This is valid syntax in GHC 9. What do the {..}
mean (as distinct from (..)
which GHC 8.10 requires here)?
ign :: forall {f :: Type -> Type} {p}. Applicative f => p -> f ()
ign _ = pure ()
See 6.4.14.1. Inferred vs. specified type variables :
..
is a specified type {..}
is an inferred type forall a.
and forall {a}.
are invisible quantifiers that GHC will instantiate automatically by unification.
This means writing const True EQ
actually instantiates a and b without the users help.
const :: forall a b. a -> b -> a
const a _ = a
If the user wants to explicitly instantiate them they can "override their visibility", using visible type applications. That's why they are specified types. (although "specifi able " might be a more accurate terminology)
>> :set -XTypeApplications
>> :t const @Bool @Ordering True EQ
const @Bool @Ordering True EQ :: Bool
If for some reason we only want to specify b (without summoning the "snail squad": @_
, umm " partial type signatures ") we can make a an inferred type. Then the first type is dropped
const2 :: forall {a} b. a -> b -> a
const2 a _ = a
>> :t const2 @Ordering True EQ
const2 @Ordering True EQ :: Bool
For your example it means ghc must infer the type of f and p . You cannot write ign @IO @Int
.
This becomes more useful when you have kind polymorphism. If you define
type Apply :: forall (k :: Type). (k -> Type) -> (k -> Type)
newtype Apply f a where
MkApply :: forall (k :: Type) (f :: k -> Type) (a :: k). f a -> Apply @k f a
you must specify the kind k when instantiating MkApply @Type @[] @Int
but this kind is implied by both []
and Int
.
so you might prefer marking k as inferred in MkApply
so you can write MkApply @[] @Int
type Apply :: forall (k :: Type). (k -> Type) -> (k -> Type)
newtype Apply f a where
MkApply :: forall {k :: Type} (f :: k -> Type) (a :: k). f a -> Apply @k f a
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.