JavaScript pushing an object in an array is what - by reference or value?

Question

When I create an object and push it to an array, is it stored by reference or value?

I see the following happening:

var abc = { a: 10, b: 20};
var def = [];
def.push(abc);

abc.a = 100;

def[0].a; // outputs 100!

// if I do this
abc = { a: 10000, b: 20000 };

def[0].a; // still 100, no change this time

Image from console:

If I use the = sign to assign an object to abc , the reference pointed by abc in the array def should also change, isn't it?. What do we call above, by value or by reference?

I have understood it like abc is a reference pointing to a value. As long as we don't use = sign, it will keep pointing to that. Please guide.

Answer 1

Objects are ALWAYS passed by reference.

When you write abc = { a: 10000, b: 20000 } , what you are overwriting is the variable abc , which was pointing to the old object, but is now pointing to the new one.

Answer 2

Let's have a look at what is happening:

var abc = { a: 10, b: 20};

A new object is created in memory and assigned to the variable abc .

var def = [];

A new array is created in memory and assigned to the variable def .

def.push(abc);

Inside the array there is now a pointer to the formerly created object.

 abc.a = 100;

 def[0].a; // outputs 100!

Obviously right. We are modifying the object, which is also referenced by the array.

 abc = { a: 10000, b: 20000 };

Again a new object is created and a reference to it is stored in abc . Now we have two objects (and an array) in memory.

 def[0].a; // still 100, no change this time

Of course, this is still 100 . The array pointer still references the first created object and not the second one.

Answer 3

as you said create an object so then you are dealing with references.

in your second statement abc = { a: 10000, b: 20000 }; what you actually did is just make variable abc pints into new object make it reference to something else than the old abc

Answer 4

The reason of the problem is when you push the object to array, it clones it self. I mean in the array there is a clone of the object. Therefore if you change the properties of the object, there is nothing will change. Because you have had 2 object already. If you want to overwrite it you have use def[0] = abc after definition of abc with new values. Except for the fact that you can not change the values without asigning it