Question

我在Haskell中尝试并行和并发，并提出了这个问题。我有2个函数来并行计算列表的每个元素：

    parMap' :: (a -> b) -> [a] -> Eval [b]
    parMap' f [] = return []
    parMap' f (a:as) = do
       b <- rpar (f a)
       bs <- parMap' f as
       return (b:bs)

    parListWHNF :: Strategy [a]
    parListWHNF xs = go xs `pseq` return xs
      where -- go :: [a] -> [a]
        go []     = []
        go (y:ys) = y `par` go ys

如果我直接使用它们，一切都很好（所有火花都被转换）：

    let solutions = runEval (parMap' solve puzzles)
result: SPARKS: 1000 (1000 converted, 0 overflowed, 0 dud, 0 GC'd, 0 fizzled)

    let solutions = map solve puzzles `using` parListWHNF
result: SPARKS: 1000 (1000 converted, 0 overflowed, 0 dud, 0 GC'd, 0 fizzled)

但是如果我将它们包装在另一个函数中：

    parList' :: (NFData a) => Strategy a -> Strategy [a]
    parList' strat xs = do
      parMap' id xs          -- case #1
      -- parListWHNF xs      -- case #2
      return xs

    let solutions = map solve puzzles `using` parList' rseq

然后，只有case #2有效（所有火花都被转换），case #1会产生很多火花GC＆＃39; d：

    SPARKS: 1000 (3 converted, 0 overflowed, 0 dud, 997 GC'd, 0 fizzled)

任何人都可以解释这个PLZ吗？

如何将火花计算为GC＆＃39; d？

0 个答案: