Class inheritance with method in javascript

Question

I am new to javascript app development.When i came with functions and all i have found a code like

function Parenizor(value) {
console.log(this.setValue());
}

Parenizor.method('setValue', function () {
console.log('am called');
});

Parenizor.method('getValue', function () {
return 1;
});

Parenizor.method('toString', function () {
return 2;
});

And when i called the function like

var a = new Parenizor(1)

a.setValue()

It throws me error like Undefined not a function..Why is it like this ??..Hope anyone here can find my mistake ..Thanz ..:)

How about Parenizor.prototype.setValue = function(){code} ?.. — orhanhenrik
– orhanhenrik, Commented Oct 10, 2014 at 17:22
@Izzey yeah i know to call via prototype ..but i just wanna call it via .method() ..is it possible like i did or is there any way to call with .method() — user3902466
– user3902466, Commented Oct 10, 2014 at 17:25
Please do not use Crockford as a reference to classical inheritance. None of his code does it correctly and he wants private variables because he's worried about encapsulation but then brakes it by modifying objects he doesn't own (Function and Object prototype) maybe this answer can help you: stackoverflow.com/questions/16063394/… — HMR
– HMR, Commented Oct 11, 2014 at 0:41

Oriol · Accepted Answer · 2014-10-10 17:25:42Z

1

It seems your code comes from Classical Inheritance in JavaScript, by Douglas Crockford.

I guess you didn't read this part:

To make the examples above work, I wrote four sugar methods. First, the method method, which adds an instance method to a class.
Function.prototype.method = function (name, func) {
    this.prototype[name] = func;
    return this;
};

answered Oct 10, 2014 at 17:25

Oriol

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6 Comments

user3902466 Over a year ago

..yeah exactly i got this code from CrickFord blog ..i didnt get what you mean by sugar methods ..can you provide the working by an example

orhanhenrik Over a year ago

Simply copy-paste the code to the top of your javascript

Oriol Over a year ago

Just include that code before attempting to use Parenizor.method.

user3902466 Over a year ago

@Oriol i have added the code above but didnt get the result..can you provide the fullcode in ur answer ..it would be really helpful.

user3902466 Over a year ago

@Oriol i have aded the code you given me in the answr ..but it still throws the errir..

|

allyourcode · Accepted Answer · 2014-10-10 17:21:43Z

1

That's not how you define a method. Do it like this:

function K() {}

K.prototype.setValue = function () {};

K.prototype.getValue = function () {};

answered Oct 10, 2014 at 17:21

allyourcode

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4 Comments

sma Over a year ago

More explanation: stackoverflow.com/questions/6918509/…

user3902466 Over a year ago

so calling like Parenizor.method('method') is invalid ??

Oriol Over a year ago

@user3902466 No, it's not invalid syntax. But it doesn't exist by default, you must define Function.prototype.method manually.

user3902466 Over a year ago

@Oriol yeah ..i got that ..but i have to ask you one thing too ..so why that .method came into action from crockfords blog ??

Rubens Pinheiro · Accepted Answer · 2014-10-10 17:29:36Z

0

Javascript is OO, not with classes, but with prototypes. First when you declare Parenizor, you're declaring a 'constructor'. You probaly want this:

function Parenizor(){
    var value = null;

    this.setValue = function(val){
        value = val;
    };

    this.getValue = function(){
        return value;
    };

    this.toString = function(){
        return String(value);
    };
}

Or also, setting into the object prototype:

function Parenizor(){}
Parenizor.prototype._value = null;
Parenizor.prototype.setValue = function(){...};
Parenizor.prototype.getValue = function(){...};
Parenizor.prototype.toString = function(){...};

answered Oct 10, 2014 at 17:29

Rubens Pinheiro

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