Question

我是Haskell的新手，我已经四处寻找下面的答案，但没有运气。

为什么这段代码不能编译？

newtype Name = Name String deriving (Show, Read)
newtype Age = Age Int deriving (Show, Read)
newtype Height = Height Int deriving (Show, Read)

data User = Person Name Age Height deriving (Show, Read)

data Characteristics a b c = Characteristics a b c

exampleFunction :: Characteristics a b c -> User
exampleFunction (Characteristics a b c) = (Person (Name a) (Age b) (Height c))

错误：

"Couldn't match expected type ‘String’ with actual type ‘a’,‘a’ is a rigid type, variable bound by the type signature"

但是，编译得很好：

exampleFunction :: String -> Int -> Int -> User
exampleFunction a b c = (Person (Name a) (Age b) (Height c))

我意识到上面有更简单的方法，但我只是测试自定义数据类型的不同用途。

更新

我倾向于编译器不喜欢'exampleFunction :: Characteristics a b c'，因为它不是类型安全的。即我不保证：a == Name String，b == Age Int，c == Height Int。

Answer 1

exampleFunction过于笼统。您声称任何类型Characteristics a b c，a和b可能需要c值。但是，a类型的值会传递给Name，而仅的值为String。解决方案是具体说明特征的实际类型。

exampleFunction :: Characteristics String Int Int -> User
exampleFunction (Characteristics a b c) = (Person (Name a) (Age b) (Height c))

但请考虑一下，你甚至可能不需要newtype;简单类型别名就足够了。

type Name = String
type Age = Int
type Height = Int

type Characteristics = (,,)

exampleFunction :: Characteristics Name Age Height -> User
exampleFunction (Charatersics n a h) = Person n a h

Answer 2

试试这个：

exampleFunction :: Characteristics String Int Int -> User
exampleFunction (Characteristics a b c) = (Person (Name a) (Age b) (Height c))

这是有效的，而你的不是，Name，Age和Height需要特定的类型，你的示例函数采用完全通用的参数。

示例的这一行中的a，b和c定义了参数的类型，而不是它们的名称。

 exampleFunction :: Characteristics a b c

Haskell - 结合数据类型？

2 个答案: