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    <p>This is the victim of bad recursive call. As you are wondering why the value of <em>cnt</em> varies, it is because the stack size depends on the platform. Java SE 6 on Windows has a default stack size of 320k in the 32-bit VM and 1024k in the 64-bit VM. You can read more <a href="http://www.oracle.com/technetwork/java/hotspotfaq-138619.html" rel="noreferrer">here</a>.</p> <p>You can run using different stack sizes and you will see different values of <em>cnt</em> before the stack overflows-</p> <blockquote> <p>java -Xss1024k RandomNumberGenerator</p> </blockquote> <p>You don't see the value of <em>cnt</em> being printed multiple times even though the value is greater than 1 sometimes because your print statement is also throwing error which you can debug to be sure through Eclipse or other IDEs.</p> <p>You can change the code to the following to debug per statement execution if you'd prefer-</p> <pre><code>static int cnt = 0; public static void main(String[] args) { try { main(args); } catch (Throwable ignore) { cnt++; try { System.out.println(cnt); } catch (Throwable t) { } } } </code></pre> <p><strong>UPDATE:</strong></p> <p>As this getting a lot more attention, let's have another example to make things clearer-</p> <pre><code>static int cnt = 0; public static void overflow(){ try { overflow(); } catch (Throwable t) { cnt++; } } public static void main(String[] args) { overflow(); System.out.println(cnt); } </code></pre> <p>We created another method named <em>overflow</em> to do a bad recursion and removed the <em>println</em> statement from the catch block so it doesn't start throwing another set of errors while trying to print. This works as expected. You can try putting <em>System.out.println(cnt);</em> statement after <em>cnt++</em> above and compile. Then run multiple times. Depending on your platform, you may get different values of <em>cnt</em>.</p> <p>This is why generally we do not catch errors because mystery in code is not fantasy.</p>
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