When it's a String use "id", otherwise use "show" - haskell

I have the following piece of code
data Showable = forall a . (Show a) => Showable a
instance Show Showable where
show (Showable a) =
show a
It works quite fine:
> show (Showable 1)
"1"
> show (Showable True)
"True"
But when it's a string, I get unwanted quotes:
> show (Showable "foo")
"\"foo\""
I know it's because of apply show over a string, so it's the same as:
> show "foo"
"\"foo\""
What I want to do, is when it's a String, use id instead of show.
Something like:
instance Show Showable where
show (Showable a) =
case a of
(String _) -> id a
_ -> show a
Is it possible? Any workarounds?

It is possible to do something along these lines but you need some boilerplate unfortunately, so it would probably be better to go about it in a different way.
Here is one way it could be done though (using something equivalent to a dependent sum):
{-# LANGUAGE ExistentialQuantification, GADTs, DataKinds, TypeFamilies #-}
type family StringP a where
StringP String = 'True
StringP a = 'False
data CheckStringness a where
IsTypeString :: CheckStringness String
NotTypeString :: (StringP a) ~ 'False => CheckStringness a
data Showable = forall a. Show a => Showable (CheckStringness a) a
instance Show Showable where
show (Showable IsTypeString str ) = str
show (Showable NotTypeString other) = show other
The difficult part is that you cannot directly reflect a type into a value in the way that you would want to for this, so you have to write a bit of boilerplate code to take care of that.
Example usage:
ghci> show (Showable NotTypeString (123 :: Int))
"123"
ghci> show (Showable NotTypeString ())
"()"
ghci> show (Showable IsTypeString "abc")
"abc"
Like I said though, I would try to approach the problem in a different way entirely (such as Luis Casillas's and ErikR's recommendations in the comments on this question), to avoid being in this situation in this first place. The main reason I demonstrated this is that things similar to this technique may at some point become nicer to work with and have more practical value than they do now, especially as the dependent Haskell initiative continues.

I don't know how to do this, but I'm pretty sure this requires some Haskell extensions. First, there is the equality constraint, like (a ~ String). The ~ notation is explained in this article: http://chrisdone.com/posts/haskell-constraint-trick
Also, there is overlapping instances: https://wiki.haskell.org/GHC/AdvancedOverlap
Here is what I would do (untested code, unlikely to compile):
class Show1 a where
show1 :: a -> String
-- requires some extension since String = [Char]
instance (a ~ String) => Show1 a where
show1 = id
instance Show a => Show1 a where
show1 a = show a
instance Show Showable where
show (Showable a) = show1 a

Related

Take action based on a type parameter's typeclass?

