JavaScript private variables with constructor prototype

Question

Given the following code:

var Test = function() {
  var self = this;
  var innerPrivate = 1;
  self.innerPublic = 1;
  self.innerShow = function() {
    console.log(innerPrivate++ + ':' + this.innerPublic++);
  };
  this.constructor.prototype.protoShow = function() {
    console.log(innerPrivate++ + ':' + this.innerPublic++);
  };
};

var t1 = new Test();
var t2 = new Test();

t2.innerShow(); //1:1
t1.innerShow(); //1:1
t2.innerShow(); //2:2
t1.innerShow(); //2:2

t2.protoShow(); //3:3 (expected: 1:3 or 3:3)
t1.protoShow(); //4:3 (expected: 2:3 or 3:3)
t2.protoShow(); //5:4 (expected: 3:4 or 4:4)
t1.protoShow(); //6:4 (expected: 4:4)

Why is the inner variable both re-used and shared?

Even if the instance is linked to the constructor, the outcome still doesn't seem right.

What am I missing?

Rajaprabhu Aravindasamy · Accepted Answer · 2016-07-16 17:16:07Z

1

Basically prototype is a live connection to the class(function). Whenever you are creating a new instances from the class, those instances will share the prototype properties. The variable innerPrivate is a closure to both the function inside, but it will be updated with the last instance's variable.

.
.
var t1 = new Test(); 
var t2 = new Test(); 
// after execution of the above line
//closure innerPrivate inside of prototype function is replaced with t2's innerPrivate. 

//So updating t1's innerPrivate via innerShow of it(t1)
//doesn't affect the closure inside of protoShow

It is always better to avoid altering the prototype inside a constructor. Because doing so will cause unnecessary confusions.

answered Jul 16, 2016 at 17:16

Rajaprabhu Aravindasamy

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