I have a type kinda data T a = T a
I want to declare Show instance for this type, something like
instance Show a => Show (T a) where
show (T smth) = ... ++ show smth
But I don't want to call show smth if a is String, because I don't need additional quotes
Is there any solution how to declare such an instance?
Thanks in advance
If you want to declare an instance of Show for T String you can use FlexibleInstances
{-# LANGUAGE FlexibleInstances #-}
data T a = T a
instance Show (T String) where
show (T smth) = smth
You would then need to provide explicit instances for other as that you wish to support.
If you want to retain your catch-all instance shown in your question too, you would then be into using OVERLAPS/OVERLAPPABLE
{-# LANGUAGE FlexibleInstances #-}
data T a = T a
instance {-# OVERLAPPABLE #-} Show a => Show (T a) where
show (T smth) = show smth
instance {-# OVERLAPPING #-} Show (T String) where
show (T smth) = smth
Related
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.
Suppose I have a simple data type in Haskell for storing a value:
data V a = V a
I want to make V an instance of Show, regardless of a's type. If a is an instance of Show, then show (V a) should return show a otherwise an error message should be returned. Or in Pseudo-Haskell:
instance Show (V a) where
show (V a) = if a instanceof Show
then show a
else "Some Error."
How could this behaviour be implemented in Haskell?
As I said in a comment, the runtime objects allocated in memory don't have type tags in a Haskell program. There is therefore no universal instanceof operation like in, say, Java.
It's also important to consider the implications of the following. In Haskell, to a first approximation (i.e., ignoring some fancy stuff that beginners shouldn't tackle too soon), all runtime function calls are monomorphic. I.e., the compiler knows, directly or indirectly, the monomorphic (non-generic) type of every function call in an executable program. Even though your V type's show function has a generic type:
-- Specialized to `V a`
show :: V a -> String -- generic; has variable `a`
...you can't actually write a program that calls the function at runtime without, directly or indirectly, telling the compiler exactly what type a will be in every single call. So for example:
-- Here you tell it directly that `a := Int`
example1 = show (V (1 :: Int))
-- Here you're not saying which type `a` is, but this just "puts off"
-- the decision—for `example2` to be called, *something* in the call
-- graph will have to pick a monomorphic type for `a`.
example2 :: a -> String
example2 x = show (V x) ++ example1
Seen in this light, hopefully you can spot the problem with what you're asking:
instance Show (V a) where
show (V a) = if a instanceof Show
then show a
else "Some Error."
Basically, since the type for the a parameter will be known at compilation time for any actual call to your show function, there's no point to testing for this type at runtime—you can test for it at compilation time! Once you grasp this, you're led to Will Sewell's suggestion:
-- No call to `show (V x)` will compile unless `x` is of a `Show` type.
instance Show a => Show (V a) where ...
EDIT: A more constructive answer perhaps might be this: your V type needs to be a tagged union of multiple cases. This does require using the GADTs extension:
{-# LANGUAGE GADTs #-}
-- This definition requires `GADTs`. It has two constructors:
data V a where
-- The `Showable` constructor can only be used with `Show` types.
Showable :: Show a => a -> V a
-- The `Unshowable` constructor can be used with any type.
Unshowable :: a -> V a
instance Show (V a) where
show (Showable a) = show a
show (Unshowable a) = "Some Error."
But this isn't a runtime check of whether a type is a Show instance—your code is responsible for knowing at compilation time where the Showable constructor is to be used.
You can with this library: https://github.com/mikeizbicki/ifcxt. Being able to call show on a value that may or may not have a Show instance is one of the first examples it gives. This is how you could adapt that for V a:
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE KindSignatures #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE UndecidableInstances #-}
import IfCxt
import Data.Typeable
mkIfCxtInstances ''Show
data V a = V a
instance forall a. IfCxt (Show a) => Show (V a) where
show (V a) = ifCxt (Proxy::Proxy (Show a))
(show a)
"<<unshowable>>"
This is the essence of this library:
class IfCxt cxt where
ifCxt :: proxy cxt -> (cxt => a) -> a -> a
instance {-# OVERLAPPABLE #-} IfCxt cxt where ifCxt _ t f = f
I don't fully understand it, but this is how I think it works:
It doesn't violate the "open world" assumption any more than
instance {-# OVERLAPPABLE #-} Show a where
show _ = "<<unshowable>>"
does. The approach is actually pretty similar to that: adding a default case to fall back on for all types that do not have an instance in scope. However, it adds some indirection to not make a mess of the existing instances (and to allow different functions to specify different defaults). IfCxt works as a a "meta-class", a class on constraints, that indicates whether those instances exist, with a default case that indicates "false.":
instance {-# OVERLAPPABLE #-} IfCxt cxt where ifCxt _ t f = f
It uses TemplateHaskell to generate a long list of instances for that class:
instance {-# OVERLAPS #-} IfCxt (Show Int) where ifCxt _ t f = t
instance {-# OVERLAPS #-} IfCxt (Show Char) where ifCxt _ t f = t
which also implies that any instances that were not in scope when mkIfCxtInstances was called will be considered non-existing.
