在his answer问题“Distinction between typeclasses MonadPlus , Alternative , and Monoid ?”中，爱德华·克梅特(Edward Kmett)说
显然，任何不是monad的应用函子都自动是Alternative而不是MonadPlus的示例，但Edward Kmett的回答暗示存在一个monad，它是Alternative但不是MonadPlus:其empty和<|>将满足Alternative律，1，但不是MonadPlus律。2我自己不能举一个例子。有人知道吗？
1我无法找到一组Alternative法则的规范引用，但是我认为它们大约在my answer到“Confused by the meaning of the Alternative type class and its relationship to other type classes”问题(搜索“正确的分配”一词)的中间位置。我认为应该遵守的四个定律是:
<*>的)右分布: (f <|> g) <*> a = (f <*> a) <|> (g <*> a) <*>): empty <*> a = empty fmap的)左分布: f <$> (a <|> b) = (f <$> a) <|> (f <$> b) fmap): f <$> empty = empty 我也很乐意接受一套更有用的
Alternative法律。2我知道there’s some ambiguity about what the
MonadPlus laws are;我对使用左分布或左捕获的答案感到满意，尽管我不太喜欢前者。 最佳答案
答案的线索在HaskellWiki about MonadPlus you linked to中:
因此，根据您的偏爱选择，Maybe不是MonadPlus(尽管有一个实例，但它不满足左分布)。让我们证明它满足替代要求。Maybe是替代方法
<*>的)右分布: (f <|> g) <*> a = (f <*> a) <|> (g <*> a) 情况1:
f=Nothing:(Nothing <|> g) <*> a = (g) <*> a -- left identity <|>
= Nothing <|> (g <*> a) -- left identity <|>
= (Nothing <*> a) <|> (g <*> a) -- left failure <*>
情况2:
a=Nothing:(f <|> g) <*> Nothing = Nothing -- right failure <*>
= Nothing <|> Nothing -- left identity <|>
= (f <*> Nothing) <|> (g <*> Nothing) -- right failure <*>
情况3:
f=Just h, a = Just x(Just h <|> g) <*> Just x = Just h <*> Just x -- left bias <|>
= Just (h x) -- success <*>
= Just (h x) <|> (g <*> Just x) -- left bias <|>
= (Just h <*> Just x) <|> (g <*> Just x) -- success <*>
<*>): empty <*> a = empty 这很容易，因为
Nothing <*> a = Nothing -- left failure <*>
fmap的)左分布: f <$> (a <|> b) = (f <$> a) <|> (f <$> b) 情况1:
a = Nothingf <$> (Nothing <|> b) = f <$> b -- left identity <|>
= Nothing <|> (f <$> b) -- left identity <|>
= (f <$> Nothing) <|> (f <$> b) -- failure <$>
情况2:
a = Just xf <$> (Just x <|> b) = f <$> Just x -- left bias <|>
= Just (f x) -- success <$>
= Just (f x) <|> (f <$> b) -- left bias <|>
= (f <$> Just x) <|> (f <$> b) -- success <$>
fmap): f <$> empty = empty 另一个简单的方法:
f <$> Nothing = Nothing -- failure <$>
Maybe不是MonadPlus让我们证明
Maybe不是MonadPlus的断言:我们需要证明mplus a b >>= k = mplus (a >>= k) (b >>= k)并不总是成立。诀窍是，与以往一样，使用一些绑定(bind)来潜入非常不同的值:a = Just False
b = Just True
k True = Just "Made it!"
k False = Nothing
现在
mplus (Just False) (Just True) >>= k = Just False >>= k
= k False
= Nothing
在这里，我使用bind
(>>=)从胜利的口中抢走失败(Nothing)，因为Just False看起来很成功。mplus (Just False >>= k) (Just True >>= k) = mplus (k False) (k True)
= mplus Nothing (Just "Made it!")
= Just "Made it!"
这里的失败(
k False)是较早计算的，因此被忽略了，我们"Made it!"。因此，
mplus a b >>= k = Nothing但mplus (a >>= k) (b >>= k) = Just "Made it!"。您可以像我一样使用
>>=打破mplus的Maybe的左偏。我的证明的有效性:
以防万一您觉得我做得不够单调乏味，我将证明我使用的身份:
首先
Nothing <|> c = c -- left identity <|>
Just d <|> c = Just d -- left bias <|>
来自实例声明
instance Alternative Maybe where
empty = Nothing
Nothing <|> r = r
l <|> _ = l
其次
f <$> Nothing = Nothing -- failure <$>
f <$> Just x = Just (f x) -- success <$>
只是来自
(<$>) = fmap和instance Functor Maybe where
fmap _ Nothing = Nothing
fmap f (Just a) = Just (f a)
第三，其他三个需要做更多的工作:
Nothing <*> c = Nothing -- left failure <*>
c <*> Nothing = Nothing -- right failure <*>
Just f <*> Just x = Just (f x) -- success <*>
来自定义
instance Applicative Maybe where
pure = return
(<*>) = ap
ap :: (Monad m) => m (a -> b) -> m a -> m b
ap = liftM2 id
liftM2 :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
liftM2 f m1 m2 = do { x1 <- m1; x2 <- m2; return (f x1 x2) }
instance Monad Maybe where
(Just x) >>= k = k x
Nothing >>= _ = Nothing
return = Just
所以
mf <*> mx = ap mf mx
= liftM2 id mf mx
= do { f <- mf; x <- mx; return (id f x) }
= do { f <- mf; x <- mx; return (f x) }
= do { f <- mf; x <- mx; Just (f x) }
= mf >>= \f ->
mx >>= \x ->
Just (f x)
因此，如果
mf或mx为Nothing，则结果也为Nothing，而如果mf = Just f和mx = Just x，则结果为Just (f x)