解析輸入錯誤'=' - Haskell

Question

當我運行這個Haskell代碼時，我得到了這個非常惱人的錯誤。出於某種原因，它不喜歡的拳頭線delLast :: (Ord a) => [a] -> [a]

delLast :: (Ord a) => [a] -> [a] 
delLast [] = [] 
delLast (x:[]) = [] 
delLast (x:xs) = [x] ++ delLast xs 

利己人知道爲什麼會即時得到這個錯誤？謝謝。

這是我的錯誤：

parse error on input `=' 
Failed, modules loaded: none. 

Answer 1

是否有一個未終止的直接定義在該行之前？

x = do 

delLast :: (Ord a) => [a] -> [a] 
delLast [] = [] 
delLast (x:[]) = [] 
delLast (x:xs) = [x] ++ delLast xs 

編譯，讓我：

test.hs:5:12: parse error on input `=' 

Answer 2

什麼

delLast [] = [] 
delLast xs = init xs 

，或者如果你想要做你的方式，但更好的

delLast :: (Ord a) => [a] -> [a] 
delLast [] = [] 
delLast [x] = []     -- Looks better IMO 
delLast (x:xs) = x : delLast xs -- : instead of ++ gives a performance boost