js private variable assigning

Question

I'm missing something in understanding how js works. here's the problem:

We declare a module like this:

ns.obj = function() { // declare private variables var test = 1, test1 = 2; // declare some private function var myFunc=function(){test=2}; return{test:test, myFunc:myFunc}; }

Every time myFunc is called, as we are not declaring test inside the function js should assume we are referring to the private variable.

The returned object makes sure we have test and myFunc visible if we have the module. So calling ns.obj.test should give us 1 at first. And after we call myFunc should give us 2. But it`s always 1. Why that happens?

here's the jsfiddle: http://jsfiddle.net/aXuwB/1/

I think your jsfiddle link is wrong

AlvaroGMJ
– AlvaroGMJ

2012-08-23 20:12:27 +00:00
Commented Aug 23, 2012 at 20:12 — AlvaroGMJ
– AlvaroGMJ, Commented Aug 23, 2012 at 20:12
Yes, you are right, updated it.

Yaroslav Yakovlev
– Yaroslav Yakovlev

2012-08-24 04:42:27 +00:00
Commented Aug 24, 2012 at 4:42 — Yaroslav Yakovlev
– Yaroslav Yakovlev, Commented Aug 24, 2012 at 4:42

pimvdb · Accepted Answer · 2012-08-25 16:14:40Z

1

In JavaScript overwriting a variable does not overwrite it anywhere else. You're effectively passing the number 1 in your return object; there is no reference to the test variable.

An option you have is returning a function. A function holds code and thus can hold a reference to a variable. Calling it would give you the variable: test: function() { return test; }.

edited Aug 25, 2012 at 16:14

answered Aug 23, 2012 at 20:13

pimvdb

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6 Comments

badunk Over a year ago

the test variable is specified in his function

pimvdb Over a year ago

@badunk: Yes, but he's retuning just the number value. Overwriting test does not overwrite it in the returned object.

Yaroslav Yakovlev Over a year ago

Watch the jsfiddle attached. Absolutely same code for array gives different result. If I call push its ok and I can see the result. If I reinit the array it doesnt show up.

Raymond Chen Over a year ago

Calling myFunc does change the value of the private variable test to 2. But it does not change the value of testObj.test. That was initialized to 1 and is still 1 because nobody changed it. Arrays are reference objects; testArr is not an array. It is a reference to the array, and you put that same reference in testObj.testArr. Therefore, the same array is accessible by two different means, and changes to the array are therefore visible in two places. TL;DR: Read about the difference between reference types and value types.

Yaroslav Yakovlev Over a year ago

thanks, Raymond, read it. But still not clear. When I make push its visible. When I make reassignment its not. Even if its a reference and I make it refer to a new instance, why isnt it visible?

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аlex · Accepted Answer · 2018-02-12 15:27:08Z

var User = function () { var age = 0; // private this.incrementAge = function () { return age++; } return this.incrementAge(); // closure }; var firstUser = new User(); console.log('firstUser', firstUser.incrementAge()); // 1 var secondUser = new User(); console.log('secondUser', secondUser.incrementAge()); // 1 console.log('secondUser', secondUser.incrementAge()); // 2

badunk · Accepted Answer · 2012-08-23 20:43:28Z

I think it has to do with the fact that you are declaring a new var test=1 everytime you execute the function. To clarify, everytime you call ns.obj(), you will be reassigning a new test variable to 1 and then returning it in your object. Thus, the value is always 1.

When you invoke the myFunc function, it will modify your private variable as you wanted, but you don't have a clean way to access the private variable (without resetting the value again).

Take a look at this and see if this demonstrates what you want:

var ns = {}; ns.obj = function(){ var test = 1; this.myFunc = function(){ test=2 }; this.getTest = function(){ return test; } } var testObj = new ns.obj(); console.log(testObj.getTest()); testObj.myFunc() console.log(testObj.getTest());

Its executed once to initialize object literal. After that Im just using it. So it`s not the case.
I am confused, when you say you are calling myFunc, it should return with 2 - but that function doesn't return any value
badunk, check the jsfiddle. It`s a module, function is called one time at start.

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