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    <p>Java always passes arguments by value NOT by reference.</p> <hr> <p>Let me explain this through an <a href="https://stackoverflow.com/a/8257486/597657">example</a>:</p> <pre><code>public class Main { public static void main(String[] args) { Foo f = new Foo("f"); changeReference(f); // It won't change the reference! modifyReference(f); // It will modify the object that the reference variable "f" refers to! } public static void changeReference(Foo a) { Foo b = new Foo("b"); a = b; } public static void modifyReference(Foo c) { c.setAttribute("c"); } } </code></pre> <p>I will explain this in steps:</p> <ol> <li><p>Declaring a reference named <code>f</code> of type <code>Foo</code> and assign it to a new object of type <code>Foo</code> with an attribute <code>"f"</code>.</p> <pre><code>Foo f = new Foo("f"); </code></pre> <p><img src="https://i.stack.imgur.com/arXpP.png" alt="enter image description here"></p></li> <li><p>From the method side, a reference of type <code>Foo</code> with a name <code>a</code> is declared and it's initially assigned to <code>null</code>.</p> <pre><code>public static void changeReference(Foo a) </code></pre> <p><img src="https://i.stack.imgur.com/k2LBD.png" alt="enter image description here"></p></li> <li><p>As you call the method <code>changeReference</code>, the reference <code>a</code> will be assigned to the object which is passed as an argument.</p> <pre><code>changeReference(f); </code></pre> <p><img src="https://i.stack.imgur.com/1Ez74.png" alt="enter image description here"></p></li> <li><p>Declaring a reference named <code>b</code> of type <code>Foo</code> and assign it to a new object of type <code>Foo</code> with an attribute <code>"b"</code>.</p> <pre><code>Foo b = new Foo("b"); </code></pre> <p><img src="https://i.stack.imgur.com/Krx4N.png" alt="enter image description here"></p></li> <li><p><code>a = b</code> is re-assigning the reference <code>a</code> NOT <code>f</code> to the object whose its attribute is <code>"b"</code>.</p> <p><img src="https://i.stack.imgur.com/rCluu.png" alt="enter image description here"></p> <hr></li> <li><p>As you call <code>modifyReference(Foo c)</code> method, a reference <code>c</code> is created and assigned to the object with attribute <code>"f"</code>.</p> <p><img src="https://i.stack.imgur.com/PRZPg.png" alt="enter image description here"></p></li> <li><p><code>c.setAttribute("c");</code> will change the attribute of the object that reference <code>c</code> points to it, and it's same object that reference <code>f</code> points to it.</p> <p><img src="https://i.stack.imgur.com/H9Qsf.png" alt="enter image description here"></p></li> </ol> <p>I hope you understand now how passing objects as arguments works in Java :)</p>
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    1. POJava pass by reference
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    1. COAWESOME explanation! I feel enlightened!
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    2. COHoly cow. This has been asked and answered many times, but this is the most elaborate and artistic solution I've seen. Bravo!
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    3. COCan you contrast what "pass by reference" would look like? What makes sense to me is with "pass by value", it is passing the memory address of the object referred to by f. In **changeReference**, **a** is a new variable that refers to the same memory address, changing **a**'s value (or referred to memory address) only changes what **a** points to and not **f**. If it were "pass by reference" **f** would get passed in so **a = f**, changing **a**'s value (or referenced memory address) would change **f**
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