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.
Suppose we have the following:
{-# LANGUAGE FlexibleInstances #-}
module Sample where
newtype A a =
A a
deriving (Show)
newtype L a =
L [a]
class ListContainer l where
getList :: l a -> [a]
instance ListContainer L where
getList (L l) = l
instance (Show a, ListContainer l) => Show (l a) where
show = const "example"
With this code, ghc complains:
warning: [-Wdeferred-type-errors]
• Overlapping instances for Show (A a)
arising from a use of ‘GHC.Show.$dmshowList’
Matching instances:
instance (Show a, ListContainer l) => Show (l a)
-- Defined at /.../src/Sample.hs:18:10
instance Show a => Show (A a)
-- Defined at /.../src/Sample.hs:7:13
• In the expression: GHC.Show.$dmshowList #(A a)
In an equation for ‘showList’:
showList = GHC.Show.$dmshowList #(A a)
When typechecking the code for ‘showList’
in a derived instance for ‘Show (A a)’:
To see the code I am typechecking, use -ddump-deriv
In the instance declaration for ‘Show (A a)’
warning: [-Wdeferred-type-errors]
• Overlapping instances for Show (A a)
arising from a use of ‘GHC.Show.$dmshow’
Matching instances:
instance (Show a, ListContainer l) => Show (l a)
-- Defined at /.../src/Sample.hs:18:10
instance Show a => Show (A a)
-- Defined at /.../src/Sample.hs:7:13
• In the expression: GHC.Show.$dmshow #(A a)
In an equation for ‘show’: show = GHC.Show.$dmshow #(A a)
When typechecking the code for ‘show’
in a derived instance for ‘Show (A a)’:
To see the code I am typechecking, use -ddump-deriv
In the instance declaration for ‘Show (A a)’
I can understand that it thinks type a can either derive Show, or derive ListContainer, which may result in Show.
How do we avoid that?
I understand that there exists a function showList, but its signature is a bit foreign. I do already have a function that I intend to use to display certain lists, which returns String directly.
I can understand that it thinks type a can either derive Show, or derive ListContainer, which may result in Show.
That is not what it thinks.
When Haskell chooses class instance, it doesn't look at instance constraints at all. All it considers when choosing an instance is the instance head (the thing that comes right after class name).
In your Show instance, the instance head is l a. This instance head matches A a (by assuming l = A). It also matches a lot of other things by the way - for example, it matches Maybe a (where l = Maybe), and Either b a (with l = Either b), and Identity a, and IO a - pretty much every type with a type parameter, come to think of it. It doesn't matter that neither A nor Maybe nor IO have an instance of ListContainer, because like I said above, Haskell doesn't look at constraints when choosing instances, only at instance heads.
It is only after finding a matching instance (by matching on its head) that Haskell will check if that instance's constraints are in fact satisfied. And will complain if they aren't. But it will never go back and try to pick another instance instead.
So coming back to your example: since A now has two matching Show instances - its own derived one and the Show (l a) one that you wrote, - the compiler complains that they are overlapping.
In your example you can just remove instance (Show a, ListContainer l) => Show (l a) and add deriving (Show) to L definition.
Alternatively you can remove deriving (Show) from A definition.
If you want you code behave as it is now remove deriving (Show) and implement it explicitly
instance {-# OVERLAPPING #-} Show a => Show (A a)
where
show (A a) = "A " ++ show a
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
I would like to implement the show method for (binary) functions and make it able to distingish endofunctions (a -> a).
Something like the pseudo-haskell code:
instance Show (a->b) where
show fun = "<<Endofunction>>" if a==b
show fun = "<<Function>>" if a\=b
How can I distinguish the two cases?
You need to enable some extensions:
{-# LANGUAGE OverlappingInstances, FlexibleInstances #-}
module FunShow where
instance Show ((->) a a) where
show _ = "<<Endofunction>>"
instance Show ((->) a b) where
show _ = "<<Function>>"
You need OverlappingInstances since the instance a -> b also matches endofunctions, so there's overlap, and you need FlexibleInstances because the language standard mandates that the type variables in instance declarations are distinct.
*FunShow> show not
"<<Endofunction>>"
*FunShow> show fst
"<<Function>>"
*FunShow> show id
"<<Endofunction>>"
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