一个没有成为应用的函子的具体类型例子？ [英] Concrete Type Example of a Functor that Fails to be an Applicative?

问题描述

从不适用的函子：

一个类型构造函数，它是一个Functor而不是一个Applicative。一个简单的例子是一对：

但是，如何定义 Applicative 实例，而不对 r 施加额外限制。特别是，如何定义 pure :: a - >是没有办法的。 （r，a）为任意 r 。

这里，纯粹无法一次为所有类型定义;然而，对于任何具体类型 T ，可以使（（，）T）

问题：是否存在一个具体仿函数（即，没有涉及的类型变量）的例子，它是一个仿函数而不是应用程序？

解决方案

我没有50个评论，所以我会尽力回答：然而，对于任何具体类型T，可以使（（，）T）成为一个应用。

$ b

...

数学中有一个定理，至少2个元素可以被制成一个monoid。因此，对于任何具体类型T，它原则上可以成为Monoid的成员，然后原则上可以适用。这个推理有什么问题？

那么这个无人居住类型的元组呢？ （，）Void

它是 Functor ，对吗？

您可以为它导出 Applicative 吗？ 纯粹如何实现？

From functors that are not applicatives:

A type constructor which is a Functor but not an Applicative. A simple example is a pair:

instance Functor ((,) r) where
    fmap f (x,y) = (x, f y)

But there is no way how to define its Applicative instance without imposing additional restrictions on r. In particular, there is no way how to define pure :: a -> (r, a) for an arbitrary r.

Here, pure fails to be definable for all types at once; however, for any concrete type T, one can make ((,) T) an applicative.

Question: Is there an example of a concrete functor (i.e., no type variables involved) that is a functor but not an applicative?

解决方案

I don't have 50 reputation to comment here, so I'll try to do it as an answer:

however, for any concrete type T, one can make ((,) T) an applicative.

...

There's a theorem in mathematics that any collection with at least 2 elements can be made into a monoid. So for any concrete type T, it could in principle be made a member of Monoid, and then could in principle be made Applicative. What's wrong with this reasoning?

What about the tuple from the uninhabited type? (,) Void

It is a Functor,right?

Could you derive Applicative for it? How would pure be implemented?

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