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Home/ Questions/Q 8245093
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T22:04:18+00:00 2026-06-07T22:04:18+00:00

In the following code snippet I declare a global variable and then check for

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In the following code snippet I declare a global variable and then check for its presence inside a function.

<script>
x = 5;
$(function() {
   var x = x || 3;
   console.log(x); // prints 3
});
</script>

This behaves differently:

<script>
x = 5;
$(function() {
   var y = x || 3;
   console.log(y); // prints 5
});
</script>

I expect that in the first example, the variable declaration in the inner scope will detect that x already exists in the global scope, and take its value. Why does the first example 3?

Specifically I recently wrote some code that checks var _gaq = _gaq || [] in a jQuery ready scope and was confused when nothing was getting pubbed to Analytics.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-07T22:04:20+00:00Added an answer on June 7, 2026 at 10:04 pm

    You’re looking for x in the wrong scope. Due to variable hoisting, var x has actually defined a local x variable with a value of undefined before your x || 3 check happens:

    var x = x || 3;
    

    is actually:

    var x = undefined;
    x = x || 3;
    

    Simply change it to look for x on the window object:

    var x = window.x || 3;
    
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