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javascript - Why does shadowed variable evaluate to undefined when defined in outside scope?

Consider the following piece of code:

<html><head></head>
<body>
    <script type="text/javascript">
        var outside_scope = "outside scope";
        function f1() {
            alert(outside_scope) ;
        }
        f1();
    </script>
</body>
</html> 

The output for this code is that the alert box displays the message "outside scope". But, if I slightly modify the code as:

<html><head></head>
<body>
    <script type="text/javascript">
        var outside_scope = "outside scope";
        function f1() {
            alert(outside_scope) ;
            var outside_scope = "inside scope";
        }
        f1();
    </script>
</body>
</html> 

the alert box displays the message "undefined". I could have understood the logic if it displays "undefined" in both the cases. But, that is not happening. It displays "undefined" only in the second case. Why is this?

Thanks in advance for your help!

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1 Answer

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Variables are subject to hoisting. This means that regardless of where a variable is placed within a function, it is moved to the top of the scope in which it is defined.

For example:

var outside_scope = "outside scope";
function f1() {
    alert(outside_scope) ;
    var outside_scope = "inside scope";
}
f1();

Gets interpreted into:

var outside_scope = "outside scope";
function f1() {
    var outside_scope; // is undefined
    alert(outside_scope) ;
    outside_scope = "inside scope";
}
f1();

Because of that, and the function only scope that JavaScript has, is recommended to declare all the variables at the top of the function, to resemble what will happen.


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