Calling method using JavaScript prototype

Is it possible to call the base method from a prototype method in JavaScript if it's been overridden?

MyClass = function(name){
    this.name = name;
    this.do = function() {
        //do somthing 
    }
};

MyClass.prototype.do = function() {  
    if (this.name === 'something') {
        //do something new
    } else {
        //CALL BASE METHOD
    }
};

Answers:

Answer

I did not understand what exactly you're trying to do, but normally implementing object-specific behaviour is done along these lines:

function MyClass(name) {
    this.name = name;
}

MyClass.prototype.doStuff = function() {
    // generic behaviour
}

var myObj = new MyClass('foo');

var myObjSpecial = new MyClass('bar');
myObjSpecial.doStuff = function() {
    // do specialised stuff
    // how to call the generic implementation:
    MyClass.prototype.doStuff.call(this /*, args...*/);
}
Answer

Well one way to do it would be saving the base method and then calling it from the overriden method, like so

MyClass.prototype._do_base = MyClass.prototype.do;
MyClass.prototype.do = function(){  

    if (this.name === 'something'){

        //do something new

    }else{
        return this._do_base();
    }

};
Answer

I'm afraid your example does not work the way you think. This part:

this.do = function(){ /*do something*/ };

overwrites the definition of

MyClass.prototype.do = function(){ /*do something else*/ };

Since the newly created object already has a "do" property, it does not look up the prototypal chain.

The classical form of inheritance in Javascript is awkard, and hard to grasp. I would suggest using Douglas Crockfords simple inheritance pattern instead. Like this:

function my_class(name) {
    return {
        name: name,
        do: function () { /* do something */ }
    };
}

function my_child(name) {
    var me = my_class(name);
    var base_do = me.do;
    me.do = function () {
        if (this.name === 'something'){
            //do something new
        } else {
            base_do.call(me);
        }
    }
    return me;
}

var o = my_child("something");
o.do(); // does something new

var u = my_child("something else");
u.do(); // uses base function

In my opinion a much clearer way of handling objects, constructors and inheritance in javascript. You can read more in Crockfords Javascript: The good parts.

Answer
function NewClass() {
    var self = this;
    BaseClass.call(self);          // Set base class

    var baseModify = self.modify;  // Get base function
    self.modify = function () {
        // Override code here
        baseModify();
    };
}
Answer

if you define a function like this (using OOP)

function Person(){};
Person.prototype.say = function(message){
   console.log(message);
}

there is two ways to call a prototype function: 1) make an instance and call the object function:

var person = new Person();
person.say('hello!');

and the other way is... 2) is calling the function directly from the prototype:

Person.prototype.say('hello there!');
Answer

An alternative :

// shape 
var shape = function(type){
    this.type = type;
}   
shape.prototype.display = function(){
    console.log(this.type);
}
// circle
var circle = new shape('circle');
// override
circle.display = function(a,b){ 
    // call implementation of the super class
    this.__proto__.display.apply(this,arguments);
}
Answer

This solution uses Object.getPrototypeOf

TestA is super that has getName

TestB is a child that overrides getName but, also has getBothNames that calls the super version of getName as well as the child version

function TestA() {
  this.count = 1;
}
TestA.prototype.constructor = TestA;
TestA.prototype.getName = function ta_gn() {
  this.count = 2;
  return ' TestA.prototype.getName is called  **';
};

function TestB() {
  this.idx = 30;
  this.count = 10;
}
TestB.prototype = new TestA();
TestB.prototype.constructor = TestB;
TestB.prototype.getName = function tb_gn() {
  return ' TestB.prototype.getName is called ** ';
};

TestB.prototype.getBothNames = function tb_gbn() {
  return Object.getPrototypeOf(TestB.prototype).getName.call(this) + this.getName() + ' this object is : ' + JSON.stringify(this);
};

var tb = new TestB();
console.log(tb.getBothNames());

Answer

If I understand correctly, you want Base functionality to always be performed, while a piece of it should be left to implementations.

You might get helped by the 'template method' design pattern.

Base = function() {}
Base.prototype.do = function() { 
    // .. prologue code
    this.impldo(); 
    // epilogue code 
}
// note: no impldo implementation for Base!

derived = new Base();
derived.impldo = function() { /* do derived things here safely */ }
Answer

If you know your super class by name, you can do something like this:

function Base() {
}

Base.prototype.foo = function() {
  console.log('called foo in Base');
}

function Sub() {
}

Sub.prototype = new Base();

Sub.prototype.foo = function() {
  console.log('called foo in Sub');
  Base.prototype.foo.call(this);
}

var base = new Base();
base.foo();

var sub = new Sub();
sub.foo();

This will print

called foo in Base
called foo in Sub
called foo in Base

as expected.

Answer

Another way with ES5 is to explicitely traverse the prototype chain using Object.getPrototypeOf(this)

const speaker = {
  speak: () => console.log('the speaker has spoken')
}

const announcingSpeaker = Object.create(speaker, {
  speak: {
    value: function() {
      console.log('Attention please!')
      Object.getPrototypeOf(this).speak()
    }
  }
})

announcingSpeaker.speak()
Answer

No, you would need to give the do function in the constructor and the do function in the prototype different names.

Answer

I know this post is from 4 years ago, but because of my C# background I was looking for a way to call the base class without having to specify the class name but rather obtain it by a property on the subclass. So my only change to Christoph's answer would be

From this:

MyClass.prototype.doStuff.call(this /*, args...*/);

To this:

this.constructor.prototype.doStuff.call(this /*, args...*/);
Answer

In addition, if you want to override all instances and not just that one special instance, this one might help.

function MyClass() {}

MyClass.prototype.myMethod = function() {
  alert( "doing original");
};
MyClass.prototype.myMethod_original = MyClass.prototype.myMethod;
MyClass.prototype.myMethod = function() {
  MyClass.prototype.myMethod_original.call( this );
  alert( "doing override");
};

myObj = new MyClass();
myObj.myMethod();

result:

doing original
doing override
Answer
function MyClass() {}

MyClass.prototype.myMethod = function() {
  alert( "doing original");
};
MyClass.prototype.myMethod_original = MyClass.prototype.myMethod;
MyClass.prototype.myMethod = function() {
  MyClass.prototype.myMethod_original.call( this );
  alert( "doing override");
};

myObj = new MyClass();
myObj.myMethod();

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