为什么二元组 Functor 实例只将函数应用于第二个元素? [英] Why does the 2-tuple Functor instance only apply the function to the second element?

问题描述

import Control.Applicative

main = print $ fmap (*2) (1,2)

产生(1,4).我希望它产生 (2,4) 但该函数仅应用于元组的第二个元素.

produces (1,4). I would expect it it to produce (2,4) but instead the function is applied only to the second element of the tuple.

更新我基本上马上就想通了.我会在一分钟内发布我自己的答案..

Update I've basically figured this out almost straight away. I'll post my own answer in a minute..

推荐答案

Functor 实例实际上来自 GHC.Base 模块，由 Control.Applicative 导入.

The Functor instance is actually from the GHC.Base module which is imported by Control.Applicative.

尝试编写我想要的实例，我可以看到它不起作用，给定元组的定义；实例只需要一个类型参数，而二元组有两个.

Trying to write the instance I want, I can see that it won't work, given the definition of tuples; the instance requires just one type parameter, while the 2-tuple has two.

一个有效的 Functor 实例至少必须在元组上，(a,a) 对于每个元素都具有相同的类型，但你不能做任何偷偷摸摸的事情，比如定义实例:

A valid Functor instance would at least have to be on tuples, (a,a) that have the same type for each element, but you cannot do anything sneaky, like define the instance on:

 type T2 a = (a,a)

因为实例类型不允许是同义词.

because instance types aren't permitted to be synonyms.

上述受限二元组同义词在逻辑上与类型相同:

The above restricted 2-tuple synonym is logically the same as the type:

data T2 a = T2 a a

哪些可以有一个 Functor 实例:

which can have a Functor instance:

instance Functor T2 where
    fmap f (T2 x y) = T2 (f x) (f y)

正如 Gabriel 在评论中所说，这对于分支结构或并发性很有用.

As Gabriel remarked in the comments, this can be useful for branching structures or concurrency.

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