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    <p>When you call:</p> <pre><code>x(a); </code></pre> <p>A few things are going on. First, the variable <code>a</code> (which simply holds a reference to an object) is passed <em>by value</em> to the function <code>x</code>. <code>x</code> now has its <em>own copy</em> of that reference, which happens to point to the same object in memory. Thus, any changes you make to properties on that referred object will affect other references to said object.</p> <p>When you call:</p> <pre><code>this.b = v; </code></pre> <p>You're again making a copy of <code>v</code> and setting it to <code>this.b</code>. Now, <code>a</code>, <code>v</code> and <code>this.b</code> are distinct variables in memory, all storing a reference to the same object.</p> <p>It seems what you're <em>attempting</em> to do is create a copy of the object itself, so you can manipulate one reference and not affect others. To do so, you'll need to create a <em>completely new</em> object in memory and copy over the properties.</p> <pre><code>var a = {"x":1} function x(v) { this.b = {}; // Create a new object this.b.x = v.x; // Copy over the x property // Copy over any other properties too this.b.x++; } x(a); alert(a.x); // The original object still has an x property of 1 </code></pre> <p>Since <code>this.b</code> is a new object and we've merely <em>copied over</em> properties from the object referred to by <code>v</code>, incrementing <code>this.b.x</code> will have no effect on whatever <code>v</code> is pointing to.</p>
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    1. POJavascript passing by reference instead of value?
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    1. USMike Christensen
    1. USMike Christensen
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    1. POJavascript passing by reference instead of value?
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    1. COBy reference means assignments are visible to caller, here just mutations are visible to caller. Caller will not see mutations in non-objects because you cannot mutate them anyway. It is always pass by value.
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    2. COBy *reference*, I meant one copy of the object will exist in memory, and a *reference* to that location will be passed around.
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      1. USMike Christensen
    3. CO@mike. In your first example, why is it, that when you pass `a.x` into `x(v)` that you can set `this.b.x = v.x` as opposed to just `this.b.x = v`? Does that question make sense?
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      1. USbobkingof12vs
 

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