Haskell - Ambiguous type variable, why?

Question

Why does the following compile:

{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE OverlappingInstances #-}

class IsList a where
  isList :: a -> Bool

instance IsList a where
  isList x = False

instance IsList [a] where
  isList x = True

main = print (isList 'a') >> print (isList ['a'])

But changing main to this:

main = print (isList 42) >> print (isList [42])

Gives the following error:

Ambiguous type variable `a0' in the constraints:
  (Num a0) arising from the literal `42' at prog.hs:13:22-23
  (IsList a0) arising from a use of `isList' at prog.hs:13:15-20
Probable fix: add a type signature that fixes these type variable(s)
In the first argument of `isList', namely `42'
In the first argument of `print', namely `(isList 42)'
In the first argument of `(>>)', namely `print (isList 42)'

isList surely isn't in the Num class is it? And if not, why the ambiguity?

mightybyte · Accepted Answer · 2013-04-17 05:25:35Z

16

The issue is not with isList but with the constant 42. The constant 'a' has a concrete type of Char. The constant 42 does not have a concrete type.

ghci> :t 42
42 :: Num a => a

The compiler needs a concrete type. It will work if you change main to the following:

main = print (isList (42 :: Int)) >> print (isList [42 :: Int])

answered Apr 17, 2013 at 5:25

mightybyte

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3 Comments

Clinton Over a year ago

Ah, thanks, that makes sense. It can compile if it doesn't know its type. But why doesn't "defaulting" work here? Why not default to Integer?

mightybyte Over a year ago

The compiler doesn't have anything indicating that it should treat that as an Int. If the same identifier was...say...passed into a function that needed an Int, then the type would be unified to Int. But it's not going to make that assumption for you automatically.

hammar Over a year ago

@Clinton: Defaulting is only done if all the constraints only refer to standard classes. IsList is not a standard class. This requirement can be relaxed with ExtendedDefaultRules.

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