获取超类的子类构造函数

Question

var Super = function() {}
Super.prototype.newOfThisKind = function() {
    return new this.myConstructor()
}
var Sub1 = function() { this.myConstructor = Sub1 }
var Sub2 = function() { this.myConstructor = Sub2 }
Sub1.prototype = new Super()
Sub2.prototype = new Super()
console.assert(new Sub1().newOfThisKind() instanceof Sub1)
console.assert(new Sub2().newOfThisKind() instanceof Sub2)

这段代码按预期工作，但是如何在不对每个子类设置myConstructor的情况下编写newOfThisKind ？

Answer 1

好吧，我不明白目标是什么，而是回答您的问题：

    var Super = function () { }
    Super.prototype.newOfThisKind = function () {
        return Object.create(this);
    }
    var Sub1 = function () { }
    var Sub2 = function () { }
    Sub1.prototype = new Super()
    Sub2.prototype = new Super()
    console.assert(new Sub1().newOfThisKind() instanceof Sub1)
    console.assert(new Sub2().newOfThisKind() instanceof Sub2)