7.10.1
Allow use of any typeclass in deriving clauses.
With DeriveAnyClass you can derive any other class. The compiler will simply generate an instance declaration with no explicitly-defined methods. This is mostly useful in classes whose minimal set is empty, and especially when writing generic functions.
As an example, consider a simple pretty-printer class SPretty, which outputs pretty strings:
{-# LANGUAGE DefaultSignatures, DeriveAnyClass #-}
class SPretty a where
sPpr :: a -> String
default sPpr :: Show a => a -> String
sPpr = show
If a user does not provide a manual implementation for sPpr, then it will default to show. Now we can leverage the DeriveAnyClass extension to easily implement a SPretty instance for a new data type:
data Foo = Foo deriving (Show, SPretty)
The above code is equivalent to:
data Foo = Foo deriving Show instance SPretty Foo
That is, an SPretty Foo instance will be created with empty implementations for all methods. Since we are using DefaultSignatures in this example, a default implementation of sPpr is filled in automatically.
Note the following details
In case you try to derive some class on a newtype, and GeneralizedNewtypeDeriving is also on, DeriveAnyClass takes precedence.
The instance context is determined by the type signatures of the derived classâs methods. For instance, if the class is:
class Foo a where bar :: a -> String default bar :: Show a => a -> String bar = show baz :: a -> a -> Bool default baz :: Ord a => a -> a -> Bool baz x y = compare x y == EQ
And you attempt to derive it using DeriveAnyClass:
instance Eq a => Eq (Option a) where ... instance Ord a => Ord (Option a) where ... instance Show a => Show (Option a) where ... data Option a = None | Some a deriving Foo
Then the derived Foo instance will be:
instance (Show a, Ord a) => Foo (Option a)
Since the default type signatures for bar and baz require Show a and Ord a constraints, respectively.
Constraints on the non-default type signatures can play a role in inferring the instance context as well. For example, if you have this class:
class HigherEq f where (==#) :: f a -> f a -> Bool default (==#) :: Eq (f a) => f a -> f a -> Bool x ==# y = (x == y)
And you tried to derive an instance for it:
instance Eq a => Eq (Option a) where ... data Option a = None | Some a deriving HigherEq
Then it will fail with an error to the effect of:
No instance for (Eq a)
arising from the 'deriving' clause of a data type declaration
That is because we require an Eq (Option a) instance from the default type signature for (==#), which in turn requires an Eq a instance, which we donât have in scope. But if you tweak the definition of HigherEq slightly:
class HigherEq f where (==#) :: Eq a => f a -> f a -> Bool default (==#) :: Eq (f a) => f a -> f a -> Bool x ==# y = (x == y)
Then it becomes possible to derive a HigherEq Option instance. Note that the only difference is that now the non-default type signature for (==#) brings in an Eq a constraint. Constraints from non-default type signatures never appear in the derived instance context itself, but they can be used to discharge obligations that are demanded by the default type signatures. In the example above, the default type signature demanded an Eq a instance, and the non-default signature was able to satisfy that request, so the derived instance is simply:
DeriveAnyClass can be used with partially applied classes, such as
data T a = MKT a deriving( D Int )
which generates
instance D Int a => D Int (T a) where {}
DeriveAnyClass can be used to fill in default instances for associated type families:
{-# LANGUAGE DeriveAnyClass, TypeFamilies #-}
class Sizable a where
type Size a
type Size a = Int
data Bar = Bar deriving Sizable
doubleBarSize :: Size Bar -> Size Bar
doubleBarSize s = 2*s
The deriving( Sizable ) is equivalent to saying
instance Sizeable Bar where {}
and then the normal rules for filling in associated types from the default will apply, making Size Bar equal to Int.
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