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Home/ Questions/Q 8502659
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T01:32:49+00:00 2026-06-11T01:32:49+00:00

This is a question for the guru of JavaScript. I’m trying to do work

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This is a question for the guru of JavaScript. I’m trying to do work with JavaScript prototype model more elegant. Here is my utility code (it provides real chain of prototypes and correct work with instanceof operator):

function Class(conf) {
  var init = conf.init || function () {};
  delete conf.init;

  var parent = conf.parent || function () {};
  delete conf.parent;

  var F = function () {};
  F.prototype = parent.prototype;
  var f = new F();
  for (var fn in conf) f[fn] = conf[fn];
  init.prototype = f;

  return init;
};

It allows me to do such thigns:

var Class_1 = new Class({
  init: function (msg) { // constructor
    this.msg = msg;
  },

  method_1: function () {
    alert(this.msg + ' in Class_1::method_1');
  },

  method_2: function () {
    alert(this.msg + ' in Class_1::method_2');
  }
});

var Class_2 = new Class({
  parent: Class_1,

  init: function (msg) { // constructor
    this.msg = msg;
  },

  // method_1 will be taken from Class_1

  method_2: function () { // this method will overwrite the original one
    alert(this.msg + ' in Class_2::method_2');
  },

  method_3: function () { // just new method
    alert(this.msg + ' in Class_2::method_3');
  }
});

var c1 = new Class_1('msg');
c1.method_1(); // msg in Class_1::method_1
c1.method_2(); // msg in Class_1::method_2

var c2 = new Class_2('msg');
c2.method_1(); // msg in Class_1::method_1
c2.method_2(); // msg in Class_2::method_2
c2.method_3(); // msg in Class_2::method_3

alert('c1 < Class_1 - ' + (c1 instanceof Class_1 ? 'true' : 'false')); // true
alert('c1 < Class_2 - ' + (c1 instanceof Class_2 ? 'true' : 'false')); // false

alert('c2 < Class_1 - ' + (c2 instanceof Class_1 ? 'true' : 'false')); // true
alert('c2 < Class_2 - ' + (c2 instanceof Class_2 ? 'true' : 'false')); // true

My question is: Is there more simple way to do this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T01:32:51+00:00Added an answer on June 11, 2026 at 1:32 am

    After some research I’ve concluded there is no more simple way to do this.

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