Haskell語法分析程序錯誤，在where子句中

Question

rs定義中的第一個where部分有什麼問題？

palindrome :: [a] -> [a] 

palindrome xs = con xs rs 
    where con a b = rev (rev a []) b 
     rs = rev xs      -- here 
     where rev [] rs = rs 
      rev (x:xs) rs = rev xs (x:rs) 

我剛剛學習Haskell，但它的語法規則讓我困惑。該錯誤消息是

[1 of 1] Compiling Main    (pelindrome.hs, interpreted) 

pelindrome.hs:5:8: parse error on input `rs' 

Answer 1

你的縮進是錯誤的，我認爲你只能有一個where在那裏（我可以非常好地錯了。我不是一個Haskell的傢伙）。還有一個說法缺少調用rev（空單）：

palindrome :: [a] -> [a] 
palindrome xs = con xs rs 
    where con a b = rev (rev a []) b 
      rs = rev xs []      -- here 
      rev [] rs = rs 
      rev (x:xs) rs = rev xs (x:rs) 

main = print (palindrome "hello") 

打印出：

"helloolleh" 

我要去嘗試，現在明白了。無論如何，玩得開心！

編輯：現在對我來說非常合適。我認爲這是正確的版本。 Haskell的縮進規則，閱讀Haskell Indentation

Answer 2

@litb：您可以在其中是如何++中拉開序幕實現方式

palindrome :: [a] -> [a] 
palindrome xs = con xs rs 
    where con [] b = b 
      con (x:xs) b = x:con xs b 
      rs = rev xs []      -- here 
      rev [] rs = rs 
      rev (x:xs) rs = rev xs (x:rs) 

改寫CON。我以前的版本是如何以尾調用方式（例如Erlang）以非惰性函數或邏輯語言編寫的。