我有标准的Vect类型:
module Vect where
data Nat = Z | S Nat
data Vect (n :: Nat) (a :: Type) where
VNil :: Vect Z a
VCons :: a -> Vect n a -> Vect (S n) a
我上面有这些功能:
foldVect :: forall (p :: Nat -> Type) n a.
p Z ->
(forall m. a -> p m -> p (S m)) ->
Vect n a -> p n
foldVect n c = go
where go :: forall l. Vect l a -> p l
go VNil = n
go (VCons x xs) = x `c` go xs
newtype FVect a n = FVect { unFVect :: Vect n a }
buildVect :: forall n a.
(forall (p :: Nat -> Type).
p Z ->
(forall m. a -> p m -> p (S m)) ->
p n
) -> Vect n a
buildVect f = unFVect $ f (FVect VNil) $ \x (FVect xs) -> FVect $ x `VCons` xs
我试图从
base重新创建(部分)允许列表融合的机器:instance Functor (Vect n) where
fmap = mapVect
{-# INLINE fmap #-}
mapVect :: forall n a b. (a -> b) -> (Vect n a -> Vect n b)
mapVect _f VNil = VNil
mapVect f (VCons x xs) = f x `VCons` mapVect f xs
mapFB :: forall (p :: Nat -> Type) n a b. (forall m. b -> p m -> p (S m)) -> (a -> b) -> a -> p n -> p (S n)
mapFB cons f = \x ys -> cons (f x) ys
{-# INLINE [0] mapFB #-}
{-# NOINLINE [0] mapVect #-}
{-# RULES "mapVect" [~1] forall f xs. mapVect f xs = buildVect (\nil cons -> foldVect nil (mapFB cons f) xs) #-}
{-# INLINE [0] foldVect #-}
-- base has this; I don't think it does anything without a "refolding" rule on mapVect
{-# INLINE [0] buildVect #-}
{-# RULES "foldVect/buildVect" forall (nil :: p Z)
(cons :: forall m. a -> p m -> p (S m))
(f :: forall (q :: Nat -> Type).
q Z ->
(forall m. a -> q m -> q (S m)) ->
q n
).
foldVect nil cons (buildVect f) = f nil cons
#-}
然后
module Test where
import Vect
test :: Vect n Int -> Vect n Int
test = fmap (*5) . fmap (+2)
没有融合发生。如果我通过
-ddump-simpl，我看到foldVect和buildVect已经被内联，但是...INLINE是分阶段的，因此它们无论如何都不会干扰融合。 (毕竟，这是base对[]的处理方式)INLINE [0]替换NOINLINE会绘制出非常漂亮的图像:test
= \ (@ (n_a141 :: Nat)) (x_X1lK :: Vect n_a141 Int) ->
buildVect
@ n_a141
@ Int
(\ (@ (p_X1jl :: Nat -> *))
(nil_X11K [OS=OneShot] :: p_X1jl 'Z)
(cons_X11M [OS=OneShot]
:: forall (m :: Nat). Int -> p_X1jl m -> p_X1jl ('S m)) ->
foldVect
@ p_X1jl
@ n_a141
@ Int
nil_X11K
(\ (@ (m_a1i5 :: Nat))
(x1_aYI :: Int)
(ys_aYJ [OS=OneShot] :: p_X1jl m_a1i5) ->
cons_X11M
@ m_a1i5
(case x1_aYI of { GHC.Types.I# x2_a1l5 ->
GHC.Types.I# (GHC.Prim.*# x2_a1l5 5#)
})
ys_aYJ)
(buildVect
@ n_a141
@ Int
(\ (@ (p1_a1i0 :: Nat -> *))
(nil1_a10o [OS=OneShot] :: p1_a1i0 'Z)
(cons1_a10p [OS=OneShot]
:: forall (m :: Nat). Int -> p1_a1i0 m -> p1_a1i0 ('S m)) ->
foldVect
@ p1_a1i0
@ n_a141
@ Int
nil1_a10o
(\ (@ (m_a1i5 :: Nat))
(x1_aYI :: Int)
(ys_aYJ [OS=OneShot] :: p1_a1i0 m_a1i5) ->
cons1_a10p
@ m_a1i5
(case x1_aYI of { GHC.Types.I# x2_a1lh ->
GHC.Types.I# (GHC.Prim.+# x2_a1lh 2#)
})
ys_aYJ)
x_X1lK)))
一切都在那里，但是简化器没有它。
如果我检查规则本身，我会看到
"foldVect/buildVect"
forall (@ (p_aYG :: Nat -> *))
(@ (n_aYJ :: Nat))
(@ a_aYH)
(nil_aYD :: p_aYG 'Z)
(cons_aYE :: forall (m :: Nat). a_aYH -> p_aYG m -> p_aYG ('S m))
(f_aYF
:: forall (q :: Nat -> *).
