與預期類型不符

如何解決此錯誤並使我的代碼有效？與預期類型不符

我的數據類型：

data ID x = ID ((x, Mark), [ ID x ]) 
    data Mark = Marked | Unmarked

顯示實例：

instance Show a => Show (ID a) where 

      show t = go " " t where 

        go aP (ID ((x, Marked), z)) = 

          aP ++ x ++ "\n"

錯誤：

Couldn't match expected type `[Char]' against inferred type `a' 
    `a' is a rigid type variable bound by 
     the instance declaration at Dictionary.hs:117:23 
    Expected type: ID [Char] 
    Inferred type: ID a 
In the second argument of `go', namely `t' 
In the expression: go "" t 
Failed, modules loaded: none.

來源

2012-04-21 john

你能描述你希望你的秀ID例如做什麼？ – Sarah 2012-04-21 10:22:48

@Sarah我試圖展示這棵樹。但是，對於這種情況，我將數字存儲爲字符。看到樹，看看http://stackoverflow.com/questions/10244343/visiting-each-node – john 2012-04-21 10:24:47

執行'go'與這個問題相關 – huon 2012-04-21 10:25:57

編輯：

我發現你的意圖難以破解，但我認爲是永恆的你想表示一個字符串的樹結構？這裏有一個演示實現，其展示你如何可以做到（在一個快速，哈克的方式。）

data ID x = ID ((x, Mark), [ID x]) 
data Mark = Marked | Unmarked 

instance Show a => Show (ID a) where 
    show (ID ((x, _), ids)) = "Val: " ++ show x ++ 
           ", Children: " ++ 
           (show $ map show ids)

來源

2012-04-21 10:25:49 Sarah

Thankssssss .... – john 2012-04-21 10:38:13

與預期類型不符

回答

相關問題