在JavaScript中,我想创建一个对象实例(通过new操作符),但将任意数量的参数传递给构造函数。这可能吗?

我想做的是这样的(但下面的代码不起作用):

function Something(){
    // init stuff
}
function createSomething(){
    return new Something.apply(null, arguments);
}
var s = createSomething(a,b,c); // 's' is an instance of Something

这个问题的答案

从这里的响应可以清楚地看出,没有内置的方法可以使用new操作符调用.apply()。然而,人们对这个问题提出了许多非常有趣的解决方案。

我更喜欢的解决方案是来自Matthew Crumley的这个(我修改了它来传递arguments属性):

var createSomething = (function() {
    function F(args) {
        return Something.apply(this, args);
    }
    F.prototype = Something.prototype;

    return function() {
        return new F(arguments);
    }
})();

当前回答

创建一个匿名原型,并使用参数将Something原型应用于它,然后创建该匿名原型的新实例。这样做的一个缺点是它不会通过s instanceof Something检查,尽管它是相同的,但它基本上是一个克隆的实例。

function Something(){
    // init stuff
}
function createSomething(){
    return new (function(){Something.apply(this, arguments)});
}
var s = createSomething(a,b,c); // 's' is an instance of Something

其他回答

看看CoffeeScript是如何做到的。

([a,b,c]…)

就变成:

var s;
s = (function(func, args, ctor) {
  ctor.prototype = func.prototype;
  var child = new ctor, result = func.apply(child, args);
  return Object(result) === result ? result : child;
})(Something, [a, b, c], function(){});

是的,我们可以,javascript在本质上更多的是原型继承。

function Actor(name, age){
  this.name = name;
  this.age = age;
}

Actor.prototype.name = "unknown";
Actor.prototype.age = "unknown";

Actor.prototype.getName = function() {
    return this.name;
};

Actor.prototype.getAge = function() {
    return this.age;
};

当我们创建一个带有"new"的对象时,我们创建的对象继承getAge(),但如果我们使用apply(…)或call(…)来调用Actor,那么我们为"this"传递了一个对象,但我们传递的对象不会继承自Actor.prototype

除非,我们直接通过apply或调用Actor。原型但是....“this”指向“Actor”。this.name将写入:actor。prototype.name。从而影响所有用Actor创建的其他对象…因为我们覆盖的是原型而不是实例

var rajini = new Actor('Rajinikanth', 31);
console.log(rajini);
console.log(rajini.getName());
console.log(rajini.getAge());

var kamal = new Actor('kamal', 18);
console.log(kamal);
console.log(kamal.getName());
console.log(kamal.getAge());

让我们试试apply

var vijay = Actor.apply(null, ["pandaram", 33]);
if (vijay === undefined) {
    console.log("Actor(....) didn't return anything 
           since we didn't call it with new");
}

var ajith = {};
Actor.apply(ajith, ['ajith', 25]);
console.log(ajith); //Object {name: "ajith", age: 25}
try {
    ajith.getName();
} catch (E) {
    console.log("Error since we didn't inherit ajith.prototype");
}
console.log(Actor.prototype.age); //Unknown
console.log(Actor.prototype.name); //Unknown

通过传递Actor。prototype to Actor.call()作为第一个参数,当Actor()函数运行时,它执行this.name=name,因为“this”将指向Actor。原型,this.name =名称;意味着Actor.prototype.name =名称;

var simbhu = Actor.apply(Actor.prototype, ['simbhu', 28]);
if (simbhu === undefined) {
    console.log("Still undefined since the function didn't return anything.");
}
console.log(Actor.prototype.age); //simbhu
console.log(Actor.prototype.name); //28

var copy = Actor.prototype;
var dhanush = Actor.apply(copy, ["dhanush", 11]);
console.log(dhanush);
console.log("But now we've corrupted Parent.prototype in order to inherit");
console.log(Actor.prototype.age); //11
console.log(Actor.prototype.name); //dhanush

回到最初的问题,如何使用新的操作符与应用,这里是我的....

Function.prototype.new = function(){
    var constructor = this;
    function fn() {return constructor.apply(this, args)}
    var args = Array.prototype.slice.call(arguments);
    fn.prototype = this.prototype;
    return new fn
};

var thalaivar = Actor.new.apply(Parent, ["Thalaivar", 30]);
console.log(thalaivar);

其实最简单的方法是:

function Something (a, b) {
  this.a = a;
  this.b = b;
}
function createSomething(){
    return Something;
}
s = new (createSomething())(1, 2); 
// s == Something {a: 1, b: 2}

假设你有一个Items构造函数,它吸收了你给它的所有参数:

function Items () {
    this.elems = [].slice.call(arguments);
}

Items.prototype.sum = function () {
    return this.elems.reduce(function (sum, x) { return sum + x }, 0);
};

你可以用Object.create()创建一个实例,然后用.apply()创建该实例:

var items = Object.create(Items.prototype);
Items.apply(items, [ 1, 2, 3, 4 ]);

console.log(items.sum());

当运行时输出10,因为1 + 2 + 3 + 4 == 10:

$ node t.js
10

这难道不行吗?半睡半醒,没有仔细读书。

var Storage = undefined;

return ((Storage = (new Something(...))) == undefined? (undefined) : (Storage.apply(...)));