haskell *>序列运算符 [英] haskell *> sequencing operator

问题描述

我的问题答案来自以下位置，我有一个问题: haskell或验证适用

I have a question arising from the answer to my question at: haskell Either and Validation Applicative

我的代码张贴在这里.

它涉及到使用 *> 排序运算符而不是< *> 应用运算符.

It concerns the use of the *> sequencing operator instead of the <*> applicative operator.

基于 https://hackage.haskell.org/package/base-4.15.0.0/docs/Control-Applicative.html#v:-42--62- ，我了解*> 对动作进行排序，丢弃第一个参数的值.因此，对于我的代码，我尝试了 fail6 = fail2 *>成功，它可以工作，但是不应该工作，因为第一个参数的值，即fail2，应该被丢弃.为什么 fail6 有效?

Based on the explanation at https://hackage.haskell.org/package/base-4.15.0.0/docs/Control-Applicative.html#v:-42--62-, I understand that *> sequences actions, discarding the value of the first argument. So for my code, I've tried fail6 = fail2 *> success, which works, but it's not supposed to work because the value of the first argument, namely fail2, should be discarded. Why does fail6 work?

fail6 的输出为 Failure [MooglesChewedWires，StackOverflow] .

推荐答案

丢弃结果"表示应用程序的结果.因此，对于 Either 来说，这意味着 Right y .因此，(*>)等同于:

With "discard the result", it means the result of an applicative. So for an Either that means a Right y. (*>) is thus equivalent to:

(*>) :: Applicative f => f a -> f b -> f b
(*>) f g = liftA2 (const id) f g

或另一种形式:

(*>) :: Applicative f => f a -> f b -> f b
(*>) f g = (const id) <$> f <*> g

它因此运行两个动作并返回第二个结果，但这在" Applicative 上下文"中.

It thus runs the two actions and returns the second result, but this in the "Applicative context".

例如，对于 Either 的任何实现，该实现都为:

For example for an Either, this is implemented as:

instance Applicative (Either a) where
    pure = Right
    Left e <*> _ = Left e
    _ <*> Left e = Left e
    Right f <*> Right x = Right (f x)

因此，这意味着(*>)为 Either 的实现方式为:

This thus means that (*>) is implemented for an Either as:

-- (*>) for Either
(*>) :: Either a b -> Either a c -> Either a c
Left x *> _ = Left x
_ *> Left x = Left x
Right _ *> Right x = Right x

或等价物:

-- (*>) for Either
(*>) :: Either a b -> Either a c -> Either a c
Left x *> _ = Left x
_ *> x = x

如果第一个操作数是 Right…，它将返回第二个操作数，如果第一个操作数是 Left x ，则将返回 Left x.

If the first operand is thus a Right …, it will return the second operand, if the first operand is Left x, it will return Left x.

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