Is it possible to call a super setter in ES6 inherited classes?

偶尔善良 提交于 2019-11-30 11:09:48
class Y extends X {
  constructor(name) {
    super(name);
  }

  set name(name) {
    super.name = name;
    this._name += "Y";
  }
}

will override the name properly with an accessor for just the setter, with no getter. That means your y.name === "hiXY" will fail because y.name will return undefined because there is no getter for name. You need:

class Y extends X {
  constructor(name) {
    super(name);
  }

  get name(){
    return super.name;
  }

  set name(name) {
    super.name = name;
    this._name += "Y";
  }
}

for this case you have a more simple solution:


class Y extends X {
    set name(name) {
        super.name = name;
        this._name += "Y";
    }
}
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!