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    <p>It's interesting in Javascript. Consider this example:</p> <p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false"> <div class="snippet-code"> <pre class="snippet-code-js lang-js prettyprint-override"><code>function changeStuff(a, b, c) { a = a * 10; b.item = "changed"; c = {item: "changed"}; } var num = 10; var obj1 = {item: "unchanged"}; var obj2 = {item: "unchanged"}; changeStuff(num, obj1, obj2); console.log(num); console.log(obj1.item); console.log(obj2.item);</code></pre> </div> </div> </p> <p>This produces the output:</p> <pre><code>10 changed unchanged </code></pre> <ul> <li>If it was pure pass by value, then changing <code>obj1.item</code> would have no effect on the <code>obj1</code> outside of the function.</li> <li>If it was pure pass by reference, then everything would have changed. <code>num</code> would be <code>100</code>, and <code>obj2.item</code> would read <code>"changed"</code>.</li> </ul> <p>Instead, the situation is that the item passed in is passed by value. But the item that is passed by value is <em>itself</em> a reference. Technically, this is called <a href="http://en.wikipedia.org/wiki/Evaluation_strategy#Call_by_sharing" rel="noreferrer">call-by-sharing</a>.</p> <p>In practical terms, this means that if you change the parameter itself (as with <code>num</code> and <code>obj2</code>), that won't affect the item that was fed into the parameter. But if you change the <strong>INTERNALS</strong> of the parameter, that will propagate back up (as with <code>obj1</code>).</p>
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    1. POIs JavaScript a pass-by-reference or pass-by-value language?
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    1. USdeworde
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    1. COThis is exactly same (or at least semantically) as C#. Object has two type: Value (primitive types) and Reference.
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      1. USPeter Lee
    2. COI think this is also used in Java: reference-by-value.
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    3. COthe real reason is that within changeStuff, num, obj1, and obj2 are references. When you change the `item` property of the object referenced by obj1, you are changing the value of the item property that was originally set to "unchanged". When you assign obj2 a value of {item: "changed"} you are changing the reference to a new object (which immediately goes out of scope when the function exits). It becomes more apparent what's happening if you name the function params things like numf, obj1f, and obj2f. Then you see that the params were hiding the external var names.
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