haskell - GHCi 调试器没有递归地命中断点 : Why, 解决方案是什么？

Question

我有一个不太有效的简单递归程序。所以我正在尝试使用 ghci 调试器来弄清楚发生了什么。我在函数 recurse、progress 和 shaded 的所有行上都设置了断点，它捕获了前几个。但是当我到达 progress 的第二行时，每次调用 :continue 都会让我回到同一行代码，即使我正在调用 recurse 和 shaded 从那里开始并期待我的断点起作用。这是代码:

import System.Environment

type Pos = (Int,Int) 
type Acc = ([[Pos]], [Pos])

main = do
  getArgs >>= putStrLn . show . length . combos . read . head

combos n = recurse n [] (allPos n) []

recurse :: Int -> [[Pos]] -> [Pos] -> [Pos] -> [[Pos]]
recurse n done avail inProg
   | length inProg == n = inProg:done
   | null avail = done
   | otherwise = fst $ foldr (progress n inProg) (done,avail) avail

progress :: Int -> [Pos] -> Pos -> Acc -> Acc
progress n inProg pos (done, avail)  =
  (recurse n done (filter (not . shaded pos) remain) (pos:inProg), remain)
  where remain = tail avail

allPos n =  [ (i,j) | i <- [0..n-1], j <- [0..n-1] ]

shaded :: Pos -> Pos -> Bool
shaded (i,j) (k,l) =
  k == i
  || l == j
  || k+l == i+j
  || k-l == i-j
  || abs (k-i) < 3 && abs (l-j) < 3

为什么 ghci 调试器不会在 progress 调用的函数中的断点处停止？是否有类似“不可重入”的东西将它们关闭？我怎样才能让调试器在每次递归调用这些函数时中断？

GHCi 版本 7.8.4

更新:我怀疑这可能与缓存的结果有关，但我并不清楚这些函数是否已使用相同的参数被调用两次。可能是我的代码中的错误？

Answer 1

我认为它按预期工作，但你必须考虑惰性求值。

如果你在 recurse 中突破警戒线(第 13 - 15 行)和递归调用(第 19 行)，当程序以 :main 2 运行时，您将看到这种模式。 :

Stopped at prog0.hs:13:6-23   - recurse
Stopped at prog0.hs:14:6-15   - recurse
Stopped at prog0.hs:15:18-67  - recurse, call to foldr progress
Stopped at prog0.hs:19:3-74   - in progress, pos = (1,1)
Stopped at prog0.hs:19:3-74   - in progress, pos = (1,0)
Stopped at prog0.hs:19:3-74   - in progress, pos = (0,1)
Stopped at prog0.hs:19:3-74   - in progress, pos = (0,0)
Stopped at prog0.hs:13:6-23   - in recurse
Stopped at prog0.hs:14:6-15
Stopped at prog0.hs:13:6-23   - in recurse
Stopped at prog0.hs:14:6-15   
Stopped at prog0.hs:13:6-23   - in recurse
Stopped at prog0.hs:14:6-15  
Stopped at prog0.hs:13:6-23   - in recurse
Stopped at prog0.hs:14:6-15

但是，如果您更改 progress强制其结果:

import Control.DeepSeq

progress n inProg pos (done, avail)  = let
  result = (recurse n done (filter (not . shaded pos) remain) (pos:inProg), remain)
  in deepseq result result
  where remain = tail avail

那么断点模式是:

Stopped at prog1.hs:14:6-23   - recurse
Stopped at prog1.hs:15:6-15   - recurse
Stopped at prog1.hs:16:18-67  - recurse, call to foldr progress
Stopped at prog1.hs:21:6-26   - progress, pos = (1,1)
Stopped at prog1.hs:14:6-23   - recurse
Stopped at prog1.hs:15:6-15
Stopped at prog1.hs:21:6-26   - progress, pos = (1,0)
Stopped at prog1.hs:14:6-23   - recurse
Stopped at prog1.hs:15:6-15
Stopped at prog1.hs:21:6-26   - progress, pos = (0,1)
Stopped at prog1.hs:14:6-23   - recurse
Stopped at prog1.hs:15:6-15
Stopped at prog1.hs:21:6-26   - progress, pos = (0,0)
Stopped at prog1.hs:14:6-23   - recurse
Stopped at prog1.hs:15:6-15

递归调用 recurse现在是交错的，因为我们立即强制他们。