Question

当我手动编写PhisicalCell数据类型的简单show实例时，程序会占用所有空间。在派生他自己的Show版本时，这不会发生。为什么呢？

这是我正在编写的代码的简化版本：

import Data.Array

type Dimensions = (Int, Int)
type Position = (Int, Int)
data PipeType = Vertical | Horizontal | UpLeft | UpRight | DownLeft | DownRight deriving (Show)

data PhisicalCell = AirCell
                  | PipeCell PipeType
                  | DeathCell
                  | RecipientCell Object
                  -- deriving (Show) SEE THE PROBLEM BELOW  

data Object = Pipe { pipeType :: PipeType  -- tipo di tubo
                   , position :: Position  -- posizione del tubo
                   , movable  :: Bool      -- se posso muoverlo
                   }
            | Bowl { position   :: Position                 -- posizione dell'angolo in alto a sinistra
                   , dimensions :: Dimensions               -- dimensioni (orizzontale, verticale)
                   , waterMax   :: Int                      -- quanta acqua puo' contenere al massimo
                   , waterStart :: Int                      -- con quanta acqua parte 
                   , hatch      :: Maybe Position           -- un eventuale casella di sbocco
                   , sourceIn   :: [Position]               -- posti da cui l'acqua entra
                   , movable    :: Bool                     -- se posso muoverlo
                   }
            | Death
            deriving (Show)

data Level = Level Dimensions [Object]
type LevelTable = Array Dimensions PhisicalCell

-- HERE IS THE PROBLEM -- 
instance Show PhisicalCell where
show AirCell = " "
show (PipeCell _) = "P"
show DeathCell = "X"
show (RecipientCell _) = "U"

both :: (a -> b) -> (a,a) -> (b,b)
both f (a,b) = (f a, f b)

levelTable :: Level -> LevelTable
levelTable (Level dim _) = initial
  where initial = array ((0,0), both (+1) dim) $
                    [((x,y), AirCell) | x <- [1..fst dim], y <- [1..snd dim] ]
                    ++ [((x,y), DeathCell) | x <- [0..fst dim + 1], y <- [0, snd dim + 1]]
                    ++ [((x,y), DeathCell) | x <- [0, fst dim + 1], y <- [0..snd dim + 1]]

main = print $ levelTable (Level (8,12) [])

Answer 1

The Show type class has mutually referencing default implementations：

class  Show a  where
    -- | Convert a value to a readable 'String'.
    --
    -- 'showsPrec' should satisfy the law
    -- ...
    ...
    showsPrec _ x s = show x ++ s
    show x          = shows x ""
    showList ls   s = showList__ shows ls s

...

shows           :: (Show a) => a -> ShowS
shows           =  showsPrec 0

因此，如果您声明Show实例而未定义任何方法

instance Show where

nextNewFunction :: Bla
...

GHC将很乐意编译所有默认值，因此不会有任何错误。但是，只要你尝试使用中的任何一个，你就会像你的Objects一样陷入一个致命的循环......并且相互递归最终会使堆栈崩溃。

现在，你的代码看起来并不像你有一个空的instance Show声明，但实际上你做了：因为错误的缩进，你定义的show被识别为一个新的免费顶级函数，它恰好与GHC.Show.show具有相同的名称。你可以添加

show :: PhisicalCell -> String

到您的文件并获得与现在相同的结果。

Answer 2

问题实际上在于Sassa NF指出的间距。当我缩进节目时，它可以工作（当我没有时，我得到堆栈溢出）。如果没有缩进，您将定义从未使用过的顶级show函数，而show Show实例的PhisicalCell函数具有未定义的show函数1}}函数，导致问题。

手动实例显示定义导致堆栈空间溢出

2 个答案: