我想實作一個簡單的RestrictedTypeclass，它有兩個函式，lowerBound每個upperBound函式都接受一個Restricted實體并回傳一個Num.
class Restricted r where
lowerBound :: (Num n) => r -> n
upperBound :: (Num n) => r -> n
我正在嘗試使用OrthogonalClass資料型別創建此型別類的實體。
instance Restricted OrthogonalClass where
lowerBound p = orthogonalLowerBound p
upperBound p = orthogonalUpperBound p
在哪里
orthogonalLowerBound :: (Num a) => OrthogonalClass -> a
orthogonalLowerBound Legendre = a
where a = (-1.0) :: Double
-- orthogonalLowerBound Hermite = NegInf
(-1.0)是 aDouble并且它應該是一個可接受的回傳值，因為 aDouble是Num. 但是，我從編譯器收到以下錯誤：
Polynomials.hs:20:33: error:
? Couldn't match expected type ‘a’ with actual type ‘Double’
‘a’ is a rigid type variable bound by
the type signature for:
orthogonalLowerBound :: forall a. Num a => OrthogonalClass -> a
at Polynomials.hs:19:1-55
? In the expression: a
In an equation for ‘orthogonalLowerBound’:
orthogonalLowerBound Legendre
= a
where
a = (- 1.0) :: Double
? Relevant bindings include
orthogonalLowerBound :: OrthogonalClass -> a
(bound at Polynomials.hs:20:1)
|
20 | orthogonalLowerBound Legendre = a
| ^
我可以通過更改為來解決此(-1.0)問題(-1)。orthogonalLowerBound但是，當我取消注釋matchHermite和 returns的第二個模式時，這對我沒有幫助，它是作為數字實體的NegInf資料類的建構式。Inf
這就是我的困惑。ADouble是 a 的一個實體Num，那么為什么我不能Double在我的函式中回傳 aorthogonalLowerBound呢？為什么我不能回傳一個Infwhich 是 a 的實體Num？
我錯過了什么？
這是完整的定義，OrthogonalClass如果Inf您需要更多資訊
data OrthogonalClass = Legendre | Laguerre | Hermite | Tchebychev
-- Playing around with infinity
data Inf = NegInf | PosInf | Undef | Finite deriving (Show)
instance Num Inf where
( ) a b = infAdd a b
(-) a b = infSub a b
(*) a b = infMult a b
signum a = infSignum a
abs a = infAbs a
fromInteger a = infFromInt a
infAdd :: Inf -> Inf -> Inf
infAdd NegInf NegInf = NegInf
infAdd PosInf PosInf = PosInf
infAdd PosInf NegInf = Undef
infAdd NegInf PosInf = Undef
infAdd Undef _ = Undef
infAdd _ Undef = Undef
infAdd NegInf Finite = NegInf
infAdd PosInf Finite = PosInf
infAdd Finite Finite = Finite
infSub :: Inf -> Inf -> Inf
infSub NegInf NegInf = Undef
infSub PosInf PosInf = Undef
infSub PosInf NegInf = PosInf
infSub NegInf PosInf = NegInf
infSub Undef _ = Undef
infSub _ Undef = Undef
infSub NegInf Finite = NegInf
infSub PosInf Finite = PosInf
infSub Finite Finite = Finite
infMult :: Inf -> Inf -> Inf
infMult NegInf NegInf = PosInf
infMult PosInf PosInf = PosInf
infMult PosInf NegInf = NegInf
infMult NegInf PosInf = NegInf
infMult Undef _ = Undef
infMult _ Undef = Undef
infMult NegInf Finite = NegInf
infMult PosInf Finite = PosInf
infMult Finite Finite = Finite
infAbs :: Inf -> Inf
infAbs NegInf = PosInf
infAbs PosInf = PosInf
infAbs Undef = Undef
infAbs Finite = Finite
infSignum :: (Num a) => Inf -> a
infSignum NegInf = (-1)
infSignum PosInf = 1
infSignum Undef = 0
infSignum Finite = 0
infFromInt :: (Integral i) => i -> Inf
infFromInt x = Finite
uj5u.com熱心網友回復：
這是關于泛型型別方向的一個非常常見的混淆。
簡而言之：選擇泛型型別的是函式的呼叫者，而不是實作者。
如果您有具有此簽名的函式：
orthogonalLowerBound :: (Num a) => OrthogonalClass -> a
那個簽名說：嘿你，誰呼叫我的函式！選擇一種型別。任意種類。讓我們稱之為a。現在確保有一個實體Num a。完畢？偉大的！現在我可以回傳一個該型別的值a。
這種型別簽名是對呼叫該函式的任何人的承諾，而您，該函式的實作者，必須履行該承諾。無論呼叫者選擇何種型別，您都必須回傳該型別的值。
呼叫者可以選擇任何具有Num實體的型別，例如：
o :: OrthogonalClass
o = ...
x :: Int
x = orthogonalLowerBound o
y :: Decimal
y = orthogonalLowerBound o
解決方案？如果您的函式應該回傳 a Double，只需在其型別簽名中說明：
orthogonalLowerBound :: OrthogonalClass -> Double
當然，在這種情況下你不能 return Inf，這應該是：一個函式不能根據引數的值回傳不同的型別。這個特性被稱為“依賴型別”，Haskell 沒有直接支持它。
如果您確實需要無窮大，則該Double型別也確實包含它，并且有多種方法可以將其作為值獲取。例如，infinity從ieee754包裝上看
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