Sunday, December 31, 2017

N-ary Functors

update: Now available on hackage as n-ary-functor.

Functor and Bifunctor are both in base, but what about Trifunctor? Quadrifunctor? There must be a better solution than creating an infinite tower of typeclasses. Here's the API I managed to implement:

> nmap <#> (+1) <#> (+2) $ (0, 0)
(1,2)

> nmap <#> (+1) <#> (+2) <#> (+3) $ (0, 0, 0)
(1,2,3)

> nmap <#> (+1) <#> (+2) <#> (+3) <#> (+4) $ (0, 0, 0, 0)
(1,2,3,4)

The implementation is really short, too:

{-# LANGUAGE RankNTypes, TypeFamilies, TypeInType #-}
import Data.Kind

newtype NMap1 k (f :: Type -> k) (g :: Type -> k) = NMap1
  { (<#>) :: forall a b. (a -> b) -> NMap k (f a) (g b) }

type family NMap k :: k -> k -> Type where
  NMap Type        = (->)
  NMap (Type -> k) = NMap1 k

class NFunctor (f :: k) where
  nmap :: NMap k f f

Of course, the hard part is not writing the code, but figuring out what to write down. Let me show you how I got there.

Computing the Type from the Kind

Since Functor instances are given to type constructors of kind * -> *, and Bifunctor instances are given to type constructors of kind * -> * -> *, my idea was to compute the type of nmap from the kind of the type constructor to which it is applied. Something like this:

class NFunctor (f :: k) where
  nmap :: NMap k f

type family NMap k (f :: k) :: *
type instance NMap (* -> *) f
  = (a -> b) -> f a -> f b
type instance NMap (* -> * -> *) f
  = (a1 -> b1) -> (a2 -> b2) -> f a1 a2 -> f b1 b2
type instance NMap (* -> * -> * -> *) f
  = (a1 -> b1) -> (a2 -> b2) -> (a3 -> b3) -> f a1 a2 a3 -> f b1 b2 b3

Except of course with some recursive definition for NMap, so we don't have to spell out the type for every kind. Thinking of it in terms of recursion made me realize that the base case is kind *, not * -> *:

type instance NMap * f
  = f -> f

This corresponds to a "nullary Functor" typeclass, whose lawful instances have no choice but to use the identity function. So this isn't particularly useful as a typeclass, but it does lead to a nice recursive definition:

type family NMap k (f :: k) (g :: k) where
  NMap Type        a b = a -> b
  NMap (Type -> k) f g = (a -> b) -> NMap k (f a) (g b)

class NFunctor (f :: k) where
  nmap :: NMap k f f

I now have to use Type instead of * for some reason, otherwise I get a "malformed head" error.

Required Newtype Wrapper

Unfortunately, GHC does not accept that recursive definition. First of all, when defining a type family, type variables aren't implicitly universally-quantified like they are in type signatures, so I need to add an explicit forall quantifier:

type family NMap k (f :: k) (g :: k) where
  NMap Type        a b = a -> b
  NMap (Type -> k) f g = forall a b. (a -> b) -> NMap k (f a) (g b)

Now GHC reveals the real problem with the definition:

• Illegal polymorphic type:
    forall a b. (a -> b) -> NMap k (f a) (g b)
• In the equations for closed type family ‘NMap’
  In the type family declaration for ‘NMap’

This is a bummer: I am simply not allowed to use forall here. The usual workaround, when forall is needed but disallowed, is to define a newtype which performs the forall for us:

newtype NMap1 k (f :: Type -> k) (g :: Type -> k) = NMap1
  { runNMap1 :: forall a b. (a -> b) -> NMap k (f a) (g b) }

type family NMap k :: k -> k -> Type where
  NMap Type        = (->)
  NMap (Type -> k) = NMap1 k

This solves the problem, and even allows me to make my NMap definition more point-free!

Ergonomics

I now have a typeclass which generalizes Functor, Bifunctor, Trifunctor, etc., but what does using this typeclass look like? Writing instances requires a bit of boilerplate, but it's not too bad:

instance NFunctor Maybe where
  nmap = NMap1 fmap

instance NFunctor (,) where
  nmap = NMap1 $ \f1
      -> NMap1 $ \f2
      -> \(x1,x2)
      -> (f1 x1, f2 x2)

instance NFunctor (,,) where
  nmap = NMap1 $ \f1
      -> NMap1 $ \f2
      -> NMap1 $ \f3
      -> \(x1,x2,x3)
      -> (f1 x1, f2 x2, f3 x3)

instance NFunctor (,,,) where
  nmap = NMap1 $ \f1
      -> NMap1 $ \f2
      -> NMap1 $ \f3
      -> NMap1 $ \f4
      -> \(x1,x2,x3,x4)
      -> (f1 x1, f2 x2, f3 x3, f4 x4)

When calling nmap, however, the extra boilerplate quickly becomes annoying:

> runNMap1 nmap (+1) $ Just (0)
Just 1

> runNMap1 (runNMap1 nmap (+1)) (+2) (0, 0)
(1,2)

> runNMap1 (runNMap1 (runNMap1 nmap (+1)) (+2)) (+3) (0, 0, 0)
(1,2,3)

> runNMap1 (runNMap1 (runNMap1 (runNMap1 nmap (+1))
|                              (+2))
|                    (+3))
|          (+4)
|          (0, 0, 0, 0)
(1,2,3,4)

The fix is really simple though: by renaming runNMap1 to some left-associative infix operator, say (<#>), the code becomes much more readable!

> nmap <#> (+1) $ Just (0)
Just 1

> nmap <#> (+1) <#> (+2) $ (0, 0)
(1,2)

> nmap <#> (+1) <#> (+2) <#> (+3) $ (0, 0, 0)
(1,2,3)

> nmap <#> (+1) <#> (+2) <#> (+3) <#> (+4) $ (0, 0, 0, 0)
(1,2,3,4)

A Tempting Overlapping Instance

Pairs have both a Bifunctor and a Functor instance. Similarly, quadruples have four NFunctor instances, five if we count the glorified identity function:

-- |
-- >>> nmap <#> (+1) <#> (+2) <#> (+3) <#> (+4) $ (0, 0, 0, 0)
-- (1,2,3,4)
instance NFunctor (,,,) where
  nmap = NMap1 $ \f1
      -> NMap1 $ \f2
      -> NMap1 $ \f3
      -> NMap1 $ \f4
      -> \(x1,x2,x3,x4)
      -> (f1 x1, f2 x2, f3 x3, f4 x4)

-- |
-- >>> nmap <#> (+1) <#> (+2) <#> (+3) $ (0, 0, 0, 0)
-- (0,1,2,3)
instance NFunctor ((,,,) a) where
  nmap = NMap1 $ \f2
      -> NMap1 $ \f3
      -> NMap1 $ \f4
      -> \(x1,x2,x3,x4)
      -> (x1, f2 x2, f3 x3, f4 x4)

-- |
-- >>> nmap <#> (+1) <#> (+2) $ (0, 0, 0, 0)
-- (0,0,1,2)
instance NFunctor ((,,,) a b) where
  nmap = NMap1 $ \f3
      -> NMap1 $ \f4
      -> \(x1,x2,x3,x4)
      -> (x1, x2, f3 x3, f4 x4)

-- |
-- >>> nmap <#> (+1) $ (0, 0, 0, 0)
-- (0,0,0,1)
instance NFunctor ((,,,) a b c) where
  nmap = NMap1 $ \f4
      -> \(x1,x2,x3,x4)
      -> (x1, x2, x3, f4 x4)

-- |
-- >>> nmap (0, 0, 0, 0)
-- (0,0,0,0)
instance NFunctor ((,,,) a b c d) where
  nmap = \(x1,x2,x3,x4)
      -> (x1, x2, x3, x4)

But if we define the following magical instance:

{-# LANGUAGE FlexibleInstances #-}

instance NFunctor (f :: * -> k) => NFunctor (f a) where
  nmap = nmap <#> id

Then we get all five instances for the price of one!

-- |
-- >>> nmap <#> (+1) <#> (+2) <#> (+3) <#> (+4) $ (0, 0, 0, 0)
-- (1,2,3,4)
-- >>> nmap <#> (+1) <#> (+2) <#> (+3) $ (0, 0, 0, 0)
-- (0,1,2,3)
-- >>> nmap <#> (+1) <#> (+2) $ (0, 0, 0, 0)
-- (0,0,1,2)
-- >>> nmap <#> (+1) $ (0, 0, 0, 0)
-- (0,0,0,1)
-- >>> nmap (0, 0, 0, 0)
-- (0,0,0,0)
instance NFunctor (,,,) where
  nmap = NMap1 $ \f1
      -> NMap1 $ \f2
      -> NMap1 $ \f3
      -> NMap1 $ \f4
      -> \(x1,x2,x3,x4)
      -> (f1 x1, f2 x2, f3 x3, f4 x4)

The big problem with that magic instance is that it overlaps with other instances we would like to define. For example, we don't want to define the NFunctor instance for State s in terms of the NFunctor instance for State, because State is not functorial in s, so it doesn't have such an instance. Oh well.

Friday, November 03, 2017

Computing with Impossible Types

Edward Kmett recently posted a puzzling gist seemingly showing that at the type level, the () kind has more than one inhabitant. The goal of this post is to explain what's going on.

Stuck Type Expressions

Here is a simple type family.

{-# LANGUAGE TypeFamilies, UndecidableInstances #-}

type family F a where
  F (Maybe a) = [F a]
  F a         = a

Since F (Maybe Int) and [F Int] both evaluate to [Int], the following type-checks.

-- |
-- >>> :kind! F (Maybe Int) -> [F Int]
-- F (Maybe Int) -> [F Int] :: *
-- = [Int] -> [Int]
runFMaybeInt :: F (Maybe Int) -> [F Int]
runFMaybeInt = id

We didn't use any Int-specific code, so let's make the type more polymorphic.

-- |
-- >>> :set -XRankNTypes
-- >>> :kind! forall b. F (Maybe b) -> [F b]
-- forall b. F (Maybe b) -> [F b] :: *
-- = [F b] -> [F b]
runFMaybe :: Proxy b -> F (Maybe b) -> [F b]
runFMaybe _ = id

Notice that F (Maybe b) and [F b] both evaluate to [F b], not to [b]! That's because we don't yet know whether b is going to be instantiated with a Maybe something or not, so unlike F Int, the type expression F b cannot be simplified further. The evaluation of F b is stuck, and will remain so until we learn more information about b. The code still type-checks though, because even though we don't know which concrete type F b will expand to, we do know that [F b] and [F b] will expand to the same type because they are the same type expression.

Pattern-Matching on the Shape

Here is another type family.

type family G a where
  G (f a) = [G a]
  G a     = Double

This time, the type family isn't pattern-matching on whether or not its input type is a Maybe something, but on whether or not it is a type which, like Maybe Int, consists of a type constructor applied to a type. Let's look at a concrete example:

-- |
-- >>> :kind! G (Maybe Int) -> [G Int]
-- G (Maybe Int) -> [G Int] :: *
-- = [Double] -> [Double]
runGMaybeInt :: G (Maybe Int) -> [G Int]
runGMaybeInt = id

No surprises there. Let's make the type more polymorphic:

-- |
-- >>> :kind! forall g b. G (g b) -> [G b]
-- forall g b. G (g b) -> [G b] :: *
-- = [G b] -> [G b]
runGMaybe :: Proxy (g b) -> G (g b) -> [G b]
runGMaybe _ = id

As before, the type expression G b is stuck because we don't yet know whether b is going to be instantiated to a type with the right shape such as Maybe Int, or to a type with a different shape such as Int. But regardless of which one it is, [G b] and [G b] will both expand to the same type, so the implementation type-checks.

One last example:

>>> :kind! forall b. G (G b) -> [G b]
forall b. G (G b) -> [G b] :: *
= G (G b) -> [G b]

Note that G (G b) did not simplify! G b might look like it has the right shape to match g b, but it doesn't, because G is a type family, not a type constructor. It's a good thing it doesn't match, because if it did, evaluating type expressions like G (G Int) wouldn't be confluent! If we evaluate the outer application first, we get [G Int] and then [Double], whereas if we evaluate the inner application first, we get G Double and then Double.

To be clear, evaluating the outer application first doesn't work because we don't yet know whether the type expression G Int will evaluate to something of the form f a or not. So the inner application is evaluated first, and G (G Int) evaluates to Double.

Two Arrow-Like Kinds

G and Maybe both seem to have kind * -> *:

>>> :kind! G
G :: * -> *
>>> :kind! Maybe
Maybe :: * -> *

But that's misleading, because there are some circumstances in which a type of kind * -> * is expected but only Maybe will be accepted:

class MyMonad (m :: * -> *)

-- ok
instance MyMonad Maybe

-- error: The type family ‘G’ should have 1 argument,
--        but has been given none
instance MyMonad G

And there are other circumstances in which both G and Maybe will be accepted:

{-# LANGUAGE DataKinds, TypeOperators #-}

-- |
-- >>> :kind! FMap Maybe '[Int, Double]
-- FMap Maybe '[Int, Double] :: [*]
-- = '[Maybe Int, Maybe Double]
--
-- >>> :kind! FMap G '[Int, Double]
-- FMap G '[Int, Double] :: [*]
-- = '[Double, Double]
type family FMap (f :: * -> *) as where
  FMap f '[]       = '[]
  FMap f (a ': as) = (f a ': FMap f as)

So I prefer to pretend that there are two different arrow-like kind constructors:

  1. (-->) for type functions which can be applied to a type argument. G and Maybe both have kind * --> *. Just like it isn't possible to pattern-match on a type variable, it is not possible to pattern-match on a type expression whose head is a (-->), we must instead apply the type function and pattern-match on the result.
  2. (->) for type constructors which can be pattern-matched on. Maybe has kind * -> *, but G does not. Clearly, (->) is a subtype of (-->).

Now we can make sense of the previous examples. Instance resolution works by pattern-matching on types, so MyMonad expects a * -> *, not a * --> *. Since G has the wrong kind, it cannot be given a MyMonad instance. FMap, on the other hand, only needs to apply its f to various as, so it expects an * --> * such as G. Since * -> * is a subtype of * --> *, FMap can also be applied to Maybe.

edit:
Apparently :kind! is misleading here. Outside of :kind!, FMap accepts Maybe but not G. So the situation is simpler than I thought: (-->) is for type families, (->) is for type constructors, and those are completely different arrow kinds, there is no subtyping relationship between the two. There is no way to ask for an argument of kind * --> *, because if we try to pass an "unsaturated" argument with that kind, G for example, GHC will complain that G is missing arguments. So MyMonad and FMap both expect an argument of kind * -> *, not * --> *.

Unusual Type Families

Here are a few surprising, yet legal type families.

-- |
-- >>> :kind! H1 ('Just '())
-- H1 ('Just '()) :: * -> *
-- = Maybe
type family H1 (a :: Maybe ()) :: * -> * where
  H1 ('Just a) = Maybe
  H1 'Nothing  = IO

H1's input has kind Maybe (), not *, and its output has kind * -> *, not *. Note that it's really * -> *, not * --> *, so G is not a valid output. Overall, the kind of H1 is thus Maybe () --> * -> *.

-- |
-- >>> :kind! H2
-- H2 :: *
-- = Int
type family H2 where
  H2 = Int

H2 has no type parameters, so it's kind is *, not * --> *. If it returned Maybe instead of Int, its kind would be * -> * instead. A type family's kind can be either * --> * or * -> * depending on how it's defined, so it's not as simple as "type constructors use (->), type families use (-->)".

Combining both ideas together:

type family J :: () -> Maybe () where
  J = 'Just

J's kind is () -> Maybe (), so it has to return a type constructor which accepts a type of kind () and produces a type of kind Maybe (). There are only two types which have the kind Maybe (): the type 'Nothing, and the type 'Just '(). 'Nothing has the wrong kind, since it doesn't accept a type of kind (), but 'Just is just right.

One last complication:

-- |
-- >>> :kind! H3 Int
-- H3 Int :: *
-- = H3 Int
type family H3 a where

H3 has no equations defining what happens when it is applied to a type argument. As a result, the type expression H3 Int remains stuck even though it doesn't contain any type variables.

Combining everything together:

type family Succ :: () -> () where

Succ pretends that it can produce a type constructor which accepts a type of kind () and produces a type of kind (). This is ridiculous! We know that '() is the only type of kind (), and like 'Nothing, it has the wrong kind because it doesn't accept a type of kind (). There are no valid types which Succ could return, so unsurprisingly it has no equations, and so Succ is a type expression which always remains stuck.

