Javascript redeclared global variable overrides old value

Answer 1

When you redeclared y with var y;, it's now undefined, so if(!undefined) evaluates to true.

Add another alert in your example to see this:

//create global variable with square bracket notation
window['y'] = 'old';

//redeclaration of the same variable
var y;
alert(y); //undefined

if (!y) y = 'new';

alert(y); // new

var won't initialize a variable twice, but it will overwrite one not initialized the first time (because it's a new, more local variable), which the window['y'] style does, adding it to the window object. Take this for example:

//create global variable with square bracket notation
window['y'] = 'old';

//redeclaration of the same variable
var y;
alert(y); //undefined

alert(window.y); //old

if (!y) y = 'new';

alert(y); //shows New instead of Old
alert(window.y);​ //still old

Answer 2

You cannot "redeclare" variables like that within the same scope in JS.

var x = "foo"
function a()
{
  alert(x); // undefined
  var x;
}

In function a, the variable x is local because it has var x. It doesn't matter if it comes before or after the usage.

Likewise:

function b()
{
  var z = 1;
  if (true)
  {
    var z = 2;
  }
  alert(z); // 2
}

because there is no such thing as "block" scope either.

Answer 3

The var statement is subject of hoisting, this means that when the code enters in execution context (just before the actual runtime), the var and function statements are made available to its enclosing scope.

Your code actually gets evaluated as this:

First example:

var y;
window['y'] = 'old';

if (!y) y = 'new';

alert(y);

Second example:

var y;
y = 'old';

if (!y) y = 'new';

alert(y);

With the var statement hoisted, you see actual behavior that the code is having.

See also:

• Variable Instantiation

Answer 4

? I just tested your code and it shows "old", and I've tested FF, Chrome, Safari (PC), and IE8.

Look here: http://jsbin.com/ifare/edit