返回后执行JavaScript代码

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?

Answer 1

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.