is an instance of specific type class? For example, can I define genericShow in the following example?

-- If type `a` is an instance of `Show`.
genericShow :: Show a => a -> String
genericShow = show
-- If type `a` is not an instance of `Show`.
genericShow :: a -> String
genericShow _ = "(Cannot be shown)"

> genericShow 3
"3"
> genericShow const
"(Cannot be shown)"

Upvotes: 1

Views: 234

Answers (1)

AntC
AntC

Reputation: 2806

No.

The closest you can get is to use Overlapping instances, with a catch-all instance for anything not having a more specific Show instance.

instance {-# OVERLAPPABLE #-} Show a  where
  show _ = "(Cannot be shown)"

Overlapping instances come with lots of caveats: see topics like 'orphan instances', 'Incoherent instances'. That's particularly awkward with Prelude classes like Show, because there's likely to be lots of instances hidden away in libraries.

As @duplode says, there are many dangers. Almost certainly there's a better way to achieve whatever it is you think you want.

Upvotes: 1

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