帮助我理解将数字附加到字符串时的这个 Haskell (GHCI) 类型错误：(Num [Char])

Question

这个周末我一直在尝试学习 Haskell，当我真正尝试编写一个递归函数（而不是仅仅从教程中输入一个）时，我得到了一个类型错误。

我非常感谢任何帮助理解（1）错误意味着什么（我不明白“修复”）； (2) 为什么会抛出错误 - 我相当确定我在传递的类型方面没有犯任何错误。

我的代码：

tell :: (Show a) => [a] -> String  
tell'in :: (Show a, Num n) => [a] -> n -> String -> (n, String)
tell [] = "The list is empty"  
tell (x:[]) = "The list has one element: " ++ show x  
tell (x:xs) = "The list has " ++ n ++ " elements: " ++ s where (n, s) = (tell'in (xs) (1) (show x))  


tell'in (x:[]) n s = ((n + 1), (s ++ " and " ++ (show x)))
tell'in (x:xs) n s = tell'in xs (n+1)  (s ++ " and " ++ show x)

当我尝试将其加载到 GHCI 时出现错误：

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

example.hs:13:88:
    Could not deduce (Num [Char]) arising from the literal `1'
    from the context (Show a)
      bound by the type signature for tell :: Show a => [a] -> String
      at example.hs:(11,1)-(13,99)
    Possible fix:
      add (Num [Char]) to the context of
        the type signature for tell :: Show a => [a] -> String
      or add an instance declaration for (Num [Char])
    In the second argument of `tell'in', namely `(1)'
    In the expression: (tell'in (xs) (1) (show x))
    In a pattern binding: (n, s) = (tell'in (xs) (1) (show x))
Failed, modules loaded: none.
Prelude>

Answer 1

tell'in 返回一个 Num n =>; n，然后在最后一个方程中将其与 String（又名 [Char]）进行 (++) 运算告诉。您可能想在那里使用 show n 而不是 n 。

Answer 2

tell'in 的类型表示其返回值的第一个坐标与其第二个参数的类型相同。在 tell 中，您使用数字作为第二个参数 (1) 调用 tell'in，然后使用其返回的第一个坐标通过将值 (n) 作为字符串附加到其他字符串。所以无论这个类型是什么，它必须同时是一个字符串和一个数字。该错误告诉您，字符串 ([Char]) 无法被视为一种数字，当它尝试解释数字文字 1 的含义时，就会出现这种情况。你可能的意思是

"The list has " ++ show n ++ " elements: " ++ ...

Answer 3

有一个通用方案来理解编译错误，如果它是某种类型不匹配：
您需要提供更具体的类型注释来缩小错误范围！

tell'in 和 tell 已经有了类型：

tell :: (Show a) => [a] -> String  
tell'in :: (Show a, Num n) => [a] -> n -> String -> (n, String)

所以，你改变这个：

tell (x:xs) = "The list has " ++ n ++ " elements: " ++ s
    where (n, s) = tell'in (xs) (1) (show x)) 

到这个：

--   vvvvvvvvvvvvvvvvvvvvvvv                      vvvvvvvv
tell ((x:xs)::(Show a=>[a])) = "The list has " ++ (n::Int) ++ " elements: " ++ s
    where (n::Int, s::String) = tell'in (xs) (1::Int) (show x)) 
--        ^^^^^^^^^^^^^^^^^^^^               ^^^^^^^^

现在，要么你已经看到错误，要么你尝试再次编译并得到一个更具体的错误消息。

---

无论如何，使用 ::Int 而不是 ::Num n=>n 是个好主意，因为后者会泛化为 ::Integer< /code>，只要未指定确切类型。 Int 不大于一个机器字（32/64 位），因此它比任意大小的 Integer 更快。