"Illegal instance declaration" when declaring instance of IsString

Question

I'm writing an application that uses UTF-16 strings, and to make use of the overloaded strings extension I tried to make an IsString instance for it:

import Data.Word ( Word16 ) import Data.String ( IsString(fromString) )  type String16 = [Word16]  instance IsString [Word16] where     fromString = encodeUTF16  encodeUTF16 :: String -> String16 

The problem is, when I try to compile the module, GHC 7.0.3 complains:

Data/String16.hs:35:10:     Illegal instance declaration for `IsString [Word16]'       (All instance types must be of the form (T a1 ... an)        where a1 ... an are *distinct type variables*,        and each type variable appears at most once in the instance head.        Use -XFlexibleInstances if you want to disable this.)     In the instance declaration for `IsString [Word16]' 

If I comment out the instance declaration, it compiles successfully.

Why is this rejected? The instance for [Char] looks pretty much like the same thing, yet it compiles fine. Is there something I've missed?

Answer 1

After having a look through the GHC manuals and around the Haskell wiki (especially the List instance page), I've got a better idea of how this works. Here's a summary of what I've learned:

Problem

The Haskell Report defines an instance declaration like this:

The type (T u₁ … u_k) must take the form of a type constructor T applied to simple type variables u₁, … u_k; furthermore, T must not be a type synonym, and the u_i must all be distinct.

The parts highlighted in bold are the restrictions that tripped me up. In English, they are:

1. Anything after the type constructor must be a type variable.
2. You can't use a type alias (using the type keyword) to get around rule 1.

So how does this relate to my problem?

[Word16] is just another way of writing [] Word16. In other words, [] is the constructor and Word16 is its argument.

So if we try to write:

instance IsString [Word16] 

which is the same as

instance IsString ([] Word16) where ... 

it won't work, because it violates rule 1, as the compiler kindly points out.

Trying to hide it in a type synonym with

type String16 = [Word16] instance IsString String16 where ... 

won't work either, because it violates part 2.

So as it stands, it is impossible to get [Word16] (or a list of anything, for that matter) to implement IsString in standard Haskell.

Enter... (drumroll please)

Solution #1: newtype

The solution @ehird suggested is to wrap it in a newtype:

newtype String16 = String16 { unString16 :: [Word16] } instance IsString String16 where ... 

It gets around the restrictions because String16 is no longer an alias, it's a new type (excuse the pun)! The only downside to this is we then have to wrap and unwrap it manually, which is annoying.

Solution #2: Flexible instances

At the expense of portability, we can drop the restriction altogether with flexible instances:

{-# LANGUAGE FlexibleInstances #-}  instance IsString [Word16] where ... 

This was the solution @[Daniel Wagner] suggested.

Solution #3: Equality constraints

Finally, there's an even less portable solution using equality constraints:

{-# LANGUAGE TypeFamilies #-}  instance (a ~ Word16) => IsString [a] where ... 

This works better with type inference, but is more likely to overlap. See Chris Done's article on the topic.

(By the way, I ended up making a foldl' wrapper around Data.Text.Internal and writing the hash on top of that.)