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

Kristoffer Føllesdal kfollesdal at gmail.com
Thu Mar 3 15:04:07 UTC 2016

```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

```