Question

我对Haskell有点新，我正在开发一个项目，我有以下代码：

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))

有没有办法可以避免将此函数的类型限制为嵌套（符号整数）？我想使用Symbol来表示许多不同的类型，并且具有以下几行的符号功能：

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

我不知道像这样的函数的类型签名是什么。我有什么办法可以获得类似的功能吗？

Answer 1

我认为不可能编写单个函数来执行此操作。一种可能的解决方案是使用类型类，它保留单个函数API：

{-# LANGUAGE FlexibleInstances #-}

...

class Token a where
    toSymbol :: String -> Nested a

instance Token (Symbol Integer) where
    toSymbol x
        |all isDigit x = Elem (Value (read x))
        |x == "+" = Elem (Transformation (\x y -> x + y))
        |otherwise = error "Wrong type"

instance Token (Symbol String) where
    toSymbol "exampleword" = Elem (Value "word")
    toSymbol "concatenate()" = Elem (Transformation (++))
    toSymbol _ = error "Wrong type"

Answer 2

您无法弄清楚类型签名的原因是因为您正在尝试执行此操作“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”，该值的值string仅在运行时可用。

所以基本上如果你试着说：toSymbol :: String -> Nested (Symbol a)，a取决于字符串的运行时值，以及编译器抱怨它的原因。

有许多方法可以优化你的类型，以便所有部分都适合在一起，一种可能的解决方案是使用一种新类型，它指定符号可能具有的不同类型的值。以下是示例：

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)

在haskell中返回不同的类型

2 个答案: