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Home/ Questions/Q 448863
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T21:42:43+00:00 2026-05-12T21:42:43+00:00

I’m hoping someone can explain to me why the below JavaScript/HTML will show door

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I’m hoping someone can explain to me why the below JavaScript/HTML will show “door #2” when the HTML is viewed in a browser:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"  "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
    <script type="text/javascript">
        function testprint() {
            alert('door #1');
        };

        window.onload = testprint;

        function testprint() {
            alert('door #2');
        };

        testprint = function() {
            alert('door #3');
        };
    </script>
    <script type="text/javascript">
        function testprint() {
            alert('door #4');
        };
    </script>
</head>
<body>
</body>
</html>

Since only the declaration testprint occurs before window.onload is set to testprint, I would expect window.onload cause ‘door #1’ to show up. Actually, onload causes ‘door #2’. Note that it will do this whether the first declaration of testprint is included or not.

The third and fourth declaration of testprint use different means of assigning the function, I tried this to see if it would override window.onload‘s behavior in the same was the second declaration of testprint does. It did not. Note that if I move the fourth declaration of testprint to the end of the first script block it would be called by window.onload.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T21:42:44+00:00Added an answer on May 12, 2026 at 9:42 pm

    Function declarations are subject of hoisting, and they are evaluated at parse time, by hoisting means that they are available to the entire scope in where they were declared, for example:

    foo(); // alerts foo
    foo = function () { alert('bar')};
    function foo () { alert('foo');}
    foo(); // alerts bar
    

    The first call to foo will execute the function declaration, because at parse time it was made available, the second call of foo will execute the function expression, declared at run-time.

    For a more detailed discussion about the differences between function expressions and function declarations, check this question and this article.

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