[英]How to create a monad in GHC 7.10 or newer
在“Haskell programming from first principles”一书中说:
如果您使用的是 GHC 7.10 或更新版本,您将在 Monad 的定义中看到一个 Applicative 约束,因为它应该是:
class Applicative m => Monad m where (>>=):: ma -> (a -> mb) -> mb (>>):: ma -> mb -> mb return:: a -> ma
我创建了以下应用函子。
data WhoCares a = ItDoesnt | Matter a | WhatThisIsCalled deriving (Eq, Show)
instance Functor WhoCares where
fmap _ ItDoesnt = ItDoesnt
fmap _ WhatThisIsCalled = WhatThisIsCalled
fmap f (Matter a) = Matter (f a)
instance Applicative WhoCares where
pure = Matter
Matter f <*> Matter a = Matter (f a)
main = do
-- fmap id == id
let funcx = fmap id "Hi Julie"
let funcy = id "Hi Julie"
print(funcx)
print(funcy)
print(funcx == funcy)
-- fmap (f . g) == fmap f . fmap g
let funcx' = fmap ((+1) . (*2)) [1..5]
let funcy' = fmap (+1) . fmap (*2) $ [1..5]
print(funcx')
print(funcy')
print(funcx' == funcy')
-- pure id <*> v = v
print(pure id <*> (Matter 10))
-- pure (.) <*> u <*> v <*> w = u <*> (v <*> w)
let appx = pure (.) <*> (Matter (+1)) <*> (Matter (*2)) <*> (Matter 10)
let appy = (Matter (+1)) <*> ((Matter (*2)) <*> (Matter 10))
print(appx)
print(appy)
print(appx == appy)
-- pure f <*> pure x = pure (f x)
let appx' = pure (+1) <*> pure 1 :: WhoCares Int
let appy' = pure ((+1) 1) :: WhoCares Int
print(appx')
print(appy')
print(appx' == appy')
-- u <*> pure y = pure ($ y) <*> u
let appx'' = Matter (+2) <*> pure 2
let appy'' = pure ($ 2) <*> Matter (+ 2)
print(appx'')
print(appy'')
print(appx'' == appy'')
由于缺少示例,我不了解如何实施>>=
和>>
。 到目前为止我想出的代码是:
instance Monad (WhoCares a) where
(>>=) :: Matter a -> (a -> Matter b) -> Matter b
(>>) :: Matter a -> Matter b -> Matter b
return :: a -> Matter a
return = pure
所以,我可以做类似的事情:
half x = if even x
then Matter (x `div` 2)
else ItDoesnt
incVal :: (Ord a, Num a) => a -> WhoCares a
incVal x
| x + 1 <= 10 = return (x + 1)
| otherwise = ItDoesnt
decVal :: (Ord a, Num a) => a -> WhoCares a
decVal x
| x - 1 >= 0 = return (x - 1)
| otherwise = ItDoesnt
main = do
print (Matter 7 >>= incVal >>= incVal >>= incVal)
print (Matter 7 >>= incVal >>= incVal >>= incVal >>= incVal)
print (Matter 7 >>= incVal >>= incVal >>= incVal >>= incVal >>= decVal >>= decVal)
print (Matter 2 >>= decVal >>= decVal >>= decVal)
print(Matter 20 >>= half >>= half)
使用 Output:
10
ItDoesnt
ItDoesnt
ItDoesnt
5
请帮忙。
如果没有有关您希望该实例如何行为的具体信息,通常无法回答如何为给定数据类型实现 monad 实例——单个类型可以有多个有效实例。
不过,对于这个例子,我只看到一种方法:让它表现得像等价形式
data ItIsCalled = ItDoesn't | WhatThisIsCalled
newtype WhoCares a = WhoCares { caresWho :: Either ItIsCalled a }
所以我建议查找Monad (Either c)
实例并以此为基础。 不同之处在于Left
案例对应于您的类型的两个不同构造函数,因此您需要为每个构造函数创建一个子句。
在这里。 我已经删除了WhatThisIsCalled
。
现在解决方案看起来像:
data WhoCares a = ItDoesnt | Matter a deriving (Eq, Show)
instance Functor WhoCares where
fmap _ ItDoesnt = ItDoesnt
fmap f (Matter a) = Matter (f a)
instance Applicative WhoCares where
pure = Matter
Matter f <*> Matter a = Matter (f a)
ItDoesnt <*> _ = ItDoesnt
_ <*> ItDoesnt = ItDoesnt
instance Monad WhoCares where
return x = Matter x
(Matter x) >>= k = k x
ItDoesnt >>= _ = ItDoesnt
half x = if even x
then Matter (x `div` 2)
else ItDoesnt
incVal :: (Ord a, Num a) => a -> WhoCares a
incVal x
| x + 1 <= 10 = return (x + 1)
| otherwise = ItDoesnt
decVal :: (Ord a, Num a) => a -> WhoCares a
decVal x
| x - 1 >= 0 = return (x - 1)
| otherwise = ItDoesnt
main = do
-- fmap id == id
let funcx = fmap id "Hi Julie"
let funcy = id "Hi Julie"
print(funcx)
print(funcy)
print(funcx == funcy)
-- fmap (f . g) == fmap f . fmap g
let funcx' = fmap ((+1) . (*2)) [1..5]
let funcy' = fmap (+1) . fmap (*2) $ [1..5]
print(funcx')
print(funcy')
print(funcx' == funcy')
-- pure id <*> v = v
print(pure id <*> (Matter 10))
-- pure (.) <*> u <*> v <*> w = u <*> (v <*> w)
let appx = pure (.) <*> (Matter (+1)) <*> (Matter (*2)) <*> (Matter 10)
let appy = (Matter (+1)) <*> ((Matter (*2)) <*> (Matter 10))
print(appx)
print(appy)
print(appx == appy)
-- pure f <*> pure x = pure (f x)
let appx' = pure (+1) <*> pure 1 :: WhoCares Int
let appy' = pure ((+1) 1) :: WhoCares Int
print(appx')
print(appy')
print(appx' == appy')
-- u <*> pure y = pure ($ y) <*> u
let appx'' = Matter (+2) <*> pure 2
let appy'' = pure ($ 2) <*> Matter (+ 2)
print(appx'')
print(appy'')
print(appx'' == appy'')
-- m >>= return = m
let monx = Matter 20 >>= return
let mony = Matter 20
print(monx)
print(mony)
print(monx == mony)
-- return x >>= f = f x
let monx' = return 20 >>= half
let mony' = half 20
print(monx')
print(mony')
print(monx' == mony')
-- (m >>= f) >>= g = m >>= (\x -> f x >>= g)
let monx'' = return 20 >>= half >>= half
let mony'' = half 20 >>= half
print(monx'')
print(mony'')
print(monx'' == mony'')
print (Matter 7 >>= incVal >>= incVal >>= incVal)
print (Matter 7 >>= incVal >>= incVal >>= incVal >>= incVal)
print (Matter 7 >>= incVal >>= incVal >>= incVal >>= incVal >>= decVal >>= decVal)
print (Matter 2 >>= decVal >>= decVal >>= decVal)
print (Matter 20 >>= half >>= half)
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