什么是无与伦比的对象克隆,由一个承诺完成?

async function clone(thingy /**/)
{
    if(thingy instanceof Promise)
    {
        throw Error("This function cannot clone Promises.");
    }
    return thingy;
}

这里是一个全面的克隆()方法,可以克隆任何JavaScript对象,它处理几乎所有情况:

function clone(src, deep) {

    var toString = Object.prototype.toString;
    if (!src && typeof src != "object") {
        // Any non-object (Boolean, String, Number), null, undefined, NaN
        return src;
    }

    // Honor native/custom clone methods
    if (src.clone && toString.call(src.clone) == "[object Function]") {
        return src.clone(deep);
    }

    // DOM elements
    if (src.nodeType && toString.call(src.cloneNode) == "[object Function]") {
        return src.cloneNode(deep);
    }

    // Date
    if (toString.call(src) == "[object Date]") {
        return new Date(src.getTime());
    }

    // RegExp
    if (toString.call(src) == "[object RegExp]") {
        return new RegExp(src);
    }

    // Function
    if (toString.call(src) == "[object Function]") {

        //Wrap in another method to make sure == is not true;
        //Note: Huge performance issue due to closures, comment this :)
        return (function(){
            src.apply(this, arguments);
        });
    }

    var ret, index;
    //Array
    if (toString.call(src) == "[object Array]") {
        //[].slice(0) would soft clone
        ret = src.slice();
        if (deep) {
            index = ret.length;
            while (index--) {
                ret[index] = clone(ret[index], true);
            }
        }
    }
    //Object
    else {
        ret = src.constructor ? new src.constructor() : {};
        for (var prop in src) {
            ret[prop] = deep
                ? clone(src[prop], true)
                : src[prop];
        }
    }
    return ret;
};

对于未来的参考,目前的 ECMAScript 6 草案将 Object.assign 引入为克隆对象的一种方式。

var obj1 = { a: true, b: 1 };
var obj2 = Object.assign(obj1);
console.log(obj2); // { a: true, b: 1 }

在编写时,支持仅限于Firefox 34在浏览器中,所以它还不能在生产代码中使用(除非您正在编写Firefox扩展)。

因为回归只是太昂贵的JavaScript,我发现的大多数答案是使用回归,而JSON方法将错过非JSON转换部分(功能等)。所以我做了一些研究,并发现这个拖拉机技术避免它。

/*
 * Trampoline to avoid recursion in JavaScript, see:
 *     https://www.integralist.co.uk/posts/functional-recursive-javascript-programming/
 */
function trampoline() {
    var func = arguments[0];
    var args = [];
    for (var i = 1; i < arguments.length; i++) {
        args[i - 1] = arguments[i];
    }

    var currentBatch = func.apply(this, args);
    var nextBatch = [];

    while (currentBatch && currentBatch.length > 0) {
        currentBatch.forEach(function(eachFunc) {
            var ret = eachFunc();
            if (ret && ret.length > 0) {
                nextBatch = nextBatch.concat(ret);
            }
        });

        currentBatch = nextBatch;
        nextBatch = [];
    }
};

/*
 *  Deep clone an object using the trampoline technique.
 *
 *  @param target {Object} Object to clone
 *  @return {Object} Cloned object.
 */
function clone(target) {
    if (typeof target !== 'object') {
        return target;
    }
    if (target == null || Object.keys(target).length == 0) {
        return target;
    }

    function _clone(b, a) {
        var nextBatch = [];
        for (var key in b) {
            if (typeof b[key] === 'object' && b[key] !== null) {
                if (b[key] instanceof Array) {
                    a[key] = [];
                }
                else {
                    a[key] = {};
                }
                nextBatch.push(_clone.bind(null, b[key], a[key]));
            }
            else {
                a[key] = b[key];
            }
        }
        return nextBatch;
    };

    var ret = target instanceof Array ? [] : {};
    (trampoline.bind(null, _clone))(target, ret);
    return ret;
};

2017年例子:

let objectToCopy = someObj;
let copyOfObject = {};
Object.defineProperties(copyOfObject, Object.getOwnPropertyDescriptors(objectToCopy));
// copyOfObject will now be the same as objectToCopy

我的场景略有不同,我有一個物體,擁有<unk>的物體以及功能,因此,Object.assign() 和 JSON.stringify() 並沒有解決我的問題。

因此,我决定做一个简单的功能,使用内置的方法来复制一个对象,其字面特性,其粘着的对象和功能。

let deepCopy = (target, source) => {
    Object.assign(target, source);
    // check if there's any nested objects
    Object.keys(source).forEach((prop) => {
        /**
          * assign function copies functions and
          * literals (int, strings, etc...)
          * except for objects and arrays, so:
          */
        if (typeof(source[prop]) === 'object') {
            // check if the item is, in fact, an array
            if (Array.isArray(source[prop])) {
                // clear the copied referenece of nested array
                target[prop] = Array();
                // iterate array's item and copy over
                source[prop].forEach((item, index) => {
                    // array's items could be objects too!
                    if (typeof(item) === 'object') {
                        // clear the copied referenece of nested objects
                        target[prop][index] = Object();
                        // and re do the process for nested objects
                        deepCopy(target[prop][index], item);
                    } else {
                        target[prop].push(item);
                    }
                });
            // otherwise, treat it as an object
            } else {
                // clear the copied referenece of nested objects
                target[prop] = Object();
                // and re do the process for nested objects
                deepCopy(target[prop], source[prop]);
            }
        }
    });
};

下面是测试代码:

let a = {
    name: 'Human', 
    func: () => {
        console.log('Hi!');
    }, 
    prop: {
        age: 21, 
        info: {
            hasShirt: true, 
            hasHat: false
        }
    },
    mark: [89, 92, { exam: [1, 2, 3] }]
};

let b = Object();

deepCopy(b, a);

a.name = 'Alien';
a.func = () => { console.log('Wassup!'); };
a.prop.age = 1024;
a.prop.info.hasShirt = false;
a.mark[0] = 87;
a.mark[1] = 91;
a.mark[2].exam = [4, 5, 6];

console.log(a); // updated props
console.log(b);

对于效率相关的担忧,我认为这是我遇到的问题的最简单和最有效的解决方案,我会欣赏任何关于这个算法的评论,这可能会使它更有效。