javascript variable assignment core

Question

suppose i have this

var x={};    //x is an address in the memory where object is stored
var z=x;     //z=x after execution is equal to z={} right?

now z has nothing to do with x or not related to x after execution so when,

x={name:"Maizere"};
z!=x        //true

but, when

x.name="maizere";
alert(z.name)//maizere why?

we are not setting the value of z but x and z relation to x shouldn't exit anymore

actual code:

 x={};
 y=x;
 x.name="maizere";
 alert(y.name)//maizere

I really have no knowledge of how this is working .Can anyone explain this in detail please?

I'm getting undefined. Please show us your exact code, and use comments properly. — Bergi
– Bergi, Commented Jan 31, 2013 at 17:45
@Bergi no u did some mistake ,u get "Maizere" .Check the code u tested plz — Maizere Pathak
– Maizere Pathak, Commented Jan 31, 2013 at 17:50

JJJ · Accepted Answer · 2013-01-31 17:51:57Z

3

Your initial assumption is wrong; z is a pointer to the same object as x.

var x = {}; 
var z = x;

alert( z === x );    // true

When you do x = { name: "Maizere" }; you're assigning a new object to x. z is still pointing to the original object.

x = { name: "Maizere" };
alert( z !== x );    // true

In the latter example you're not creating a new object but changing the original object's property.

var x = {}; 
var z = x;

x.name = "maizere";
alert( z === x );    // true

A further example of where the confusion might stem from: the bracket syntax creates a new object instead of modifying the original.

var x = { name: "Maizere" };
var y = { name: "Zaimere" };

x = { age: 20 };
y.age = 30;

console.log( x );  // {age: 20}                  <-- original object is replaced
console.log( y );  // {name: "Zaimere", age: 30} <-- original object is modified

answered Jan 31, 2013 at 17:51

JJJ

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2 Comments

Maizere Pathak Over a year ago

according to u if x={} and z={} x==z //true so if x.name="maizere" will z have the property "name"

JJJ Over a year ago

No. The objects are equal, but not the same object. If you change one of them they're not equal anymore. (Just like if you have x=1; z=1; x==z // true but if you change z=2 then x != z.)

Ja͢ck · Accepted Answer · 2013-02-01 00:30:31Z

3

After these two statements:

x={};
y=x;

The internal representation is like this:

      +---- x
      |
{} <--+
      |
      +---- y

So any changes to x are reflected in y:

 x.name="maizere";
 alert(y.name)//maizere

Updated:

                     +---- x
                     |
{name: 'maizere'} <--+
                     |
                     +---- y

This goes away once you assign either variable to something else:

x = { name: "Maizere" }

Representation:

{name: 'Maizere'} <------- x

{name: 'maizere'} <------- y

edited Feb 1, 2013 at 0:30

answered Jan 31, 2013 at 17:53

Ja͢ck

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3 Comments

Maizere Pathak Over a year ago

but what if i do x = { name: "Maizere" }; after z value is set

Barmar Over a year ago

That creates a new object and changes x to point to it, it has no effect on z.

Ja͢ck Over a year ago

@MaizerePathak I've updated the answer with what I think you meant with the comment, let me know if that makes sense.

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