Default values for objects in Javascript

Question

Javascript Code

var a = {};
a.test += 1; //NaN
++a.test; //NaN

Instead

var a = {};
a.test = 0;
++a.test; //1
a.test += 1; //2

I wonder if there could be anyway that can make first code sample work the same as second, i.e without an explicit assignment to 0. As in assigning default value for any property of an object to 0 instead undefined. I'm trying to do this in node.js. So, no problem of cross browser things and old ECMA Specs.

var i;
for(i = 0; i<10; i++) {
   if(a.test) {
     ++a.test;
   } else {
     a.test = 0;
     ++a.test;
   }
   //a.test = a.test || 0; (1)
   //++a.test;
}

If it is possible then the inner if/else or the assignment statement(1) in the above code can be eliminated.

@outis properties are dynamically generated. You can even consider var a = []; a[0]++; Here size of the array can't be determined upfront — Tamil
– Tamil, Commented May 1, 2012 at 9:13
what should happen if a.test === false or ""? it would be safer to do a.hasOwnProperty("test") if you are trying to establish whether or not the property is present — Rune FS
– Rune FS, Commented May 1, 2012 at 9:23

Ivan Drinchev · Accepted Answer · 2012-05-01 09:05:49Z

4

Javascript by default defines all new variables as undefined ( if not explicitly defined ) , which is different from Number object, which you are trying to define. So you should use smth like :

for (i=0; i<10; i++) {
    a.test = a.test || 0;
    ++a.test
}

answered May 1, 2012 at 9:05

Ivan Drinchev

19.6k4 gold badges69 silver badges97 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Tamil Over a year ago

I have mentioned ur answer in my code only. See the comment in the 3rd code snippet

Ivan Drinchev Over a year ago

You should use prototype/class object if you want your default a.test always defined, as I mention if something isn't defined explicitly as Number, String, etc. it is undefined

Alnitak · Accepted Answer · 2012-05-01 10:30:44Z

3

There's no way to do this for any arbitrary undefined property.

For a known property name, there is a way, but DO NOT USE IT !! ¹

Object.prototype.test = 0;

This will give every object an implicit .test property with that value in it, and the first time you attempt to modify it the result will be stored in your own object:

> Object.prototype.test = 0
> a = {}
> a.test++
> a.test
1

¹ Adding stuff to Object.prototype will break stuff, including for (key in obj)

edited May 1, 2012 at 10:30

answered May 1, 2012 at 9:06

Alnitak

341k72 gold badges420 silver badges503 bronze badges

Comments

ninjagecko · Accepted Answer · 2012-05-01 09:07:26Z

1

This is where prototypes come in useful in javascript:

function Demo() {
}
Demo.prototype.x = 0;

> a = new Demo();
> a.x += 1
> a.x
1

> b = new Demo()
> b.x
0

answered May 1, 2012 at 9:07

ninjagecko

91.5k24 gold badges143 silver badges153 bronze badges

1 Comment

Alnitak Over a year ago

@mihai yup, just like my answer, since there's no way do it if you don't know the name of the property

Joseph · Accepted Answer · 2012-05-01 09:10:06Z

0

In the first code, you can't. Any number added to a non-number or NaN value will always result in NaN

var a = {};
a.test += 1; // undefined + 1 = NaN
++a.test;    // ++(undefined) = NaN

as for the inner if

for(i = 0; i<10; i++) {
   a.test = a.test || 0; //use existing value, or if undefined, 0
   ++a.test;             //increment
}

edited May 1, 2012 at 9:10

answered May 1, 2012 at 9:03

Joseph

120k31 gold badges186 silver badges238 bronze badges

Comments

Shahbaz Chishty · Accepted Answer · 2012-05-01 09:16:41Z

0

I think this would do:

var a = {"test":0};
a.test += 1;

answered May 1, 2012 at 9:16

Shahbaz Chishty

5121 gold badge4 silver badges9 bronze badges

1 Comment

Rune FS Over a year ago

OP does not want to explicitly declare/initialize test

Koenyn · Accepted Answer · 2012-05-01 09:07:09Z

-1

Using standard JS it's not possible to do what you're asking. Unless you prototype I guess? But I'm not experienced with Prototype.And I don't think prototyping works with Node.js

The reason for this is because js is not a typed language (i.e. we only use var to declare a variable, not int a or string b), so in order to use the ++ operator, the variable needs to be given a type through assignment.

hope that helps

answered May 1, 2012 at 9:07

Koenyn

7045 silver badges16 bronze badges

Collectives™ on Stack Overflow

Default values for objects in Javascript

6 Answers 6

2 Comments

Comments

1 Comment

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

2 Comments

Comments

1 Comment

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related