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?

Answer 1

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