计数反转：Frege中的StackOverflowError在Haskell中工作正常

Question

我试图计算一个数字列表的倒数。 以下Frege程序适用于少量数字，但抛出StackOverflowError为100000个数字。

import frege.IO

inversionCount [] _ = (0, [])
inversionCount [x] _ = (0, [x])
inversionCount xs n = (count, sorted) where
  count = lcount + rcount + mergecount
  (lcount, lsorted) = inversionCount left lsize
  (rcount, rsorted) = inversionCount right rsize
  (mergecount, sorted) = inversionMergeCount lsorted lsize rsorted rsize (0, [])
  (left, right) = splitAt mid xs
  mid = n `quot` 2
  lsize = mid
  rsize = n - mid

inversionMergeCount xs _ [] _ (acc,sorted) = (acc, reverse sorted ++ xs)
inversionMergeCount [] _ ys _ (acc,sorted) = (acc, reverse sorted ++ ys)
inversionMergeCount (xs@(x:restx)) m (ys@(y:resty)) n (acc, sorted)
  | x < y = inversionMergeCount restx (m - 1) ys n (acc, x:sorted)
  | x > y = inversionMergeCount xs m resty (n - 1) (acc + m, y:sorted)
  | otherwise = inversionMergeCount restx (m - 1) resty (n - 1) (acc, x:y:sorted)

main (fileName:_) = do
  input <- readFile fileName
  let xs = map atoi (lines input)
  println . fst $ inversionCount xs 100000

--Haskell's readFile using Java's Scanner
type JScanner = JScannerT RealWorld

data JScannerT s = native java.util.Scanner where
  native new :: File -> ST s (Exception JScanner)
  native useDelimiter :: JScanner -> String -> ST s JScanner
  native next :: JScanner -> ST s String

readFile f = do
  file <- File.new f
  exceptionScanner <- JScanner.new file
  let scanner = either throw id exceptionScanner
  scanner.useDelimiter "\\Z"
  scanner.next

Haskell中的相同代码工作正常：

import System.Environment

inversionCount [] _ = (0, [])
inversionCount [x] _ = (0, [x])
inversionCount xs n = (count, sorted) where
  count = lcount + rcount + mergecount
  (lcount, lsorted) = inversionCount left lsize
  (rcount, rsorted) = inversionCount right rsize
  (mergecount, sorted) = inversionMergeCount lsorted lsize rsorted rsize (0, [])
  (left, right) = splitAt mid xs
  mid = n `quot` 2
  lsize = mid
  rsize = n - mid

inversionMergeCount xs _ [] _ (acc,sorted) = (acc, reverse sorted ++ xs)
inversionMergeCount [] _ ys _ (acc,sorted) = (acc, reverse sorted ++ ys)
inversionMergeCount (xs@(x:restx)) m (ys@(y:resty)) n (acc, sorted)
  | x < y = inversionMergeCount restx (m - 1) ys n (acc, x:sorted)
  | x > y = inversionMergeCount xs m resty (n - 1) (acc + m, y:sorted)
  | otherwise = inversionMergeCount restx (m - 1) resty (n - 1) (acc, x:y:sorted)


main = do
  (fileName: _) <- getArgs
  contents <- readFile fileName
  let xs :: [Int]
      xs = map read (lines contents)
  print . fst $ inversionCount xs 100000

堆栈溢出的原因是什么？ 这是一些函数不是尾递归的吗？

Answer 1

Haskell很可能有一个更好的严格性分析器，或者尾部递归的实现方式不同，或者运行时只有更多的堆栈可用。

我要尝试的第一件事是设置-Xss8m，甚至是16m。

如果这没有帮助，请记住，使用诸如（ - ），（+）等严格函数的应用程序更新的延迟参数构建了有时候必须立即评估的thunk。 这与foldl问题相同，看起来像inversionMergeCount和inversionCount的第二个参数受此影响。

如果它看到这个，Frege编译器应该警告这个，但它现在不是。

另一点是，为什么你传递元组中的最后两个参数？ 你也可以使acc严格。

Answer 2

根据Ingo的建议，我做了更改，代码现在正在运行。 我还必须将计数作为Integer而不是Int来避免int溢出。 这是带有严格注释和Integer转换的更新代码：

inversionCount [] _ = (zero, [])
inversionCount [x] _ = (zero, [x])
inversionCount xs n = (count, sorted) where
  count = lcount + rcount + mergecount
  (lcount, lsorted) = inversionCount left lsize
  (rcount, rsorted) = inversionCount right rsize
  (mergecount, sorted) = inversionMergeCount lsorted lsize rsorted rsize zero []
  (left, right) = splitAt mid xs
  mid = n `quot` 2
  lsize = mid
  rsize = n - mid


inversionMergeCount xs _ [] _ !acc sorted = (acc, reverse sorted ++ xs)
inversionMergeCount [] _ ys _ !acc sorted = (acc, reverse sorted ++ ys)
inversionMergeCount (xs@(x:restx)) !m (ys@(y:resty)) n !acc sorted
  | x < y = inversionMergeCount restx (pred m) ys n acc (x:sorted)
  | x > y = inversionMergeCount xs m resty (pred n) (acc + m.big) (y:sorted)
  | otherwise = inversionMergeCount restx (pred m) resty (pred n) acc (x:y:sorted)