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    <p>It's testing <code>$target</code> (which I assume is a jQuery object) for elements and if empty, assigning a default value.</p> <p>See <a href="https://developer.mozilla.org/en/JavaScript/Reference/Operators/Logical_Operators" rel="nofollow">Logical Operators</a> for a detailed explanation.</p> <h2>Update</h2> <p>To explain (in case you don't feel like reading the MDN docs), JavaScript logical comparison operators work left-to-right.</p> <pre><code>expr1 &amp;&amp; expr2 </code></pre> <p>This will return <code>expr1</code> if it can be converted to <code>false</code>; otherwise, returns <code>expr2</code></p> <pre><code>expr1 || expr2 </code></pre> <p>This will return <code>expr1</code> if it can be converted to <code>true</code>; otherwise, returns <code>expr2</code></p> <p>To break down the line in question, think of it this way</p> <pre><code>$target = ($target.length &amp;&amp; $target) || $('[name=' + this.hash.slice(1) +']'); </code></pre> <p>Looking at the first operation...</p> <pre><code>$target.length &amp;&amp; $target </code></pre> <p>This will return <code>$target.length</code> if <code>$target.length</code> can be converted to <code>false</code> (ie <code>0</code>), otherwise it will return <code>$target</code>.</p> <p>The second operation looks like this...</p> <pre><code>(operation1) || $('[name=' + this.hash.slice(1) +']') </code></pre> <p>If the result of the <code>operation1</code> can be converted to <code>true</code> (ie <code>$target</code>), then it will be returned, otherwise <code>$('[name=' + this.hash.slice(1) +']')</code>.</p>
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    1. POCan anyone explain what is happening in this bit of Javascript?
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    1. USPhil
    1. USPhil
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    1. COI've seen this syntax a bit too. Why doesn't this statement evaluate to a boolean?
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      1. USJeff
    2. COPerfect, thanks
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      1. USJeff
    3. COActually it should be -- if ($target.length && $target) {$target = $target;} else {$target = $('[name=' + this.hash.slice(1) +']');} . I cannot think of a situation where $target.length is true and $target is false but still it is the precise translation.
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      1. USNarendra Yadala
 

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