了解正确的申请

Question

对于List,为什么right apply (*>)表现为重复并附加第二个参数n次数,n第一个参数的长度在哪里？

ghci> [1,2,3] *> [4,5]
[4,5,4,5,4,5]

Answer 1

该*>运营商的定义,默认情况下,作为

xs *> ys = id <$ xs <*> ys

默认情况下,它会转换为

const id <$> xs <*> ys

也就是说,它将xswith的每个元素替换id为get xs'然后计算xs' <*> ys.[]是一个Monad实例,在哪里(=<<) = concatMap.Applicative规定Applicative与Monad实例之间关系的法则之一:

pure = return
fs <*> as = fs `ap` as = fs >>= \f -> as >>= \a -> f a

对于列表,这是

fs <*> as = [f a | f <- fs, a <- as]

因此*>,列表最终由Monad实例确定.

请注意,列表还有另一个非常明智的Applicative实例,可以通过以下新类型获得Control.Applicative:

newtype ZipList a = ZipList [a]
instance Applicative ZipList where
  pure = repeat
  (<*>) = zipWith ($)