I suspect I have a fundamental misunderstanding to be corrected, so will start with the general concept and then zoom in on the particular instance that lead me to think this way.
Generally speaking, is it possible to write a function whose type signature has a parameterised type, and take different action depending on whether the type parameter belongs to a typeclass?
So for example if you had
data MyTree a = Node { val :: a, left :: Maybe (MyTree a), right :: Maybe (MyTree a) }
prettyPrint :: MyTree a -> String
prettyPrint (Show a => ...) t = show (val t)
prettyPrint t = show "?"
where prettyPrint $ Node 'x' Nothing Nothing would print x while prettyPrint $ Node id Nothing Nothing would print ?.
What lead me here is a few instances where I'm working on a complex, parameterised data type (eg. MyTree), which is progressing fine until I need to do some debugging. When I insert trace statements I find myself wishing my data type parameter derived Show when I use test (Showable) data. But I understand one should never add typeclass constraints in data declarations as the wonderfully enlightening LYAH puts it. That makes sense, I shouldn't have to artificially restrict my data type simply because I want to debug it.
So I end up adding the typeclass constraints to the code I'm debugging instead, but quickly discover they spread like a virus. Every function that calls the low level function I'm debugging also needs the constraint added, until I've basically just temporarily added the constraint to every function so I can get enough test coverage. Now my test code is polluting the code I'm trying to develop and steering it off course.
I thought it would be nice to pattern match instead and leave the constraint out of the signature, or use polymorphism and define debug versions of my function, or otherwise somehow wrap my debug traces in a conditional that only fires if the type parameter is an instance of Show. But in my meandering I couldn't find a way to do this or a sensible alternative.
A good mindset is that from the compiler's point of view, every type is potentially an instance of every class. When a type is not an instance of Show, it just means the instance has not been found yet, possibly not been written yet, but not that it doesn't exist.
Approach 1
...Therefore, trying to make a decision based on whether or not a type is an instance of a class is indeed quite fundamentally flawed. However, what you can do is to write a class that explicitly makes this distinction. For Show this could simply be
class MaybeShow a where
showIfPossible :: a -> Maybe a
A generalizable version is to wrap the following around the Show class:
{-# LANGUAGE GADTs #-}
data ShowDict a where
ShowDict :: Show a => ShowDict a
class MaybeShow a where
maybeShowDict :: Maybe (ShowDict a)
and then
{-# LANGUAGE TypeApplications, ScopedTypeVariables, UnicodeSyntax #-}
showIfPossible :: ∀ a . MaybeShow a => Maybe (a -> String)
showIfPossible = fmap (\ShowDict -> show) (maybeShowDict #a)
Either way, this would still mean you have the MaybeShow constraint polluting your codebase – which is in a sense better than Show as it doesn't preclude unshowable types, but in a sense also worse because it requires adding instance for all the types you need to use (even if they already have a Show instance).
Approach 2
You already seem to have considered adding the constraint to the data type instead. And although the old syntax data Show a => MyTree a = ... should indeed never be used, it is possible to encapsulate instances in data. In fact I already did it above with ShowDict. Rather than obtaining that implicitly via a MaybeShow constraint, you can also just add it optionally to your data type:
data MyTree a = Node { val :: a
, showable :: Maybe (ShowDict a)
, left :: Maybe (MyTree a)
, right :: Maybe (MyTree a) }
Of course, if all you're using the Show instance for is for showing the val of this specific node, then you could instead also just put the result right there:
data MyTree a = Node { val :: a
, valDescription :: Maybe (String)
, left :: Maybe (MyTree a)
, right :: Maybe (MyTree a) }
Now of course you're polluting your codebase in a different way: every function that generates a MyTree value needs to procure a Show instance, or decide it can't. This likely has less of an impact though, and especially not if MyTree is only an example and you have many more functions that just work on abstract containers instead.
Approach 3
At least for the specific case of debugging, but also some other use cases, it might be best use a separate means of turning the Show requirement on and off. The most brute-force way is a good old preprocessor flag:
{-# LANGUAGE CPP #-}
#define DEBUGMODE
-- (This could be controlled from your Cabal file)
prettyPrint ::
#ifdef DEBUGMODE
Show a =>
#endif
MyTree a -> String
#ifdef DEBUGMODE
prettyPrint (Show a => ...) t = show (val t)
#else
prettyPrint t = show "?"
#endif
A bit more refined is a constraint synonym and fitting debug function, that can be swapped out in just a single place:
{-# LANGUAGE ConstraintKinds #-}
#ifdef DEBUGMODE
type DebugShow a = Show a
debugShow :: DebugShow a => a -> String
debugShow = show
#else
type DebugShow a = ()
debugShow :: DebugShow a => a -> String
debugShow _ = "?"
#else
PrettyPrint :: DebugShow a => MyTree a -> String
