Javascript OOP Methods

Question

Just wondering, having the following simple code:

var object1 = {
    name: function (){
        return 'myName';
    },
    surname: function (){
        return 'mySurname';
    }
};

Why does JS returns function() in this case object1.name ?

Why does JS returns the expected result myName if I call object1.name() ?

score 2 · Accepted Answer · edited May 23 '17 at 12:12

2

Referencing name returns what name is–in this case, a function.
Calling name by appending (), i.e., name(), returns a value–the string "myName".

My answer to When do I use parenthesis and when do I not? provides more details.

edited May 23 '17 at 12:12

Community

1
1

answered Jun 19 '12 at 01:30

Dave Newton

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Thanks, your answer there was complete. – Jackie Chan Jun 19 '12 at 01:43

score 1 · Answer 2 · answered Jun 19 '12 at 01:30

object1.name;//Returns the function declaration
object1.name();//Calls the function and returns its value

It works like the following code:

var myFn = function(){
    return "This is the return value of this function";
}
alert(myFn);//Alerts the myFn function's declaration
alert(myFn());//Alerts "This is the return value of this function"

score 1 · Answer 3 · answered Jun 19 '12 at 01:31

1

because in object1.name you are calling the function declaration

and in object1.name() you are calling the function

answered Jun 19 '12 at 01:31

Ibu

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Javascript OOP Methods

3 Answers3