q 'Z -> (forall (m :: Nat). a_aYH -> q m -> q ('S m)) -> q n_aYJ).
foldVect @ p_aYG
@ n_aYJ
@ a_aYH
nil_aYD
cons_aYE
(buildVect
@ n_aYJ
@ a_aYH
(\ (@ (p1_a156 :: Nat -> *))
(ds_d1io :: p1_a156 'Z)
(ds1_d1ip
:: forall (m :: Nat). a_aYH -> p1_a156 m -> p1_a156 ('S m)) ->
f_aYF @ p1_a156 ds_d1io ds1_d1ip))
= f_aYF @ p_aYG nil_aYD cons_aYE
看来问题在于
buildVect的参数需要是一种非常特定形式的lambda抽象，并且在构建最终发生重写的系统时遇到了麻烦。如何融合才能工作？
(我不知道这是否有用或什至正确；我只是在这样做以查看是否可以。)
最佳答案
像往常一样，每当编译器受到威胁时，newtype都会节省一天的时间:
module Vect where
-- everything else the same...
newtype VectBuilder n a = VectBuilder { runVectBuilder :: forall (p :: Nat -> Type).
p Z ->
(forall m. a -> p m -> p (S m)) ->
p n
}
buildVect' :: forall n a. VectBuilder n a -> Vect n a
buildVect' f = unFVect $
runVectBuilder f (FVect VNil) $ \x (FVect xs) -> FVect $ x `VCons` xs
{-# INLINE [0] buildVect' #-}
buildVect :: forall n a.
(forall (p :: Nat -> Type).
p Z ->
(forall m. a -> p m -> p (S m)) ->
p n
) -> Vect n a
buildVect f = buildVect' (VectBuilder f)
{-# INLINE buildVect #-}
{-# RULES "foldVect/buildVect'" forall (nil :: p Z)
(cons :: forall m. a -> p m -> p (S m))
(f :: VectBuilder n a).
foldVect nil cons (buildVect' f) = runVectBuilder f nil cons
#-}
-- compiler no longer has a chance to muck up the LHS by eta expanding f because
-- f "isn't" a function anymore
-- rule for mapVect goes unchanged, so I guess that's evidence that this is totally transparent
module Test where
import Vect
test :: Vect n Int -> Vect n Int
test = fmap (*5) . fmap (+2)
Rec {
-- RHS size: {terms: 19, types: 31, coercions: 13, joins: 0/0}
Test.test_go [Occ=LoopBreaker]
:: forall (l :: Nat). Vect l Int -> FVect Int l
[GblId, Arity=1, Caf=NoCafRefs, Str=<S,1*U>]
Test.test_go
= \ (@ (l_a14W :: Nat)) (ds_d1jk :: Vect l_a14W Int) ->
case ds_d1jk of {
VNil co_a14Y -> (Vect.$WVNil @ Int) `cast` <Co:4>;
VCons @ n2_a151 co_a152 x_aYE xs_aYF ->
(Vect.VCons
@ ('S n2_a151)
@ Int
@ n2_a151
@~ <Co:2>
(case x_aYE of { GHC.Types.I# x1_a1xr ->
GHC.Types.I# (GHC.Prim.*# (GHC.Prim.+# x1_a1xr 2#) 5#) -- success!
})
((Test.test_go @ n2_a151 xs_aYF) `cast` <Co:3>))
`cast` <Co:4>
}
end Rec }
-- RHS size: {terms: 4, types: 5, coercions: 0, joins: 0/0}
Test.test1 :: forall (n :: Nat). Vect n Int -> FVect Int n
[GblId,
Arity=1,
Caf=NoCafRefs,
Str=<S,1*U>,
Unf=Unf{Src=<vanilla>, TopLvl=True, Value=True, ConLike=True,
WorkFree=True, Expandable=True,
Guidance=ALWAYS_IF(arity=1,unsat_ok=True,boring_ok=True)}]
Test.test1
= \ (@ (n_a1wd :: Nat)) (x_X1xa :: Vect n_a1wd Int) ->
Test.test_go @ n_a1wd x_X1xa
-- RHS size: {terms: 1, types: 0, coercions: 9, joins: 0/0}
test :: forall (n :: Nat). Vect n Int -> Vect n Int
[GblId,
Arity=1,
Caf=NoCafRefs,
Str=<S,1*U>,
Unf=Unf{Src=<vanilla>, TopLvl=True, Value=True, ConLike=True,
WorkFree=True, Expandable=True,
Guidance=ALWAYS_IF(arity=0,unsat_ok=True,boring_ok=True)}]
test = Test.test1 `cast` <Co:9>
故事的寓意:排名靠前的类型会使
RULES系统崩溃，因此即使没有其他必要，也可以向GHC提供有关newtype的帮助。关于haskell - 融合以编制长度索引的链表，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/50844444/