haskell - 为什么 `Vector.length (Vector.replicate n 0)"没有融合？

Question

以下代码意外地(至少对我而言)产生了一个中间向量:

import qualified Data.Vector as Vector

main :: IO ()
main =
  print (test n)

n :: Int
n = 1000000

test :: Int -> Int
test n = Vector.length (Vector.replicate n (0 :: Int))

Core 的相关部分在这里(注意 newArray# 1000000 调用):

Main.main4
  :: forall s_a38t.
     GHC.Prim.State# s_a38t
     -> (# GHC.Prim.State# s_a38t, Vector.Vector Int #)
[GblId,
 Arity=1,
 Str=DmdType,
 Unf=Unf{Src=<vanilla>, TopLvl=True, Value=True, ConLike=True,
         WorkFree=True, Expandable=True, Guidance=IF_ARGS [0] 399 30}]
Main.main4 =
  \ (@ s_a38t) (s1_a38u [OS=OneShot] :: GHC.Prim.State# s_a38t) ->
    case GHC.Prim.newArray#
           @ Int
           @ (Control.Monad.Primitive.PrimState (GHC.ST.ST s_a38t))
           1000000
           (Data.Vector.Mutable.uninitialised @ Int)
           (s1_a38u
            `cast` ((GHC.Prim.State#
                       (Sym (Control.Monad.Primitive.TFCo:R:PrimStateST[0] <s_a38t>_N)))_R
                    :: GHC.Prim.State# s_a38t
                       ~R# GHC.Prim.State#
                             (Control.Monad.Primitive.PrimState (GHC.ST.ST s_a38t))))
    of _ [Occ=Dead] { (# ipv_a5RG, ipv1_a5RH #) ->
    letrec {
      $wa_s609 [InlPrag=[0], Occ=LoopBreaker]
        :: GHC.Types.SPEC
           -> GHC.Prim.Int#
           -> Bool
           -> GHC.Prim.State# s_a38t
           -> (# GHC.Prim.State# s_a38t, Int #)
      [LclId, Arity=4, Str=DmdType <S,1*U><L,U><S,1*U><L,U>]
      $wa_s609 =
...

同时如果我更换 length与 sum , 融合正确发生:

test n = Vector.sum (Vector.replicate n (0 :: Int))

核:

Rec {
Main.main_$s$wfoldlM'_loop [Occ=LoopBreaker]
  :: GHC.Prim.Int# -> GHC.Prim.Int# -> GHC.Prim.Int#
[GblId, Arity=2, Caf=NoCafRefs, Str=DmdType <L,U><L,U>]
Main.main_$s$wfoldlM'_loop =
  \ (sc_s6bx :: GHC.Prim.Int#) (sc1_s6by :: GHC.Prim.Int#) ->
    case GHC.Prim.tagToEnum# @ Bool (GHC.Prim.<=# sc1_s6by 0)
    of _ [Occ=Dead] {
      False ->
        Main.main_$s$wfoldlM'_loop sc_s6bx (GHC.Prim.-# sc1_s6by 1);
      True -> sc_s6bx
    }
end Rec }

Main.main2 :: String
[GblId,
 Str=DmdType,
 Unf=Unf{Src=<vanilla>, TopLvl=True, Value=False, ConLike=False,
         WorkFree=False, Expandable=False, Guidance=IF_ARGS [] 100 30}]
Main.main2 =
  case Main.main_$s$wfoldlM'_loop 0 1000000 of ww_s67W { __DEFAULT ->
  case GHC.Show.$wshowSignedInt 0 ww_s67W (GHC.Types.[] @ Char)
  of _ [Occ=Dead] { (# ww5_a5Vq, ww6_a5Vr #) ->
  GHC.Types.: @ Char ww5_a5Vq ww6_a5Vr
  }
  }

