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    <p>Ok, I'm late on this and an answer is already accepted, but I wonder why nobody has used a clean and simple counter yet?</p> <pre><code>public class Counter { static Counter[] vtab = new Counter[] { new Counter(), new Counter() { public void print( int first, int last ) {} } }; public void print( int first, int last ) { vtab[ ( last - first - 1 ) &gt;&gt;&gt; 31 ].print( first, last - 1 ); System.out.println( last ); } public static void main( String[] args ) { vtab[ 0 ].print( 1, 100 ); } } </code></pre> <p>Thread safe, configurable, no exceptions, no dependance on API side effects, just plain OOP and some trivial math.</p> <hr> <p>For those not familiar with binary operators here is how it works:</p> <ul> <li><p>The <code>( x &gt;&gt;&gt; n )</code> expression moves all bits of the integer value <code>x</code> to the right by <code>n</code> places. Lower bits simply fall off the right side by this operation and new bits that come in from the left side are always <code>0</code>.</p></li> <li><p>So the effect of <code>( x &gt;&gt;&gt; 31 )</code> is to move the highest bit of <code>x</code> to the lowest place and to set all other bits of <code>x</code> to <code>0</code>. The result is now always either <code>0</code> or <code>1</code> for all possible values of <code>x</code>.</p></li> <li><p>As the highest bit of an <code>int</code> is the sign bit which is <code>0</code> for positive values and <code>1</code> for negative values, the expression <code>( x &gt;&gt;&gt; 31 )</code> evaluates to <code>0</code> for all positve values of <code>x</code> and to <code>1</code> for all negative values of <code>x</code>.</p></li> <li><p>Now if both <code>first</code> and <code>last</code> are positive numbers and if <code>last</code> is greater than <code>first</code>, the result of <code>( last - first - 1 )</code> will be <code>&gt;= 0</code> and if <code>last == first</code> it will be <code>-1</code>.</p></li> <li><p>So <code>( ( last - first - 1 ) &gt;&gt;&gt; 31 )</code> evaluates to <code>0</code> if <code>last</code> is greater than <code>first</code> and becomes <code>1</code> if they are equal.</p></li> </ul> <p>Now this value <code>0/1</code> is used to switch between the 2 implementations of <code>print( int first, int last )</code> based on the comparision of <code>first</code> and <code>last</code>. At first the recursion takes place without printing anything. <code>print( 1, 100 )</code> calls <code>print( 1, 99 )</code> and so on... until <code>last</code> equals <code>first</code> which causes a switch to the other implementation of <code>print</code> which in turn does nothing at all. So now the stack unwinds again and the values are printed on the way down in ascending order and the invocation of <code>vtab[ 0 ].print( 1, 100 )</code> finishes normally.</p>
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