Haskell解析错误输入`|' [英] Haskell parse error on input `|'

问题描述

Haskell不断给这个，我看不出为什么。试图用if / else重写，但后来抱怨... ...

错误信息; 2006grafen.hs：40：12：parse error on input`|'

代码：

 数据Edge a = Boog aa派生Show 
 data Graph a = Graaf [a] [Edge a]派生Show 
 
 vindBuren :: Eq a => a  - >图表a  - > [a] 
 vindBuren _（Graaf list []）= [] 
 vindBuren knoop（Graaf list（（Boog x y）：rest））
 | knoop == x = y：（vindBuren knoop（Graaf list rest））
 | knoop == y = x：（vindBuren knoop（Graaf list rest））
 |否则= vindBuren knoop（Graaf列表休息）
 
 graadKnoop ::等式a =>图表a  - > a  - > Int 
 graadKnoop graaf el = length（vindBuren el graaf）
 
 
 buurgraadKnoop :: Eq a =>图表a  - > a  - > [Int] 
 buurgraadKnoop graaf el = 
 let 
 buren = vindBuren el graaf 
 in 
 reverse $ sort $ map（graadKnoop graaf）buren 
 
 buurgraad :: Eq a =>图表a  - > [（[Int]，[a]）] 
 buurgraad graaf = hulp graaf [] 
 
 
 hulp :: Eq a =>图表a  - > [（[Int]，[a]）]  - > （Graaf（fstN：restN）l）soFar 
 | elem bgraden（map fst soFar）= hulp（Graaf restN l）（mergeG fstN bgraden soFar）
 |否则= hulp（Graaf restN l）[（bgraden，[fstN]）]：soFar 
其中
 bgraden = buurgraadKnoop g fstN 
 
 mergeG :: a  - > [Int]  - > [（[Int]，[a]）]  - > [（[Int]，[a]）] 
 mergeG n ngraden（（gradenL，nodenL）：rest）
 | ngraden == gradenL =（（gradenL，sort（nodenL：n））：rest 
 | otherwise = mergeG n ngraden rest 
  

$ b $你忘了关闭（（gradenL，sort（nodenL：n））中的圆括号。

前面的括号是多余的，所以你可以简单地写（gradenL，sort $ nodenL：n）

Haskell keeps giving this and I can't see why. Tried to rewrite with if/else but then it complains about that...

Error message; 2006grafen.hs:40:12: parse error on input `|'

Code:

data Edge a = Boog a a deriving Show
data Graph a = Graaf [a] [Edge a] deriving Show

vindBuren :: Eq a => a -> Graph a -> [a]
vindBuren _ (Graaf list []) = []
vindBuren knoop (Graaf list ((Boog x y):rest))
       | knoop == x = y : (vindBuren knoop (Graaf list rest))
       | knoop == y = x : (vindBuren knoop (Graaf list rest))
       | otherwise = vindBuren knoop (Graaf list rest)

graadKnoop :: Eq a => Graph a -> a -> Int
graadKnoop graaf el = length (vindBuren el graaf) 


buurgraadKnoop :: Eq a => Graph a -> a -> [Int]
buurgraadKnoop graaf el = 
       let
           buren = vindBuren el graaf
       in
          reverse $ sort $ map (graadKnoop graaf) buren

buurgraad :: Eq a => Graph a -> [([Int],[a])]
buurgraad graaf = hulp graaf []


hulp :: Eq a => Graph a -> [([Int],[a])] -> [([Int],[a])]
hulp (Graaf [] bgL) soFar = soFar
hulp g@(Graaf (fstN:restN) l) soFar
       | elem bgraden (map fst soFar) = hulp (Graaf restN l) (mergeG fstN bgraden soFar)
       | otherwise = hulp (Graaf restN l) [(bgraden,[fstN])]:soFar
       where
           bgraden = buurgraadKnoop g fstN

mergeG :: a -> [Int] -> [([Int],[a])] -> [([Int],[a])]
mergeG n ngraden ((gradenL,nodenL):rest)
           | ngraden == gradenL = ((gradenL, sort(nodenL:n)):rest
           | otherwise = mergeG n ngraden rest

解决方案

You forgot to close the parenthesis in ((gradenL, sort(nodenL:n)).

The outher parentheses are redundant anyway, so you could simply write (gradenL, sort $ nodenL : n)

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