JavaScript code executing after return

Question

In the following example, JavaScript seems to be completely ignoring my return statement, and just carrying on executing code.

var x = 1;
(function() {
  x = 2;
  return;
  var x = 3;
})();
console.log(x); // Outputs 1 in both Chrome and FF

Surely the code should output 2? If I remove the var keyword from var x = 3, it outputs 2 as expected. Is there some strange compiler optimization at work here?

I think the x is different inside the function.. easy fix would be to do window.x = 2 — Madushan
– Madushan, Commented Jul 29, 2013 at 10:16
As @dystroy said variable declarations get hoisted to the top of the function scope. So because you have the var x = 3, var x get's hoisted to the top and makes the x = 2 a local variable. If you just put x = 3 (without var) then the console.log will output 2 — Pete D
– Pete D, Commented Jul 29, 2013 at 10:17

Denys Séguret · Accepted Answer · 2013-07-29 10:10:23Z

8

No, the code shouldn't output 2 because variable declarations are hoisted so your code is equivalent to

var x = 1;
(function() {
  var x;
  x = 2; // changes the internal x variable
  return;
  x = 3; // does nothing because it's not reached
})();
console.log(x); // Outputs the outside x, which is still 1

The line

x = 2;

only changes the internal x variable which shadows the outside one.

The scope of a non global variable is the entire function in which it is declared. From the start of this function to its end.

answered Jul 29, 2013 at 10:10

Denys Séguret

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