Question

I was reading the following topic How do JavaScript closures work? and found this code:

function foo(x) {
  var tmp = 3;
  return function (y) {
    alert(x + y + tmp);
    x.memb = x.memb ? x.memb + 1 : 1;
    alert(x.memb);
  }
}
var age = new Number(2);
var bar = foo(age); // bar is now a closure referencing age.
bar(10);

The author commented:

As expected, each call to bar(10) will increment x.memb. What might
not be expected, is that x is simply referring to the same object as
the age variable! After a couple of calls to bar, age.memb will be 2!

I am confused why it will return 2 always. Can you explain me how it will come 2 always?

Answer 1

It will not “always” be 2.

Author says: “After a couple of calls to bar, age.memb will be 2” – and after 2 calls it will be “2”.

After 4 calls it will be 4.

Just paste this to firebug console:

function foo(x) {
  var tmp = 3;
  return function (y) {
    //console.log(x + y + tmp);
    x.memb = x.memb ? x.memb + 1 : 1;
    console.log(x.memb);
  }
}
var age = new Number(2);
var bar = foo(age); // bar is now a closure referencing age.
bar(10);
bar(10);
bar(10);
bar(10);

console.log('result: ', age.memb)    // result: 4