haskell - 在haskell中返回不同的类型

Question

我是 Haskell 的新手，正在从事一个项目，其中包含以下代码:

data Nested a = Elem a | Nested [Nested a] deriving (Eq, Show)
data Symbol a = Value a | Transformation (a -> a -> a) deriving (Show)


toSymbol :: [Char] -> Nested (Symbol Integer)
toSymbol x  
|all isDigit x = Elem (Value (strToInt x))
|x == "+" = Elem (Transformation (\x y -> x + y))

有什么方法可以避免将此函数的类型限制为 Nested (Symbol Integer)？我想使用 Symbol 来表示许多不同的类型，并有一个函数 toSymbol something along the lines of the following:

toSymbol x  
|x == "1" = Elem (Value 1)
|x == "+" = Elem (Transformation (\x y -> x + y))
|x == "exampleword" = Elem (Value "word")
|x == "concatenate()" = Elem (Transformation concatTwoStrings)

我不知道这样的函数的类型签名是什么。我可以做些什么来获得与此类似的功能吗？

Answer 1

你无法弄清楚类型签名的原因是因为你试图做的是“使函数的返回类型取决于传递的字符串的值，也就是依赖类型编程”，字符串将仅在运行时可用。

所以基本上，如果您尝试说:toSymbol::String -> Nested (Symbol a)，a 取决于字符串的运行时值，这就是为什么编译器提示它。

有很多方法可以优化您的类型，以便所有部分组合在一起，一个可能的解决方案是使用一种新类型，它指定一个符号可能具有的不同类型的值。下面是示例:

data Nested a = Elem a | Nested [Nested a] deriving (Eq, Show)
data Symbol a = Value a | Transformation (a -> a -> a)
data SymbolType = SInteger Integer | SString String

addSymbols :: SymbolType -> SymbolType -> SymbolType
addSymbols (SInteger a) (SInteger b) = SInteger (a+b)
addSymbols _ _ = error "Not possible"

concatSymbols :: SymbolType -> SymbolType -> SymbolType
concatSymbols (SString a) (SString b) = SString (a++b)
concatSymbols _ _ = error "Not possible"

toSymbol :: String -> Nested (Symbol SymbolType)
toSymbol x  
  |x == "1" = Elem (Value (SInteger 1))
  |x == "+" = Elem (Transformation addSymbols)
  |x == "exampleword" = Elem (Value (SString "word"))
  |x == "concatenate()" = Elem (Transformation concatSymbols)