函数的Haskell作用域

我正在尝试做一些将在“where”子句之后定义的输入函数作为输入函数并输出由它组成的字符串的操作。提到它会将“where”子句之后列出的函数范围限定为 writeOut 函数。我想带什么（它不起作用）：

writeOut :: [Char] -> [Char]
writeOut x = x
 where
 number :: Char -> [Char]
 number n = [n]
 multiplication :: [Char] -> [Char] -> [Char]
 multiplication a b = a++"*"++b
 addition :: [Char] -> [Char] -> [Char]
 addition a b = "("++a++"+"++b++")"
 subtraction :: [Char] -> [Char] -> [Char]
 subtraction a b = "("++a++"-"++b++")"
 ...

因此，例如，对于输入：

writeOut multiplication (addition (number '1') (subtraction (number '2') (number '3'))) (number '3')

会给我一个输出：

(1+(2-3))*3

当我删除 writeOut 函数并只保留 'where' 关键字之后的内容时，它工作正常，但我真的想在 'writeOut' 函数下“锁定它”。

你可能会问我为什么要这样做，因为我想要另一个函数来为我计算表达式：

writeOut multiplication (addition (number '1') (subtraction (number '2') (number '3'))) (number '3')

Output: (1+(2-3))*3

calc multiplication (addition (number '1') (subtraction (number '2') (number '3'))) (number '3')

Ouput: 0

以防万一你想看看什么是“正常”工作的：

number :: Char -> [Char]
number n = [n]
multiplication :: [Char] -> [Char] -> [Char]
multiplication a b = a++"*"++b
addition :: [Char] -> [Char] -> [Char]
addition a b = "("++a++"+"++b++")"
subtraction :: [Char] -> [Char] -> [Char]
subtraction a b = "("++a++"-"++b++")"

回答

data Expr = Number Int
          | Mult Expr Expr
          | Add Expr Expr
          | Subtract Expr Expr


writeOut :: Expr -> String
writeOut (Number x) = show x
writeOut (Add x y) = "(" ++  writeOut x ++ " + " ++ writeOut y ++ ")"
writeOut (Subtract x y) = "(" ++  writeOut x ++ " - " ++ writeOut y ++ ")"
writeOut (Mult x y) = "(" ++ writeOut x ++ " * " ++ writeOut y ++ ")"


print $ writeOut (Mult (Add (Number 1) (Subtract (Number 2) (Number 3))) (Number 3)

evaluate :: Expr -> Int
evaluate = undefined