Question

I was told on another thread that after the function exits, any modifications made to its operands from within the function will persist. I always thought that it made a temporary copy of all the values passed into it, and then the only things that persisted were return values and implicitly more broadly scoped variables that were modified.

I suppose thinking back to all the jquery plugins whose source code I’ve peaked at, they all use the construct:

(function($){
  $.fn.foo = function(){ console.log('foo'); };
})(jQuery);

which imply that modifications to the jQuery object even by the internal-scope $ identifier persist after the function exits, or the jQuery plugins would not work. So, this works as the above snippet does:

var x = {n:0};

(function addOneTo(p) {
  p.n = p.n + 1;
})(x);

console.log(x);

But this:

var x = 0;

(function addOneTo(p) {
  p = p + 1;
})(x);

console.log(x);

doesn’t, leaving x unmodified with a value of 0.

Could someone just explain how argument passing works? I thought I knew how it did, but I guess I don’t. Thanks

Answer 1

Actually in your first example

var x = {n:0};
(function addOneTo(p) {
    p.n = p.n + 1;
})(x);
console.log(x);

The x is an object and when you used p.n = p.n + 1; it modified the original object’s property because it passed as a reference (p is the same object as x) but in your second example

var x = 0;
(function addOneTo(p) {
    p = p + 1;
})(x);
console.log(x);

x is passed by a value in to the function and p = p + 1; here p is a new private variable inside that function and it has only the function scope.