Haskell 应用程序的左星和右星测序操作符应该做什么？

我查看了实现，它更神秘：

-- | Sequence actions, discarding the value of the first argument.
(*>) :: f a -> f b -> f b
a1 *> a2 = (id <$ a1) <*> a2
-- This is essentially the same as liftA2 (flip const), but if the
-- Functor instance has an optimized (<$), it may be better to use
-- that instead. Before liftA2 became a method, this definition
-- was strictly better, but now it depends on the functor. For a
-- functor supporting a sharing-enhancing (<$), this definition
-- may reduce allocation by preventing a1 from ever being fully
-- realized. In an implementation with a boring (<$) but an optimizing
-- liftA2, it would likely be better to define (*>) using liftA2.

-- | Sequence actions, discarding the value of the second argument.
(<*) :: f a -> f b -> f a
(<*) = liftA2 const

——|对操作进行排序，丢弃第一个参数的值。
（*>）：：FA->FB->FB
a1*>a2=（id这些运算符对两个应用程序操作进行排序，并提供箭头指向的操作的结果。例如

> Just 1 *> Just 2
Just 2

> Just 1 <* Just 2
Just 1

>只有1个*>只有2个
只有两个
>仅1 p）
与
（>>）
相同，但只需要
应用程序
约束，而不是
Monad
约束

我甚至不明白为什么
liftA2（翻转常量）fa fb
=
fmap（\a b->b）fa fb
=
fmap（常量id）fa fb
=
常量id fa fb
（=
id和
fa a）fa fb
。将其与使用
（，）
=
\ab->（a，b）
时进行比较，如上所述。确实，递归类型需要在编写
fmap
时谨慎，以允许它内联到默认的
数据中。Sequence.Seq
是一个具有特别有效
的类型的好例子
brackets p = char '(' *> p <* char ')'