Question

在学校，我们的任务是建立lodash method flowRight!

的实施

在它提到的规范中：

采用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（第一个除外）是由函数返回值确定。电话到flowRight返回的函数求值为返回值最左边的功能。

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')

我觉得我理解静态示例中发生的事情就像这个示例演示一样。

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

猜猜我很难将我的大脑环绕在一个循环中，我认为你需要它。这是我到目前为止的实施。

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

无需循环。如果允许，则使用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)

我已经评论了下面的代码，试图解释这一切是如何协同工作的。

＆＃13;

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'));

＆＃13;

Answer 3

下面写的函数中的想法是返回一个函数，该函数将遍历函数列表并存储每个调用的结果并在结束时返回它。

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

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

        return value;   
    };
}

Answer 4

从右到左的构图

const compose = (f,...fs) => x =>
  f === undefined ? x : f(compose(...fs)(x))

从左到右的构图

const compose = (f,...fs) => x =>
  f === undefined ? x : compose(...fs)(f(x))

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

4 个答案: