CODEWITHSUNDEEP
javascriptfunctionparameter-passing
Ashish Negi
Ashish Negi

Reputation: 5301

Understanding functions in javascript

I played with javascript and this is it :

> var obj = new Object();

> obj
{}
> obj.x = 0;
0
> function change_x(o) { o.x = o.x + 1; }

> change_x(obj);

> obj
{ x: 1 }
> function change_obj(o) { o = null; }

> change_obj(obj);

> obj
{ x: 1 }

function change_obj_x(o) { console.log(o); o.x = o.x + 1; o = null; console.log(o); }

> change_x(obj)

> change_obj_x(obj);
{ x: 2 }
null

> obj
{ x: 3 }

When i passed obj to change_x, it made changes into the obj itself, but when i tried to make obj null by passing it to change_obj, it did not changed the obj. Nor change_obj_x did what i expected.

Please explain this and give me some links to know everything about functions.

Upvotes: 1

Views: 120

Answers (2)

lastr2d2
lastr2d2

Reputation: 3968

See Functions

If you pass an object (i.e. a non-primitive value, such as Array or a user-defined object) as a parameter, and the function changes the object's properties, that change is visible outside the function

Note that assigning a new object to the parameter will not have any effect outside the function, because this is changing the value of the parameter rather than the value of one of the object's properties

There is a great explanation:

Imagine your house is white and you give someone a copy of your address and say, "paint the house at this address pink." You will come home to a pink house.

That's what you did in 

> function change_x(o) { o.x = o.x + 1; }
> change_x(obj);

And
Imagine you give someone a copy of your address and you tell them, "Change this to 1400 Pennsylvania Ave, Washington, DC." Will you now reside in the White House? No. Changing a copy of your address in no way changes your residence.

That's what 

> function change_obj(o) { o = null; }
> change_obj(obj);

do.

Upvotes: 0

sebastian_oe
sebastian_oe

Reputation: 7320

When you assign something to o in a function like in 

function change_obj(o) { o = null; }

you don't change the parameter but just assign null to the variable. As the o variable doesn't exists outside the function, nothing happens.

In contrast, 

function change_x(o) { o.x = o.x + 1; }

changes the parameter itself. As the parameter is passed by reference, the value of the x property is also changed outside the function.

In your function function change_obj_x(o), you combine these two effects. At first, you change the x property of o (which references to your obj), and then you assign null to o. The latter doesn't influence obj.

Upvotes: 3

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