haskell - parTraversable 不会产生任何 Spark

Question

当我使用 -s 运行该程序时:

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)

Spark 产生:

SPARKS: 1000 (993 converted, 0 overflowed, 0 dud, 6 GC'd, 1 fizzled)

如果我将 parList 更改为 parTraversable

x  = product (xs `using` parTraversable rseq)

没有产生 Spark :

SPARKS: 0 (0 converted, 0 overflowed, 0 dud, 0 GC'd, 0 fizzled)

如果我将 rseq 更改为 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 Spark

SPARKS: 0 (0 converted, 0 overflowed, 0 dud, 0 GC'd, 0 fizzled)

我使用的是parallel-3.2.1.1，在源代码中，parList是使用parTraversable定义的!

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

我错过了什么？

Answer 1

我可以重现您的行为(ghc-8.2.1、parallel-3.2.1.1)。

稍后，Strategies.hs 是一个 RULES 编译指示，用于特殊情况 parList rseq。我猜这是一个错误，它与 parTraversable 具有不同的行为(我对内部结构了解不够，无法确定错误所在)。我建议在并行问题跟踪器中提交票证:https://github.com/haskell/parallel/issues

这里是有问题的代码，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
 #-}