Question

Why the following code does not increase the variable a for 1 ?

var a =5;

function abc(y){
    y++;
}

abc(a);

//a is 5 not 6 why?

but this does

var a = 5;

function abc(){
a++;
}

abc();

//a is 6

Answer 1

Because primitive values are passed by value in JavaScript.

To get the value to be updated, you could put a on an object and take advantage of the fact that objects are passed by reference (well, mostly, really a copy of the reference is passed, but we won’t worry about that):

var obj = { a: 5 };

function  abc(o){
   o.a++;
} 

abc(obj);