此外，如果我根据一元流组合器重写原始函数，则也不会分配中间向量:

import qualified Data.Vector.Fusion.Stream.Monadic as Stream
import Data.Functor.Identity

test n = runIdentity $ Stream.length (Stream.replicate n (0 :: Int))

核:

Rec {
Main.main_$s$wfoldlM'_loop [Occ=LoopBreaker]
  :: GHC.Prim.Int# -> GHC.Prim.Int# -> GHC.Prim.Int#
[GblId, Arity=2, Caf=NoCafRefs, Str=DmdType <L,U><L,U>]
Main.main_$s$wfoldlM'_loop =
  \ (sc_s5lE :: GHC.Prim.Int#) (sc1_s5lF :: GHC.Prim.Int#) ->
    case GHC.Prim.tagToEnum# @ Bool (GHC.Prim.<=# sc1_s5lF 0)
    of _ [Occ=Dead] {
      False ->
        Main.main_$s$wfoldlM'_loop
          (GHC.Prim.+# sc_s5lE 1) (GHC.Prim.-# sc1_s5lF 1);
      True -> sc_s5lE
    }
end Rec }

Main.main2 :: String
[GblId,
 Str=DmdType,
 Unf=Unf{Src=<vanilla>, TopLvl=True, Value=False, ConLike=False,
         WorkFree=False, Expandable=False, Guidance=IF_ARGS [] 100 30}]
Main.main2 =
  case Main.main_$s$wfoldlM'_loop 0 1000000 of ww_s5ke { __DEFAULT ->
  case GHC.Show.$wshowSignedInt 0 ww_s5ke (GHC.Types.[] @ Char)
  of _ [Occ=Dead] { (# ww5_a5gi, ww6_a5gj #) ->
  GHC.Types.: @ Char ww5_a5gi ww6_a5gj
  }
  }

为什么 Vector.length打破融合？

我正在使用 ghc-7.10.3和 vector-0.11.0.0 .

添加:
这是一个问题: https://github.com/haskell/vector/issues/111

Answer 1

我用了sum和 length来自 Data.Vector.Generic而不是 Data.Vector因为后者只是定义为前者。

这是长度的代码(来自 Data.Vector.Generic )...

-- | /O(1)/ Yield the length of the vector.
length :: Vector v a => v a -> Int
{-# INLINE length #-}
length = Bundle.length . stream

嗯..让我们看看“总和”

-- | /O(n)/ Compute the sum of the elements
sum :: (Vector v a, Num a) => v a -> a
{-# INLINE sum #-}
sum = Bundle.foldl' (+) 0 . stream

但是如果我运行 ghc -ddump-inlinings -ddump-rule-firings -O2总和我看到

Rule fired: SPEC Data.Vector.$fVectorVectora [GHC.Types.Int]
Inlining done: System.IO.print
Inlining done: System.IO.print1
Inlining done: Data.Vector.Generic.sum
Rule fired: Class op +
Rule fired: Class op fromInteger
Inlining done: GHC.Num.$fNumInt_$cfromInteger
Rule fired: integerToInt
Inlining done: Data.Vector.Fusion.Util.unId
Inlining done: Data.Vector.Fusion.Util.unId1
Inlining done: Data.Vector.replicate
Inlining done: Data.Vector.Generic.replicate

如果我用 length 运行它我懂了:

Rule fired: SPEC Data.Vector.$fVectorVectora [GHC.Types.Int]
Inlining done: System.IO.print
Inlining done: System.IO.print1
Inlining done: Data.Vector.replicate
Inlining done: Data.Vector.Generic.replicate
Rule fired: SPEC Data.Vector.$fVectorVectora [GHC.Types.Int]

所以 sum被内联和 length没有，我不明白为什么。甚至将展开的阈值提高到 absurd 的数量也不会改变这一点。

也就是说，如果我手动替换 Vector.length与 Bundle.length . Vector.stream , stream/unstream规则会触发，如 sum在这种情况下，生成了一个非常整洁的核心，没有分配数组。