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copied!<p>If the <code>return</code> in the <code>try</code> block is reached, it transfers control to the <code>finally</code> block, and the function eventually returns normally (not a throw).</p> <p>If an exception occurs, but then the code reaches a <code>return</code> from the <code>catch</code> block, control is transferred to the <code>finally</code> block and the function eventually returns normally (not a throw).</p> <p>In your example, you have a <code>return</code> in the <code>finally</code>, and so regardless of what happens, the function will return <code>34</code>, because <code>finally</code> has the final (if you will) word.</p> <p>Although not covered in your example, this would be true even if you didn't have the <code>catch</code> and if an exception were thrown in the <code>try</code> block and <em>not</em> caught. By doing a <code>return</code> from the <code>finally</code> block, you suppress the exception entirely. Consider:</p> <pre><code>public class FinallyReturn { public static final void main(String[] args) { System.out.println(foo(args)); } private static int foo(String[] args) { try { int n = Integer.parseInt(args[0]); return n; } finally { return 42; } } } </code></pre> <p>If you run that without supplying any arguments:</p> <pre>$ java FinallyReturn</pre> <p>...the code in <code>foo</code> throws an <code>ArrayIndexOutOfBoundsException</code>. But because the <code>finally</code> block does a <code>return</code>, that exception gets suppressed.</p> <p>This is one reason why it's best to avoid using <code>return</code> in <code>finally</code>.</p>

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