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我正在学习 YouTube 上的 Bartosz Milewski 范畴理论课程。他将 Const 和 Identity 函子描述为可以派生的“基础”函子(可能是我的自由解释)。

一旦我开始与地图和管道的 Sanctuary 库集成,我的问题就出现了,已经实现了 ES6+ / Fantasy-land(不重要)版本的函子。

实现非常简单

const {map: flMap, extract } = require('fantasy-land');

const getInstance = (self, constructor) =>
    (self instanceof constructor) ?
        self :
        Object.create(constructor.prototype) ;

const Identity = function(x){
    const self = getInstance(this, Identity)

    self[flMap] = f => Identity(f(x))
    self[extract] = () => x

    return Object.freeze(self)
}

这是一些简单的用法(因为我也在使用离子衍生透镜)

// USAGE
const {map, pipe, curry} = require("sanctuary")

const extractFrom = x => x[extract]()

const setter = (f, x) => (pipe([
    Identity,
    map(f),
    extractFrom
])(x))

const double = x => x + x

console.log(Identity(35)) //=> 35
console.log(map(double, Identity(35))) // ERROR Should be Identity(70)
console.log(setter(double, 35)) // ERROR Should be: 70
TypeError: Type-variable constraint violation

map :: Functor f => (a -> b) -> f a -> f b
                     ^            ^
                     1            2

1)  35 :: Number, FiniteNumber, NonZeroFiniteNumber, Integer,
    NonNegativeInteger, ValidNumber

2)  () => x :: Function, (c -> d)
    f => Identity(f(x)) :: Function, (c -> d)

Since there is no type of which all the above values are members, the
type-variable constraint has been violated.

但是 Const 仿函数工作得更好一些(地图中没有调用 f)

const Const = function(x) {
    const self = getInstance(this, Const)

    self[map] = _ =>  Const(x)
    self[extract] = () => x

    return Object.freeze(self)
}

const getter = (f, x) => (pipe([
    Const,
    map(f),
    extractFrom
])(x))

console.log(getter(double, 35)) //=> 35

此外,通过删除类型检查证明一切都是“逻辑上合理的”

const {create, env} = require('sanctuary'); 
const {map, pipe, curry} = create({checkTypes: false, env: env});

或用 ramda 替换 sanctuary。所以它看起来像是身份映射函数的某种类型一致性问题。

问题是我如何让所有这些部分以一种快乐的方式一起玩。

4

1 回答 1

2

您需要为您的类型 ( IdentityType :: Type -> Type) 定义一个类型构造函数,并按照文档中的说明将其包含IdentityType ($.Unknown)在您的 Sanctuary 环境中S.create。具体来说,你需要这样的东西:

//    IdentityType :: Type -> Type
const IdentityType = $.UnaryType
  ('my-package/Identity')
  ('http://example.com/my-package#Identity')
  (x => type (x) === Identity['@@type'])
  (identity => [Z.extract (identity)]);

const S = create ({
  checkTypes: process.env.NODE_ENV !== 'production',
  env: env.concat ([IdentityType ($.Unknown)]),
});

在上面的代码片段中,$指的是sanctuary-defZ指的是sanctuary-type-classes,并type指的是sanctuary-type-identifiers

于 2018-06-07T17:29:38.267 回答