The proxy cxt argument is used to pass a Constraint to the function, the (cxt => a) argument (I had no idea RankNTypes allowed that) is an argument that can use the constraint cxt, but as long as that argument is unused, the constraint doesn't need to be solved. This is similar to:
f :: (Show (a -> a) => a) -> a -> a
f _ x = x
The proxy argument supplies the constraint, then the IfCxt constraint is solved to either the t or f argument, if it's t then there is some IfCxt instance where this constraint is supplied which means it can be solved directly, if it's f then the constraint is never demanded so it gets dropped.
This solution is imperfect (as new modules can define new Show instances which won't work unless it also calls mkIfCxtInstances), but being able to do that would violate the open world assumption.
Even if you could do this, it would be a bad design. I would recommend adding a Show constraint to a:
instance Show a => Show (V a) where ...
If you want to store members in a container data type that are not an instance of Show, then you should create a new data type fore them.
Headline: I would like to provide a default implementation for a class method parametrised over a constraint, which uses the default instance for that constraint.
Consider the following:
{-# LANGUAGE ConstraintKinds #-}
{-# LANGUAGE TypeFamilies #-}
import GHC.Exts (Constraint)
class Foo a where
type Ctx a :: Constraint
type Ctx a = Show a
foo :: (Ctx a) => a -> String
foo = show
main :: IO ()
main = putStrLn "Compiles!"
This fails to compile with the error of:
Could not deduce (Show a) arising from a use of ‘show’
from the context (Foo a)
From my perspective, it should be using the default constraint of Show, which would let this compile. Is there any reason this doesn't work, or can anyone suggest a good way to achieve this?
You can achieve this using DefaultSignatures:
{-# LANGUAGE ConstraintKinds #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE DefaultSignatures #-}
import GHC.Exts (Constraint)
class Foo a where
type Ctx a :: Constraint
type Ctx a = Show a
foo :: (Ctx a) => a -> String
default foo :: Show a => a -> String
foo = show
main :: IO ()
main = putStrLn "Compiles!"
From my perspective, it should be using the default constraint of Show, which would let this compile.
The reason your approach doesn't work is that the user of your class should be able to override any number of defaults. Your code would break if someone tried to override Ctx but not foo.
I'm working on an applicative functor that contains a monoid to "view" the execution. However, sometimes I don't care about this part at all, so the choice of monoid is irrelevant as it will never be consumed. I've simplified what I have into:
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE ConstraintKinds #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE TypeFamilies #-}
import GHC.Exts
class Render a b where render :: a -> b
instance Render a () where render = const ()
class Merge a where
type Renderer a b :: Constraint
merge :: Renderer a b => a -> b
data Foo = Foo Bool
instance Merge Foo where
type (Renderer Foo) m = (Render Bool m)
merge (Foo b) = render b
Render is used to transform various as into a single b. Merge is a big simplification of my actual functor, but the point is it contains a type family/constraint and my intention of that is to specify exactly what Renderers a Merge requires.
Now, I might want to "run" the Merge, but discard the view, which is akin to something like:
runFoo :: Merge a => a -> Int
runFoo x = case merge x of () -> 5
But this will fail because:
Could not deduce (Renderer a ()) arising from a use of merge
I chose () as my monoid because forall a, we have an instance of Render a (). So if there was a way to say that Merge a just means a collection Render constraints then this would work fine. Of course, Merge a is more general than that - it could add arbitrary constraints, which explains the compilation error.
Is there anyway to achieve what I want without changing the signature of runFoo?
This might not scale if you have a lot of these cases, but this works:
class Renderer a () => Merge a where
...
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>>"