[Haskell-cafe] Function that put elements in Left or Right side of Either depending on type

[Kristoffer Føllesdal]

I want a function fun :: q -> Either a b where q is of type a or b.  fun shall work in the following way

fun x    gives    Left x     if x :: a
fun x    gives    Right x   if x :: b

Following is a more precise description of what I want. I have function fun1 and fun2

newtype Vec k b = V [(b,k)] deriving (Eq,Ord,Show)

fun1 :: (Num k) => a -> Either a (Vec k a)
fun1 = Left

fun2 :: (Num k) => Vec k a -> Either a (Vec k a)
fun2 = Right

that work in the following way

{-
>>> fun1 6
Left 6

>>> fun2 (V[(6,1)])
Right (V [(6,1)])
-}

I want a overloaded function ‘fun' such that 

{-
>>> fun 6
Left 6

>>> fun (V[(6,1)])
Right (V[(6,1)])
-}

I have tried to use a type class to do this (see code below). But when I try
>>> fun 6

I get the following error

 Could not deduce (Num a0)
    from the context (Num a, Fun a b v)
      bound by the inferred type for ‘it’:
                 (Num a, Fun a b v) => Either b (v b)
      at <interactive>:70:1-5
    The type variable ‘a0’ is ambiguous
    When checking that ‘it’ has the inferred type
      it :: forall a b (v :: * -> *).
            (Num a, Fun a b v) =>
            Either b (v b)
    Probable cause: the inferred type is ambiguous

Is the someone that have know how I can solve this?

Kristoffer

——— CODE ——— 

{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FlexibleInstances #-}

newtype Vec k b = V [(b,k)] deriving (Eq,Ord,Show)

fun1 :: (Num k) => a -> Either a (Vec k a)
fun1 = Left

fun2 :: (Num k) => Vec k a -> Either a (Vec k a)
fun2 = Right

class Fun a b v  where
  fun :: a -> Either b (v b)

instance (Num k) => Fun t t (Vec k) where
  fun = fun1

instance (Num k) => Fun (Vec k t) t (Vec k) where
  fun = fun2