Ignoring Impossible Types

The type expression Succ '() is stuck, but well-kinded. It has kind ().

That's the kind which 'Just :: () -> Maybe () expects. Thus, 'Just (Succ '()) is also stuck and well-kinded. It has kind Maybe ().

That's the kind which our H1 :: Maybe () --> * -> * type family from earlier expects. Is H1 ('Just (Succ '())) stuck as well?

>>> :kind! H1 ('Just (Succ '()))
H1 ('Just (Succ '())) :: * -> *
= Maybe

Not stuck! That's because H1 ignores the part which is stuck. Its pattern is ('Just a), so it pattern-matches on the 'Just constructor, but it ignores its argument. If its pattern was ('Just '()) instead, it would have been stuck.

Here comes the clever part: it is possible to write a type family which pattern-matches on the '() but ignores the stuck Succ part.

-- |
-- >>> :kind! IsSucc (Succ '())
-- IsSucc (Succ '()) :: Bool
-- = 'True
--
-- >>> :kind! IsSucc '()
-- IsSucc '() :: Bool
-- = 'False
type family IsSucc (a :: ()) :: Bool where
  IsSucc (succ '()) = 'True
  IsSucc '()        = 'False

The trick is to do like G and pattern-match on the shape, not the contents.

Computing with Impossible Types

It is also possible to distinguish the two inhabitants using a typeclass instead of a type family:

{-# language FlexibleInstances #-}
import Data.Proxy

-- |
-- >>> isSucc (Proxy :: Proxy (Succ '()))
-- True
-- >>> isSucc (Proxy :: Proxy '())
-- False
class IsSucc (a :: ()) where
  isSucc :: Proxy a -> Bool

instance IsSucc (succ '()) where
  isSucc _ = True

instance IsSucc '() where
  isSucc _ = False

The fact that this works is surprising, because () is supposed to be a closed kind with only one inhabitant, '(), and yet here we seemingly have a second inhabitant, Succ '(), which can be distinguished from '() even though it is stuck. And as you might surmise from its name, we can manufacture many more inhabitants: Succ (Succ '()), Succ (Succ (Succ '())), etc.

{-# language ScopedTypeVariables #-}

-- |
-- >>> countSuccs (Proxy :: Proxy '())
-- 0
-- >>> countSuccs (Proxy :: Proxy (Succ '()))
-- 1
-- >>> countSuccs (Proxy :: Proxy (Succ (Succ (Succ '()))))
-- 3
class CountSuccs (a :: ()) where
  countSuccs :: Proxy a -> Int

instance CountSuccs '() where
  countSuccs _ = 0

instance CountSuccs a => CountSuccs (succ a) where
  countSuccs _ = 1 + countSuccs (Proxy :: Proxy a)

Those examples show how to compute booleans and integers from a stuck type expression containing Succs. Using polymorphic recursion, it is also possible to go the other way, from an integer to a stuck type expression containing that many Succs:

{-# language RankNTypes #-}

-- |
-- >>> mkSuccs 42 countSuccs
-- 42
mkSuccs :: Int -> (forall a. CountSuccs a => Proxy a -> r) -> r
mkSuccs 0 cc = cc (Proxy :: Proxy '())
mkSuccs n cc = mkSuccs (n - 1) $ \(Proxy :: Proxy a)
            -> cc (Proxy :: Proxy (Succ a))

Since Haskell doesn't have dependent types, the output type is independent of the integer, so we cannot directly return the stuck type as an output. Instead, we use continuation-passing-style to accept a polymorphic continuation which produces an r regardless of which stuck type we instantiate it at.

When we use countSuccs as the continuation, this r is an integer, and the integer it computes is the number of Succs. So we start with n, we convert it to a stuck type containing n Succs, and then we count those Succs and get n back. This is a very simple example of a computation which relies on the existence of those seemingly-impossible non-'() inhabitants of () in order to compute its result: if there was only one type of kind (), the integer would be lost during the conversion to Proxy (a :: ()), and we would not be able to get that same integer back at the end.

Full Circle

Now that we have seen and understood each of the pieces individually, we are now ready to marvel at Kmett's creation:

{-# language PolyKinds #-}

import Data.Proxy

class KnownUnit (k :: ()) where
   reflect :: Proxy k -> Int

instance KnownUnit '() where
  reflect _ = 0

instance KnownUnit x => KnownUnit (f x) where
  reflect _ = 1 + reflect (Proxy :: Proxy x)

type family Succ :: k -> k

-- |
-- >>> reify 23 reflect
-- 23
reify :: Int -> (forall k. KnownUnit k => Proxy k -> r) -> r
reify 0 f = f (Proxy :: Proxy '())
reify n f = reify (n - 1) $ \(Proxy :: Proxy k)
         -> f (Proxy :: Proxy (Succ k))

Neat!

...or worrisome?

Accepting Impossible Types

We Haskellers like to use precise types in order to make illegal states unrepresentable. We accept, reluctantly, that ⊥ inhabits all types, so () doesn't really have exactly one possible value. But it does have exactly one possible total value, and if we write a function whose type signature expects a (), that's the value which this function expects to receive. And so, most functions don't document what their behaviour is on ⊥ inputs, and nobody complains, because they know they're not supposed to use ⊥ inputs.

DataKinds allows us to use precise kinds, and thus to make illegal types unrepresentable. We don't often think about them, but stuck type expressions also inhabit all kinds, so there isn't really only one type of kind (). Today we saw that some of those extra inhabitants are really weird. That's an interesting quirk of Haskell's type system, but ultimately, I don't think those weird inhabitants are any more worrisome than their less exotic cousins, the stuck type expressions which contain type variables. After all, there is only one total type of kind (), and when we write a type-level function (or an instance) which expects a (), that's the type we expect.

Saturday, October 14, 2017

Composing Declarations in Template Haskell

I have recently tried to use Template Haskell to generate both a datatype and lenses for accessing the fields of this datatype, and it was harder than it should have been. In this post, I will demonstrate the problem, I will pinpoint its cause, and I will propose a solution.

The Problem

Consider the following code. I'm using a simple, contrived example instead of a more realistic one because it will be easier to write Template Haskell code for generating this silly code than it would be to write Template Haskell code generating lenses and such.

class Default a where
  def :: a


data X = X
data Y = Y X

data DoubledX = DoubledX
data DoubledY = DoubledY X X

instance Default X        where def = X
instance Default Y        where def = Y def
instance Default DoubledX where def = DoubledX
instance Default DoubledY where def = DoubledY def def

Most of that code is boilerplate, and I would like to generate that boilerplate using Template Haskell. I hasten to note that Template Haskell is a tool of last resort, to be used only when none of Haskell's many other abstraction facilities would have sufficed. In this example, I would probably use some generics library to define a default implementation of def for any algebraic datatype:

{-# LANGUAGE DefaultSignatures, DeriveGeneric, FlexibleContexts #-}
import Generics.Eot

class Default a where
  def :: a
  
  default def :: (HasEot a, Default (Eot a)) => a
  def = fromEot def

instance Default () where
  def = ()

instance (Default a, Default b) => Default (a, b) where
  def = (def, def)

instance Default a => Default (Either a b) where
  def = Left def


instance Default X
instance Default Y
instance Default DoubledX
instance Default DoubledY

This works fine, but today I want to talk about one of Template Haskell's limitations, so let's write a Template Haskell implementation instead.

{-# LANGUAGE TemplateHaskell #-}
import Data.List
import Language.Haskell.TH

-- > data Foo = Bar Int String
-- > generateDefault ''Foo
--
-- generates
--
-- > instance Default Foo where def = Bar def def
generateDefault :: Name -> Q [Dec]
generateDefault name = do
  --      data Foo =             Bar   Int...
  TyConI (DataD _ _ _ _ (NormalC cname cargs:_) _) <- reify name
  
  -- Bar def...
  let expr = foldl' (\c _ -> [|$c def|]) (conE cname) cargs
  
  [d|instance Default $(conT name) where def = $expr|]


data X = X
data Y = Y X

data DoubledX = DoubledX
data DoubledY = DoubledY X X

generateDefault ''X
generateDefault ''Y
generateDefault ''DoubledX
generateDefault ''DoubledY

In addition to the Default instances, we can also generate the Doubled datatypes, they are a version of the original datatype which has two copies of each field:

