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copied!<blockquote> <p>As I understand it, it's because a and b are two different references to the same object (apple2e).</p> </blockquote> <p>Because of string interning, and <strong>only</strong> because of string interning <code>a</code> and <code>b</code> are different references to the same <code>String</code> object.</p> <hr> <p>Unfortunately, your code does not do what you think it does. Try this:</p> <pre><code>String a = "apple2e"; String b = "apple2e"; System.out.println("a==b? " + a == b); // "false" System.out.println("a==b? " + (a == b)); // "a==b? true" </code></pre> <p>Java automatically interns all string literals. <strong>That</strong> is why the second sysout prints what it does. The first sysout prints only <code>"false"</code> because string concatenation (<code>+</code>) has higher precedence than <code>==</code>, so it's equivalent to this:</p> <pre><code>System.out.println("a==b? apple2e" == "apple2e"); </code></pre> <p>I don't think that's what you meant to test!</p> <p>This, on the other hand, will give you two separate <code>String</code> instances:</p> <pre><code>String a = new String("apple2e"); String b = new String("apple2e"); System.out.println("a==b? " + (a == b)); // "a==b? false" </code></pre> <p>Which would schematically look like</p> <pre><code>a (reference_id 123) --------------- "apple2e" b (reference_id 456) --------------- "apple2e" </code></pre> <p>and can be reduced to the original situation using <code>String#intern()</code>:</p> <pre><code>String a = new String("apple2e").intern(); String b = new String("apple2e").intern(); System.out.println("a==b? " + (a == b)); // "a==b? true" </code></pre> <p>e.g.</p> <pre><code>a (reference_id 123) ------+ +--------- "apple2e" b (reference_id 456) ------+ </code></pre>

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