这是一个通用的解决方案，可以调用任何构造函数(除了本机构造函数，当作为函数调用时行为不同，如String, Number, Date等)与参数数组:

function construct(constructor, args) {
    function F() {
        return constructor.apply(this, args);
    }
    F.prototype = constructor.prototype;
    return new F();
}

通过调用construct(Class，[1,2,3])创建的对象将与用new Class(1,2,3)创建的对象相同。

您还可以创建一个更具体的版本，这样就不必每次都传递构造函数。这也稍微更有效，因为它不需要每次调用时都创建内部函数的新实例。

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

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

这样创建和调用外部匿名函数的原因是为了防止函数F污染全局命名空间。它有时被称为模块模式。

(更新)

对于那些想在TypeScript中使用它的人，因为如果F返回任何东西，TS会给出一个错误:

function construct(constructor, args) {
    function F() : void {
        constructor.apply(this, args);
    }
    F.prototype = constructor.prototype;
    return new F();
}

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);

Matthew Crumley在CoffeeScript中的解决方案:

construct = (constructor, args) ->
    F = -> constructor.apply this, args
    F.prototype = constructor.prototype
    new F

or

createSomething = (->
    F = (args) -> Something.apply this, args
    F.prototype = Something.prototype
    return -> new Something arguments
)()

从ES6开始，这可以通过扩展操作符实现，请参见https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_operator#Apply_for_new

这个答案已经在https://stackoverflow.com/a/42027742/7049810评论中给出了，但似乎被大多数人忽略了

@jordancpaul的答案修改后的解决方案。

var applyCtor = function(ctor, args)
{
    var instance = new ctor();
    ctor.prototype.constructor.apply(instance, args);
    return instance;
}; 

你不能像new操作符那样调用带有可变数量参数的构造函数。

你能做的就是稍微改变构造函数。而不是:

function Something() {
    // deal with the "arguments" array
}
var obj = new Something.apply(null, [0, 0]);  // doesn't work!

你可以这样做:

function Something(args) {
    // shorter, but will substitute a default if args.x is 0, false, "" etc.
    this.x = args.x || SOME_DEFAULT_VALUE;

    // longer, but will only put in a default if args.x is not supplied
    this.x = (args.x !== undefined) ? args.x : SOME_DEFAULT_VALUE;
}
var obj = new Something({x: 0, y: 0});

或者如果你必须使用数组:

function Something(args) {
    var x = args[0];
    var y = args[1];
}
var obj = new Something([0, 0]);