PrettyPrint t = debugShow (val t)
The latter again pollutes the codebase with constraints, but you never need to write any new instances for these.
CPP is quite a blunt tool, in that it requires selecting globally during compilation whether or not you want to require Show. But it can also be done more confined, with a dedicated type-level flag:
{-# LANGUAGE TypeFamilies, DataKinds #-}
data DebugMode = NoDebug | DebugShowRequired
type family DebugShow mode a where
DebugShow 'NoDebug a = ()
DebugShow 'DebugShowRequired a = Show a
class KnownDebugMode (m :: DebugMode) where
debugShow :: DebugShow m a => a -> String
instance KnownDebugMode 'NoDebug where
debugShow _ = "?"
instance KnownDebugMode 'DebugShowRequired where
debugShow = show
{-# LANGUAGE AllowAmbiguousTypes #-}
prettyPrint :: ∀ m a . DebugShow m a => MyTree a -> String
prettyPrint t = debugShow (val t)
This looks a lot like approach 1, but the nice thing is that you don't need any new instances for individual a types.
The way to use prettyPrint now is to specify the debug mode with a type application. For example you could extract debug- and production-specific versions thus:
prettyPrintDebug :: Show a => MyTree a -> String
prettyPrintDebug = prettyPrint #('DebugShowRequired)
prettyPrintProduction :: MyTree a -> String
prettyPrintProduction = prettyPrint #('NoDebug)
I think the simplest approach is to explicitly define overlapping instances for the unshowable types you want. As #leftaroundabout pointed out this solution forces you to define instances for potencially many many types, for example a -> b, IO a, State s a, Maybe (a -> b), etc...
I am assuming that you mostly want to show a tree of type MyTree (a -> b). If that's the case this might do the trick
{-# LANGUAGE DeriveFunctor #-}
{-# LANGUAGE FlexibleInstances #-}
data MyTree a =
Node { val :: a
, left :: Maybe (MyTree a)
, right :: Maybe (MyTree a)
} deriving (Show, Functor) -- The functor instance is just a easy way to map every val to "?", but is not strictly necessary for this problem
-- Create a class for pretty printing. The is a package which already provides it
class Pretty a where
prettyprint :: a -> String
-- Define an instance when the inner type is showable. (here is simply show, but that's up to you)
instance Show a => Pretty (MyTree a) where
prettyprint = show
-- Define an instance for the function type.
-- Notice that this isn't an instance for "non-showable" types,
-- but only for the function type.
-- The overlapping is necessary to distinguish from the previous instance
instance {-# OVERLAPPING #-} Pretty (MyTree (a -> b)) where
prettyprint = show . fmap (const "?")
main = do
putStrLn
$ prettyprint
$ Node (1 :: Int)
(Just $ Node 2 Nothing Nothing)
Nothing
putStrLn
$ prettyprint
$ Node id
(Just $ Node (+ 1) Nothing Nothing)
Nothing
-- outputs
> Node {val = 1, left = Just (Node {val = 2, left = Nothing, right = Nothing}), right = Nothing}
> Node {val = "?", left = Just (Node {val = "?", left = Nothing, right = Nothing}), right = Nothing}
See the plugin if-instance: https://www.reddit.com/r/haskell/comments/x9k5fl/branching_on_constraints_ifinstance_applications/
{-# Options_GHC -fplugin=IfSat.Plugin #-}
import Data.Constraint.If (IfSat, ifSat)
prettyPrint :: IfSat (Show a) => a -> String
prettyPrint x = ifSat #(Show a) (show x) "?"
This is rarely what you want and if used incorrectly can be used to write unsafeCoerce, but this plugin is a recent development and it's good to keep in your back pocket. Previous solutions required a lot more boilerplate.
OP here. The other answers resoundingly answer the question I asked. After quite some time digesting them and experimenting, I've arrived at a particular solution to my particular fundamental goal, which satisfies me.
It certainly not general or sophisticated. But for me it's a great workaround, so I wanted to leave some breadcrumbs for others:
First I use the CPP trick to define two different trace wrappers, so I don't need to use show in the non-debug code:
{-# LANGUAGE CPP #-}
#define DEBUG
#ifdef DEBUG
import Debug.Trace ( trace )
type Traceable = Char
dTrace :: (Show a) => a -> b -> b
dTrace traceable expr = trace (show traceable) expr
#else
dTrace :: a -> b -> b
dTrace _ expr = expr
#endif
Similarly, I then define two different data types. Both are deriving (Show) but only the debug version actually results in something that will satisfy show.
data MyTree a = Node {
#ifdef DEBUG
val :: Traceable
#else
val :: a
#endif
, left :: Maybe (MyTree a)
, right :: Maybe (MyTree a)
} deriving (Show)
And that's it, the pollution stops there. Everything is controlled by the DEBUG define and the rest of the code remains unperturbed:
workOnTree :: MyTree a -> MyTree a
workOnTree t = dTrace t $ t{left=Just t}
go = workOnTree $ Node 'x' Nothing Nothing
main :: IO ()
main = putStrLn [val go]
If I combine the three code sections and compile with #define DEBUG, it outputs:
Node {val = 'x', left = Nothing, right = Nothing}
x
And with #define DEBUG commented out (and no other changes!), I get:
x
and Node will happily accept non-showable values for val.
Even without the CPP stuff (which, even as a long time fan of the C preprocessor, I can understand is not to all tastes), this is pretty manageable. At the least you could just manually swap a few lines to switch between testing and production.

Return `show a` if (Show a) exists, otherwise its type representation if (Typeable a)

I would like to write
class Described a where
describe :: a -> String
instance {-# OVERLAPPING #-} (Show a) => Described a where
describe = show
instance {-# OVERLAPPABLE #-} (Typeable a) => Described a where
describe = show . typeOf
This won't work because the right hand side of each instance is the same. I thought would be solved by having a look at https://wiki.haskell.org/GHC/AdvancedOverlap but it seems that I need to define instances for many existing types to make any of these solutions work. What would be the best solution here?
The standard trick for guiding instance selection is to make a new type. So:
newtype DescribeViaTypeable a = DVT a
newtype DescribeViaShow a = DVS a
instance Show a => Described (DescribeViaShow a) where describe (DVS x) = show x
instance Typeable a => Described (DescribeViaTypeable a) where describe (DVT x) = show (typeOf x)
Now callers may choose which kind of description they like if both are available, and data types can be explicit about which kind of description they expect to be available for their fields, eliminating any magic.

Haskell show pair of string

How can I define an instance for showing (String, String) structure
instance Show (String, String) where
show (a, b) = show a ++ show b
Thanks!
The short answer is that you can't without a bunch of language extensions that really are better suited to other tasks.
There is already an instance for (Show a, Show b) => Show (a, b), meaning that defining it for (String, String) would overlap with the already existing one. A better choice would be to write your own showStrTuple as
showStrTuple :: (String, String) -> String
showStrTuple (a, b) = show a ++ show b
Alternatively, if you really want to use show on it, make a newtype (which are designed for defining new typeclasses that would otherwise conflict with existing ones):
newtype StrTuple = StrTuple { unStrTuple :: (String, String) } deriving (Eq)
instance Show StrTuple where
show (StrTuple (a, b)) = show a ++ show b
Then you just construct it with
show $ StrTuple ("hello", "world")
If you'd used proper indentation, and switched on the altogether harmless -XFlexibleInstances
{-# LANGUAGE FlexibleInstances #-}
instance Show (String, String) where
show (a, b) = show a ++ show b
then this instance would, in itself, work (you need to switch on -XFlexibleInstances). However, it won't compile because a strictly more general instance
instance (Show a, Show b) => Show (a, b) where
show (a, b) = "(" ++ show a ++ "," ++ show b ++ ")"
is already defined in the prelude. If you're determined to override that one then you also need to switch on -XOverlappingInstances. But this one is not so harmless; in fact it's evil: overlapping instances can lead to all kinds of trouble, and for your specific definition the instance also doesn't comply with the requirement that read . show ≡ id.

Why constraints on data are a bad thing?

I know this question has been asked and answered lots of times but I still don't really understand why putting constraints on a data type is a bad thing.
For example, let's take Data.Map k a. All of the useful functions involving a Map need an Ord k constraint. So there is an implicit constraint on the definition of Data.Map. Why is it better to keep it implicit instead of letting the compiler and programmers know that Data.Map needs an orderable key.
Also, specifying a final type in a type declaration is something common, and one can see it as a way of "super" constraining a data type.
For example, I can write
data User = User { name :: String }
and that's acceptable. However is that not a constrained version of
data User' s = User' { name :: s }
After all 99% of the functions I'll write for the User type don't need a String and the few which will would probably only need s to be IsString and Show.
So, why is the lax version of User considered bad:
data (IsString s, Show s, ...) => User'' { name :: s }
while both User and User' are considered good?
I'm asking this, because lots of the time, I feel I'm unnecessarily narrowing my data (or even function) definitions, just to not have to propagate constraints.
Update
As far as I understand, data type constraints only apply to the constructor and don't propagate. So my question is then, why do data type constraints not work as expected (and propagate)? It's an extension anyway, so why not have a new extension doing data properly, if it was considered useful by the community?
TL;DR:
Use GADTs to provide implicit data contexts.
Don't use any kind of data constraint if you could do with Functor instances etc.
Map's too old to change to a GADT anyway.
Scroll to the bottom if you want to see the User implementation with GADTs
Let's use a case study of a Bag where all we care about is how many times something is in it. (Like an unordered sequence. We nearly always need an Eq constraint to do anything useful with it.
I'll use the inefficient list implementation so as not to muddy the waters over the Data.Map issue.
GADTs - the solution to the data constraint "problem"
The easy way to do what you're after is to use a GADT:
Notice below how the Eq constraint not only forces you to use types with an Eq instance when making GADTBags, it provides that instance implicitly wherever the GADTBag constructor appears. That's why count doesn't need an Eq context, whereas countV2 does - it doesn't use the constructor:
{-# LANGUAGE GADTs #-}
data GADTBag a where
GADTBag :: Eq a => [a] -> GADTBag a
unGADTBag (GADTBag xs) = xs
instance Show a => Show (GADTBag a) where
showsPrec i (GADTBag xs) = showParen (i>9) (("GADTBag " ++ show xs) ++)
count :: a -> GADTBag a -> Int -- no Eq here
count a (GADTBag xs) = length.filter (==a) $ xs -- but == here
countV2 a = length.filter (==a).unGADTBag
size :: GADTBag a -> Int
size (GADTBag xs) = length xs
ghci> count 'l' (GADTBag "Hello")
2
ghci> :t countV2
countV2 :: Eq a => a -> GADTBag a -> Int
Now we didn't need the Eq constraint when we found the total size of the bag, but it didn't clutter up our definition anyway. (We could have used size = length . unGADTBag just as well.)
Now lets make a functor:
instance Functor GADTBag where
fmap f (GADTBag xs) = GADTBag (map f xs)
oops!
DataConstraints_so.lhs:49:30:
Could not deduce (Eq b) arising from a use of `GADTBag'
from the context (Eq a)
That's unfixable (with the standard Functor class) because I can't restrict the type of fmap, but need to for the new list.
Data Constraint version
Can we do as you asked? Well, yes, except that you have to keep repeating the Eq constraint wherever you use the constructor:
{-# LANGUAGE DatatypeContexts #-}
data Eq a => EqBag a = EqBag {unEqBag :: [a]}
deriving Show
count' a (EqBag xs) = length.filter (==a) $ xs
size' (EqBag xs) = length xs -- Note: doesn't use (==) at all
Let's go to ghci to find out some less pretty things:
ghci> :so DataConstraints
DataConstraints_so.lhs:1:19: Warning:
-XDatatypeContexts is deprecated: It was widely considered a misfeature,
and has been removed from the Haskell language.
[1 of 1] Compiling Main ( DataConstraints_so.lhs, interpreted )
Ok, modules loaded: Main.
ghci> :t count
count :: a -> GADTBag a -> Int
ghci> :t count'
count' :: Eq a => a -> EqBag a -> Int
ghci> :t size
size :: GADTBag a -> Int
ghci> :t size'
size' :: Eq a => EqBag a -> Int
ghci>
So our EqBag count' function requires an Eq constraint, which I think is perfectly reasonable, but our size' function also requires one, which is less pretty. This is because the type of the EqBag constructor is EqBag :: Eq a => [a] -> EqBag a, and this constraint must be added every time.
We can't make a functor here either:
instance Functor EqBag where
fmap f (EqBag xs) = EqBag (map f xs)
for exactly the same reason as with the GADTBag
Constraintless bags
data ListBag a = ListBag {unListBag :: [a]}
deriving Show
count'' a = length . filter (==a) . unListBag
size'' = length . unListBag
instance Functor ListBag where
fmap f (ListBag xs) = ListBag (map f xs)
Now the types of count'' and show'' are exactly as we expect, and we can use standard constructor classes like Functor:
ghci> :t count''
count'' :: Eq a => a -> ListBag a -> Int
ghci> :t size''
size'' :: ListBag a -> Int
ghci> fmap (Data.Char.ord) (ListBag "hello")
ListBag {unListBag = [104,101,108,108,111]}
ghci>
Comparison and conclusions
The GADTs version automagically propogates the Eq constraint everywhere the constructor is used. The type checker can rely on there being an Eq instance, because you can't use the constructor for a non-Eq type.
The DatatypeContexts version forces the programmer to manually propogate the Eq constraint, which is fine by me if you want it, but is deprecated because it doesn't give you anything more than the GADT one does and was seen by many as pointless and annoying.
The unconstrained version is good because it doesn't prevent you from making Functor, Monad etc instances. The constraints are written exactly when they're needed, no more or less. Data.Map uses the unconstrained version partly because unconstrained is generally seen as most flexible, but also partly because it predates GADTs by some margin, and there needs to be a compelling reason to potentially break existing code.
What about your excellent User example?
I think that's a great example of a one-purpose data type that benefits from a constraint on the type, and I'd advise you to use a GADT to implement it.
(That said, sometimes I have a one-purpose data type and end up making it unconstrainedly polymorphic just because I love to use Functor (and Applicative), and would rather use fmap than mapBag because I feel it's clearer.)
{-# LANGUAGE GADTs #-}
import Data.String
data User s where
User :: (IsString s, Show s) => s -> User s
name :: User s -> s
name (User s) = s
instance Show (User s) where -- cool, no Show context
showsPrec i (User s) = showParen (i>9) (("User " ++ show s) ++)
instance (IsString s, Show s) => IsString (User s) where
fromString = User . fromString
Notice since fromString does construct a value of type User a, we need the context explicitly. After all, we composed with the constructor User :: (IsString s, Show s) => s -> User s. The User constructor removes the need for an explicit context when we pattern match (destruct), becuase it already enforced the constraint when we used it as a constructor.
We didn't need the Show context in the Show instance because we used (User s) on the left hand side in a pattern match.
Constraints
The problem is that constraints are not a property of the data type, but of the algorithm/function that operates on them. Different functions might need different and unique constraints.
A Box example
As an example, let's assume we want to create a container called Box which contains only 2 values.
data Box a = Box a a
We want it to:
be showable
allow the sorting of the two elements via sort
Does it make sense to apply the constraint of both Ord and Show on the data type? No, because the data type in itself could be only shown or only sorted and therefore the constraints are related to its use, not it's definition.
instance (Show a) => Show (Box a) where
show (Box a b) = concat ["'", show a, ", ", show b, "'"]
instance (Ord a) => Ord (Box a) where
compare (Box a b) (Box c d) =
let ca = compare a c
cb = compare b d
in if ca /= EQ then ca else cb
The Data.Map case
Data.Map's Ord constraints on the type is really needed only when we have > 1 elements in the container. Otherwise the container is usable even without an Ord key. For example, this algorithm:
transf :: Map NonOrd Int -> Map NonOrd Int
transf x =
if Map.null x
then Map.singleton NonOrdA 1
else x
Live demo
works just fine without the Ord constraint and always produce a non empty map.
Using DataTypeContexts reduces the number of programs you can write. If most of those illegal programs are nonsense, you might say it's worth the runtime cost associated with ghc passing in a type class dictionary that isn't used. For example, if we had
data Ord k => MapDTC k a
then #jefffrey's transf is rejected. But we should probably have transf _ = return (NonOrdA, 1) instead.
In some sense the context is documentation that says "every Map must have ordered keys". If you look at all of the functions in Data.Map you'll get a similar conclusion "every useful Map has ordered keys". While you can create maps with unordered keys using
mapKeysMonotonic :: (k1 -> k2) -> Map k1 a -> Map k2 a
singleton :: k2 a -> Map k2 a
But the moment you try to do anything useful with them, you'll wind up with No instance for Ord k2 somewhat later.

Haskell Existential Types

I'm trying to wrap my brain around Haskell's existential types, and my first example is a heterogeneous list of things that can be shown:
{-# LANGUAGE ExistentialQuantification #-}
data Showable = forall a. Show a => Showable a
showableList :: [Showable]
showableList = [Showable "frodo", Showable 1]
Now it seems to me that the next thing I would want to do is make Showable an instance of Show so that, for example, my showableList could be displayed in the repl:
instance Show Showable where
show a = ...
The problem I am having is that what I really want to do here is call the a's underlying show implementation. But I'm having trouble referring to it:
instance Show Showable where
show a = show a
picks out Showable's show method on the RHS which runs in circles. I tried auto-deriving Show, but that doesn't work:
data Showable = forall a. Show a => Showable a
deriving Show
gives me:
Can't make a derived instance of `Show Showable':
Constructor `Showable' does not have a Haskell-98 type
Possible fix: use a standalone deriving declaration instead
In the data type declaration for `Showable'
I'm looking for someway to call the underlying Show::show implementation so that Showable does not have to reinvent the wheel.
instance Show Showable where
show (Showable a) = show a
show a = show a doesn't work as you realized because it recurses infinitely. If we try this without existential types we can see the same problem and solution
data D = D Int
instance Show D where show a = show a -- obviously not going to work
instance Show D where show (D a) = "D " ++ (show a) -- we have to pull out the underlying value to work with it

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