parTraversable没有产生任何火花

Question

当我运行此程序时-s:

module Main where

import Control.Parallel.Strategies

main :: IO ()
main = do let xs = take 1000 $ product . take 1000 . repeat <$> [1..]
              x  = product (xs `using` parList rseq)
          putStrLn (show x)

火花创建:

SPARKS:1000(993转换,0溢出,0哑,6 GC,1失败)

如果我换parList到parTraversable

x  = product (xs `using` parTraversable rseq)

没有产生火花:

SPARKS:0(0转换,0溢出,0哑,0 GC'd,0失败)

如果我rseq改为rdeepseq:

main = do let xs = (take 1000 $ product . take 1000 . repeat <$> [1..]) :: [Integer]
              x  = product (xs `using` parList rdeepseq)
          putStrLn (show x)

没有产生火花

SPARKS:0(0转换,0溢出,0哑,0 GC'd,0失败)

我使用parallel-3.2.1.1并在源代码parList中定义使用parTraversable!

parList :: Strategy a -> Strategy [a]
parList = parTraversable

我错过了什么？

Answer 1

我可以重现你的行为(ghc-8.2.1,parallel-3.2.1.1).

后来,这Strategies.hs是一个RULES特殊情况的pragma parList rseq.我想这是一个错误,它有不同的行为parTraversable(我不太了解内部,以确定错误所在的位置).我建议在parallel问题跟踪器上提交一张票:https://github.com/haskell/parallel/issues

这是有问题的代码,从文件的第505行开始:

-- Non-compositional version of 'parList', evaluating list elements
-- to weak head normal form.
-- Not to be exported; used for optimisation.

-- | DEPRECATED: use @'parList' 'rseq'@ instead
parListWHNF :: Strategy [a]
parListWHNF xs = go xs `pseq` return xs
  where -- go :: [a] -> [a]
           go []     = []
           go (y:ys) = y `par` go ys

-- The non-compositional 'parListWHNF' might be more efficient than its
-- more compositional counterpart; use RULES to do the specialisation.

{-# NOINLINE [1] parList #-}
{-# NOINLINE [1] rseq #-}
{-# RULES
 "parList/rseq" parList rseq = parListWHNF
 #-}