how to call parent constructor?

Question

Let's say I have the following code snippet.

function test(id) { alert(id); }

testChild.prototype = new test();

function testChild(){}

var instance = new testChild('hi');

Is it possible to get alert('hi')? I get undefined now.

Solution 1

That's how you do this in CoffeeScript:

class Test
  constructor: (id) -> alert(id)

class TestChild extends Test

instance = new TestChild('hi')

Nope, I'm not starting a holy war. Instead, I'm suggesting to take a look at resulting JavaScript code to see how subclassing could be implemented:

// Function that does subclassing
var __extends = function(child, parent) {
  for (var key in parent) {
    if (Object.prototype.hasOwnProperty.call(parent, key)) {
      child[key] = parent[key];
    }
  }
  function ctor() { this.constructor = child; }
  ctor.prototype = parent.prototype;
  child.prototype = new ctor;
  child.__super__ = parent.prototype;
  return child;
};

// Our code
var Test, TestChild, instance;

Test = function(id) { alert(id); };

TestChild = function() {
  TestChild.__super__.constructor.apply(this, arguments);
}; __extends(TestChild, Test);

instance = new TestChild('hi');

// And we get an alert

See it in action at http://jsfiddle.net/NGLMW/3/.

To stay correct, the code is slightly modified and commented to be more readable, compared to CoffeeScript output.

Solution 2

JS OOP ...

// parent class
var Test = function(id) {
    console.log(id);
};

// child class
var TestChild = function(id) {
    Test.call(this, id); // call parent constructor
};

// extend from parent class prototype
TestChild.prototype = Object.create(Test.prototype); // keeps the proto clean
TestChild.prototype.constructor = TestChild; // repair the inherited constructor

// end-use
var instance = new TestChild('foo');

Solution 3

You already have many answers, but I'll throw in the ES6 way, which IMHO is the new standard way to do this.

class Parent { 
  constructor() { alert('hi'); } 
}
class Child extends Parent { 
  // Optionally include a constructor definition here. Leaving it 
  // out means the parent constructor is automatically invoked.
  constructor() {
    // imagine doing some custom stuff for this derived class
    super();  // explicitly call parent constructor.
  }
}

// Instantiate one:
var foo = new Child();  // alert: hi

Solution 4

By taking advantage of variable arguments and the apply() method, you could do it this way. Here's a fiddle for this example.

function test(id) { alert(id); }
function testChild() {
  testChild.prototype.apply(this, arguments);
  alert('also doing my own stuff');
}
testChild.prototype = test;
var instance = new testChild('hi', 'unused', 'optional', 'args');

Solution 5

You need to declare the function testChild() before you set its prototype. Then you need to call testChild.test to call the method. I believe you want to set testChild.prototype.test = test, then you can call testChild.test('hi') and it should resolve properly.

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