x = 1;
alert(x);
var y = function() {
alert(x);
var x = 2;
alert(x);
}
y();
The result of the 3 alerts is: 1
, undefined
, 2
(Chrome 25)
My question is: why the second alert is undefined? Why not 1? Isn't there a global variable x?
Due to hoisting , this is what gets executed:
x = 1;
alert(x);
var y = function() {
var x; // <-- this gets hoisted up from where it was.
alert(x);
x = 2;
alert(x);
}
y();
At the start of function y()
, the local variable x
is declared but not initialized.
The variable declaration in the function is hoisted to the top. So it technically looks like this:
var y = function() {
var x;
alert(x);
x = 2;
};
The local variable overshadows the global one. That is why the alert returns undefined
.
Since scope in JavaScript is a function object. When you execute some code in a function(your code sample), "alert(x)" will find if there's any definition of "x" in the function. So, there's a "var x = 2" in this function. But the JavaScript runtime will explain your code like this:
x = 1;
alert(x);
var y = function() {
var x;
alert(x);
x = 2;
alert(x);
}
y();
So, the x in the second alert is "undefined" not "1". So when you declare some variable in a function, I recommend you to declare the variables in the top of your function.
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