-- > data Foo = Bar Int String
-- > generateDoubled ''Foo
--
-- generates
--
-- > data DoubledFoo = DoubledBar Int String Int String
generateDoubled :: Name -> Q [Dec]
generateDoubled name = do
  --    data Foo =             Bar   Int...
  TyConI (DataD _ _ _ _ (NormalC cname cargs:_) _) <- reify name
  
  let cons = [NormalC (doubledName cname) (cargs ++ cargs)]
  pure [DataD [] (doubledName name) [] Nothing cons []]

doubledName :: Name -> Name
doubledName = mkName . ("Doubled" ++) . nameBase


data X = X
data Y = Y X

generateDoubled ''X
generateDoubled ''Y

generateDefault ''X
generateDefault ''Y
generateDefault ''DoubledX
generateDefault ''DoubledY

So, we can write a Template Haskell function which generates a datatype, and we can write one which generates an instance for that datatype. But can we write one which generates both the datatype and its instance? Both of our functions are Q actions which produce a [Dec], so it looks like a no brainer: we can simply run both Q actions one after the other and concatenate the resulting lists.

generateBoth :: Name -> Q [Dec]
generateBoth name = (++) <$> generateDoubled name
                         <*> generateDefault (doubledName name)


data X = X

-- error: ‘DoubledX’ is not in scope at a reify
generateBoth ''X

Sadness, it does not work :(

The Cause

The reason DoubledX is not in scope when generateDefault calls reify ''DoubledX is that the DoubledX datatype is not created as a side-effect of the generateDoubled ''X action, but as a side-effect of splicing the resulting [Dec] into the surrounding code. When concatenating the two lists, this doesn't happen until after both lists have been generated, and so DoubledX cannot be "reified" while generating the second list.

I didn't thought I'd ever say something like this, but: this pure API was a mistake, an API based on side-effects would be better! I'll qualify that statement though: since reify obtains information about datatypes (and other named entities) via a side-effect, namely reading from some global symbol table, I think there should be a corresponding action for adding new names to this table. As we have seen, with the current API in which names are added by returning a pure [Dec] value, declaration templates don't compose, so I think that API was a mistake.

I should note that there is, in fact, an action for adding new declarations as a side-effect:

addTopDecls :: [Dec] -> Q ()

Unfortunately, as of this writing, addTopDecls is unable to add datatypes:

import Language.Haskell.TH.Syntax

generateBoth :: Name -> Q [Dec]
generateBoth name = do
  decs <- generateDoubled name
  addTopDecls decs
  generateDefault (doubledName name)


data X = X

-- error: Only function, value, annotation, and foreign import 
--        declarations may be added with addTopDecl
-- error: ‘DoubledX’ is not in scope at a reify
generateBoth ''X

So the real solution would be to fix the implementation so it also supports datatypes, but until then, I have a simpler solution.

Simple Solution

We can't change declaration-splicing's Q [Dec] API, but if we create our own API, we can write an adapter which turns it into a Q [Dec].

{-# LANGUAGE GeneralizedNewtypeDeriving #-}

newtype LocalQ a = LocalQ { unLocalQ :: StateT [Dec] Q a }
  deriving (Functor, Applicative, Monad, MonadFail)

runLocalQ :: LocalQ () -> Q [Dec]
runLocalQ = flip execStateT [] . unLocalQ

The idea is that in addition to Q's global symbol table, LocalQ also has access to a local symbol table holding the declarations which have been added within the LocalQ computation.

addLocalDecls :: [Dec] -> LocalQ ()
addLocalDecls decls = LocalQ $ modify (++ decls)

reifyLocallyFirst :: Name -> LocalQ Info
reifyLocallyFirst name = do
  decls <- LocalQ get
  case find (hasName name) decls of
    Just dec -> pure $ TyConI dec
    Nothing  -> liftQ $ reify name

-- for this simple example, I'm only interested
-- in datatype declarations
hasName :: Name -> Dec -> Bool
hasName expected (DataD _ actual _ _ _ _) = actual == expected
hasName _        _                        = False

liftQ :: Q a -> LocalQ a
liftQ = LocalQ . lift

If we reimplement our declaration templates to use this new, better API...

locallyGenerateDefault :: Name -> LocalQ ()
locallyGenerateDefault name = do
  info <- reifyLocallyFirst name
  let TyConI (DataD _ _ _ _ (NormalC cname cargs:_) _) = info
  let expr = foldl' (\c _ -> [|$c def|]) (conE cname) cargs
  decls <- liftQ [d|instance Default $(conT name) where def = $expr|]
  addLocalDecls decls

locallyGenerateDoubled :: Name -> LocalQ ()
locallyGenerateDoubled name = do
  info <- reifyLocallyFirst name
  let TyConI (DataD _ _ _ _ (NormalC cname cargs:_) _) = info
  let cons = [NormalC (doubledName cname) (cargs ++ cargs)]
  addLocalDecls [DataD [] (doubledName name) [] Nothing cons []]

...then this time we can compose them just fine:

locallyGenerateBoth :: Name -> LocalQ ()
locallyGenerateBoth name = do
  locallyGenerateDoubled name
  locallyGenerateDefault (doubledName name)


data X = X
data Y = Y X

runLocalQ $ locallyGenerateDefault ''X
runLocalQ $ locallyGenerateDefault ''Y
runLocalQ $ locallyGenerateBoth ''X
runLocalQ $ locallyGenerateBoth ''Y

Happiness, it works! Now all that's left is to convince everybody to rewrite their declaration templates using LocalQ instead of Q, and we'll finally be able to reuse each other's code.

Final Solution

Okay, so that last part is probably not going to happen. If only there was a way to monkey-patch existing Q code so it would use reifyLocallyFirst instead of reify...

Well, here's a little-known fact about Q:

newtype Q a = Q { unQ :: forall m. Quasi m => m a }

That's right, Q isn't some magic Monad which only the compiler can execute! It's a concrete type, which we can examine and manipulate however we want. The finally-tagless encoding might be a bit intimidating, but in practice, a Q a value is basically an AST listing which Q actions need to be performed in order to produce an a. So we should be able to dive in and replace all the reify calls with reifyLocallyFirst calls, no problem.

The finally-tagless way to do that is to write a Quasi instance which instantiates reify with reifyLocallyFirst, and delegates all the other operations to some other Quasi instance:

instance Quasi LocalQ where
  qReify              = reifyLocallyFirst
  qAddTopDecls        = addLocalDecls
  
  -- Most of those aren't being exercised by my simple example,
  -- so I can afford to use 'undefined' for the trickier methods.
  qGetQ                   = undefined
  qPutQ                   = undefined
  qRecover                = undefined
  qAddDependentFile   x   = liftQ $ qAddDependentFile   x
  qAddModFinalizer    x   = liftQ $ qAddModFinalizer    x
  qExtsEnabled            = liftQ $ qExtsEnabled
  qIsExtEnabled       x   = liftQ $ qIsExtEnabled       x
  qLocation               = liftQ $ qLocation
  qLookupName         x y = liftQ $ qLookupName         x y
  qNewName            x   = liftQ $ qNewName            x
  qReifyAnnotations   x   = liftQ $ qReifyAnnotations   x
  qReifyConStrictness x   = liftQ $ qReifyConStrictness x
  qReifyFixity        x   = liftQ $ qReifyFixity        x
  qReifyInstances     x y = liftQ $ qReifyInstances     x y
  qReifyModule        x   = liftQ $ qReifyModule        x
  qReifyRoles         x   = liftQ $ qReifyRoles         x
  qReport             x y = liftQ $ qReport             x y
  qRunIO              x   = liftQ $ qRunIO              x

With this instance, I can now transform Q [Dec] templates into LocalQ () templates, and the transformed version will use reifyLocallyFirst instead of reify.

localize :: Q [Dec] -> LocalQ ()
localize declarationTemplate = do
  decls <- unQ declarationTemplate
  addLocalDecls decls

generateBoth :: Name -> Q [Dec]
generateBoth name = runLocalQ $ do
  localize $ generateDoubled name
  localize $ generateDefault (doubledName name)


data X = X
data Y = Y X

generateDefault ''X
generateDefault ''Y
generateBoth ''X
generateBoth ''Y

Notice that I am reusing the original generateDefault and generateDoubled implementations, those which caused the reify error when I first tried to implement generateBoth. I am not using the locallyGenerateDefault and locallyGenerateDoubled reimplementations from the previous section. This means that (with a more fleshed out implementation of LocalQ), I should be able to reuse any existing declaration template out there, including Control.Lens.TH.makeLenses! :D