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haskell - 长度索引链表的融合

问题描述

我有标准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

我试图重新创建（部分）base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
  #-}

进而

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了，我会看到foldVect并且buildVect已经被内联了，但是......

  1. s 是分阶段的INLINE，因此它们无论如何都不会干扰融合。（毕竟，这是怎么base回事[]）

  2. 用 s替换INLINE [0]sNOINLINE绘制了一个相当惊人的图像：

    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，但简化器只是没有它。

如果我检查规则本身，我会看到这个

"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

看来问题在于参数buildVect需要是非常特定形式的 lambda 抽象，而我在构建最终发生的重写系统时遇到了麻烦。

如何让融合发挥作用？

（我不知道这是否有用甚至正确；我只是这样做看看我是否可以。）

标签： haskelloptimization

解决方案

像往常一样，newtype只要编译器被牛逼，就可以节省一天的时间：

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：排名足够高的类型会使RULES系统崩溃，因此请给 GHC 一些帮助newtypes，即使它们不是必需的。

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