Question

使用此功能：

getK x y
    | y < 1 = error "Index can not be less than 1!"
    | y > length x = error "Index greater than array length"
    | otherwise    = x !! (y-1)

如果我没有定义类型，GHC就可以了。

*Main> :info getK
 getK :: [a] -> Int -> a    -- Defined at main.hs:7:1

但是假设我将类型约束写为：

getK :: (Int b) => [a] -> b -> a

据我所知，以上是GHC推断的内容。但不，它会产生以下错误：

main.hs:6:10:
`Int' is applied to too many type arguments
In the type signature for `getK': getK :: Int b => [a] -> b -> a
Failed, modules loaded: none.

我的直接反应是，我没有将类型分配给＆＃39; a一定存在问题。（虽然不应该因为GHC认识到＆＃39; a＆＃39;可以是任何类型。所以我尝试了：

getK :: (Num a, Ord a, Int b) => [a] -> b -> a

另一个错误......

main.hs:6:24:
`Int' is applied to too many type arguments
In the type signature for `getK':
  getK :: (Num a, Ord a, Int b) => [a] -> b -> a

请帮助我理解这里的行为。

Answer 1

Int是一种类型，而不是类型类。只有像Num和Ord这样的类型才能用作类型约束。这就是你的签名的写法：

getK :: [a] -> Int -> a

GHCI有一个:info命令（可以缩写为:i），它会告诉您有关类型和类型类的信息，并比较它对Int与其所说内容的说法Num：

λ> :i Int
data Int = GHC.Types.I# GHC.Prim.Int#   -- Defined in ‘GHC.Types’
instance Bounded Int -- Defined in ‘GHC.Enum’
instance Enum Int -- Defined in ‘GHC.Enum’
instance Eq Int -- Defined in ‘GHC.Classes’
instance Integral Int -- Defined in ‘GHC.Real’
instance Num Int -- Defined in ‘GHC.Num’
instance Ord Int -- Defined in ‘GHC.Classes’
instance Read Int -- Defined in ‘GHC.Read’
instance Real Int -- Defined in ‘GHC.Real’
instance Show Int -- Defined in ‘GHC.Show’
λ> 
λ> 
λ> :i Num
class Num a where
  (+) :: a -> a -> a
  (*) :: a -> a -> a
  (-) :: a -> a -> a
  negate :: a -> a
  abs :: a -> a
  signum :: a -> a
  fromInteger :: Integer -> a
    -- Defined in ‘GHC.Num’
instance Num Integer -- Defined in ‘GHC.Num’
instance Num Int -- Defined in ‘GHC.Num’
instance Num Float -- Defined in ‘GHC.Float’
instance Num Double -- Defined in ‘GHC.Float’

定义函数：使用多个变量键入约束

1 个答案: