了解如何在 vanilla JavaScript 中实现 lodash 的 _.flowRight

Question

在学校，我们的任务是构建 lodash method flowRight! 的实现

在它提到的规范中：

接受任意数量的函数并返回一个新函数 使用它的参数并从右到左调用提供的函数 （最后到第一个）。每个函数的参数（第一个函数除外）是 由其右侧函数的返回值决定。通话 到 flowRight 返回的函数计算返回值 最左边的函数。

这是他们给出的一个例子：

e.g.

var sayHello = function (name) {
    return 'Hello, ' + name;
},

addExclamation = function (s) {
    return s + '!';
},

smallTalk = function (s) {
    return s + ' Nice weather we are having, eh?';
};

var greetEnthusiastically = flowRight(addExclamation, sayHello);

greetEnthusiastically('Antonio');
// --> returns 'Hello, Antonio!'
//(sayHello is called with 'Antonio', 
//  addExclamation is called with 'Hello, Antonio')

我觉得我理解了这个示例所演示的静态示例中发生了什么。

function (func1, func2) {
    return function(value) {
        return func1(func2(value));
    }
}

我想我很难把我的大脑围绕在一个循环中，我认为你会需要它。这是我目前的实现。

var flowRight = function (...args) {
    var Func;
    for(var i = args.length - 2; 0 > i; i--) {
        function Func(value) {
            return args[i](args[i + 1](value));
        }
    }
    return Func;
};

任何帮助将不胜感激！

Answer 1

不需要循环。如果允许，这将使用 ES6。

这使用spread、rest和reduce

const flowRight = (...functions) => functions.reduce((a, c) => (...args) => a(c(...args)));

以下示例

var sayHello = function (name) {
  return 'Hello, ' + name;
 },

addExclamation = function (s) {
  return s + '!';
},

smallTalk = function (s) {
  return s + ' Nice weather we are having, eh?';
}

const flowRight = (...functions) => functions.reduce((a, c) => (...args) => a(c(...args)))

var greetEnthusiastically = flowRight(smallTalk, addExclamation, sayHello)

console.log(greetEnthusiastically('Antonio'));

Answer 2

要从右向左流动，您可以使用...spread 和.reduceRight(x, y)

我已经对下面的代码进行了注释，试图解释这一切是如何协同工作的。

const sayHello = function (name) {
  return 'Hello, ' + name;
 };

const addExclamation = function (s) {
  return s + '!';
};

const smallTalk = function (s) {
  return s + ' Nice weather we are having, eh?';
}

// function that takes functions and then
// returns a function that takes  a value to apply to those functions in reverse
const flowRight = (...fns) => val => fns.reduceRight((val, fn) => {
  // return the function and pass in the seed value or the value of the pervious fn.
  // You can think of it like the following.
  // 1st pass: sayHello(value) -> "Hello, " + value;
  // 2nd pass: addExclamation("Hello,  $value") -> "Hello,  $value" + "!";
  // 3rd pass: smallTalk("Hello,  $value!") -> "Hello,  $value!" + ' Nice weather we are having, eh?'
  // ... and so on, the reducer will keep calling the next fn with the previously returned value
  return fn(val)
// seed the reducer with the value passed in
}, val);

var greetEnthusiastically = flowRight(smallTalk, addExclamation, sayHello);

console.log(greetEnthusiastically('Antonio'));

Answer 3

从右到左的构图

const flowRight = (f, ...more) => x =>
  f == null ? x : f(flowRight(...more)(x))

const upper = s =>
  s.toUpperCase()

const greeting = s =>
  `Hello, ${s}`

const addQuotes = s =>
  `"${s}"`

const sayHello =
  flowRight(addQuotes, greeting, upper)

console.log(sayHello("world"))
// "Hello, WORLD"

从左到右的构图

const flowLeft = (f, ...more) => x =>
  f == null ? x : flowLeft(...more)(f(x))

const upper = s =>
  s.toUpperCase()

const greeting = s =>
  `Hello, ${s}`

const addQuotes = s =>
  `"${s}"`

const sayHello =
  flowLeft(addQuotes, greeting, upper)

console.log(sayHello("world"))
// HELLO, "WORLD"

---

使用 reduceRight

我们可以使用reduceRight 轻松实现flowRight -

const flowRight = (...fs) => init =>
  fs.reduceRight((x, f) => f(x), init)

const upper = s =>
  s.toUpperCase()

const greeting = s =>
  `Hello, ${s}`

const addQuotes = s =>
  `"${s}"`

const sayHello =
  flowRight(addQuotes, greeting, upper)

console.log(sayHello("world"))
// "Hello, WORLD"

使用reduce

或者我们可以使用reduce 轻松实现flowRight -

const flowLeft = (...fs) => init =>
  fs.reduce((x, f) => f(x), init)

const upper = s =>
  s.toUpperCase()

const greeting = s =>
  `Hello, ${s}`

const addQuotes = s =>
  `"${s}"`

const sayHello =
  flowLeft(addQuotes, greeting, upper)

console.log(sayHello("world"))
// HELLO, "WORLD"

Answer 4

下面编写的函数的想法是返回一个函数，该函数将遍历函数列表并存储每次调用的结果并在最后返回。

function flowRight(...args) {
    return function (initial) {
        let value = initial;

        for (let i = args.length - 1; i >= 0; i--) {
            value = args[i](value);
        }

        return value;   
    };
}