haskell - 应用左星测序算子和右星测序算子预期做什么？

Question

我查了一下实现，它更加神秘:

-- | 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

我什至不明白为什么 <$值得在类型类中占有一席之地。看起来有一些共享增强效果fmap . const可能没有，a1可能没有“完全实现”。这与 Applicative 的含义有何关系？排序运算符？

Answer 1

这些运算符对两个应用操作进行排序，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

编写解析器组合器的另一个例子是

brackets p = char '(' *> p <* char ')'

这将是一个匹配 p 的解析器包含在括号中并给出解析结果 p .

事实上，(*>)与 (>>) 相同但只需要 Applicative约束而不是 Monad约束。

I don't even understand why does <$ deserve a place in a typeclass.

答案由 the Functor documentation 给出:(<$)有时可以比默认的 fmap . const 更有效的实现.

How is that related to the meaning of Applicative sequencing operators?

在 (<$) 的情况下更高效，您希望在 (*>) 的定义中保持该效率.