Question

我查看了实现，这更加神秘：

-- | 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

我甚至不明白为什么<$应该在类型类中占有一席之地。看起来有fmap . const可能没有的共享增强效应，a1可能没有“完全实现”。这与Applicative排序操作符的含义有什么关系？

Answer 1

这些操作符对两个应用操作进行排序，并提供箭头指向的操作结果。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 ')'

将是一个与括号中包含的p匹配的解析器，并提供解析p的结果。

实际上，(*>)与(>>)相同，但只需要Applicative约束而不是Monad约束。

我甚至不明白为什么<$应该在类型类中占有一席之地。

答案由the Functor documentation给出：(<$)有时可以提供比默认值更高效的实现，即fmap . const。

这与Applicative测序运算符的含义有何关系？

如果(<$)效率更高，您希望在(*>)的定义中保持这种效率。

什么是应用左右明星测序操作员应该做什么？

1 个答案: