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    <p>I think there is a slight misunderstanding from the background text:</p> <blockquote> <p>In Java, when we pass a java object argument to method myMethod((MyObject) myObject), it is basically like passing a reference to object in C++ like myFunction(&amp; myObject).</p> </blockquote> <p>Java is <strong>not</strong> a "pass-by-reference" language. It always passes by value. When you pass an object, you are passing a reference to that object <em>by value</em>. This is analogous in C++ when you pass a pointer by value. </p> <p>For example, in Java:</p> <pre><code>public void myMethod(MyObject myObject) { myObject = null; } </code></pre> <p>The object passed as "MyObject" is only a reference by value. So setting it to null does not affect the original object that was passed. Correspondingly:</p> <pre><code>public void myMethod(MyObject myObject) { myObject.myIntProperty = 15; } </code></pre> <p>Accessing members of the object, as above, does so to the actual object, since it is still a reference to that object. If it helps, think of the passed in <code>myObject</code> as having a copy of the memory location that is the same as the memory location of the original object that was passed:</p> <pre><code>public void myMethod(MyObject myObject) {} /* Example memory address: 0x009FC12A in memory, a *copy* of the same address of the object that was passed */ </code></pre> <hr> <p>With this mindset, let's take a look at your question. What does this yield:</p> <pre><code>myObjectsFour[0] = myObjectsOne[0]; </code></pre> <p>Well, as Java assigns the reference of the element <em>by value</em>, <code>myObjectsFour[0]</code> now points to precisely the same object. This means:</p> <pre><code>myObjectsFour[0] = myObjectsOne[0]; myObjectsFour[0].myIntProperty = 15; </code></pre> <p>Also changes the value of <code>myObjectsOne[0].myIntProperty</code> to be 15, just as above. However doing this:</p> <pre><code>myObjectsFour[0] = myObjectsOne[0]; myObjectsFour[0] = null; // (or anything else, for that matter) </code></pre> <p>has no affect on <code>myObjectsOne[0]</code>. It's not the <em>same</em> variable by the statement, <code>myObjectsFour[0] = myObjectsOne[0];</code>. It's simply two variables that have a copy of the same reference. </p> <p>Does that make sense?</p>
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    1. COOfcourse it does, you wrote a decent answer. Thinking of passed object as of "a copy" of the reference to its memory address and not direct reference is vivid and makes very good sense why `=null` or `=new Obj()` on passed object won't affect the original one. Pass-by-value it is, but it would be way more clear if it was called pass-by-reference-value (for non-primitives).
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