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POcast from double to int is not always just dropping decimal part
 

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    <p>I'm experimenting with the code I found here <a href="http://www.javaspecialists.eu/archive/Issue197.html" rel="nofollow noreferrer">The Java Specialists' Newsletter</a>.</p> <pre><code>public class MeaningOfLife { public static String findOutWhatLifeIsAllAbout() { int meaning = 0; for (int i = 0; i &lt; 10; i++) { for (int j = 0; j &lt; 20; j++) { for (int k = 0; k &lt; 300; k++) { for (int m = 0; m &lt; 7000; m++) { meaning += Math.random() + 1; } } } } return String.valueOf(meaning).replaceAll("0*$", ""); } public static void main(String[] args) { System.out.println(findOutWhatLifeIsAllAbout()); } } </code></pre> <p>The answer to the question "what does it print" seemed obvious once I realized that there is an implicit cast with the <a href="http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.26.2" rel="nofollow noreferrer">compound assignment operator</a> +=.</p> <p>But it printed something like: 420000006 or 420000007, instead of (the expected) 420000000 (or "42", after removing trailing zeros).</p> <p>So that was showing that casting from double to int is not always just dropping the decimal part of the double as stated here: <a href="https://stackoverflow.com/questions/2143476/how-to-cast-a-double-to-an-int-in-java">How to cast a double to an int in Java?</a></p> <p>So I made some trials and here is an example of what I found:</p> <pre><code>System.out.println((int) (131070.99999999999)); // -&gt; 131070 System.out.println((int) (131071.99999999999)); // -&gt; 131071 System.out.println((int) (131072.99999999999)); // -&gt; 131073 !!! System.out.println((int) (131073.99999999999)); // -&gt; 131074 !!! System.out.println((int) (16382.999999999999)); // -&gt; 16382 System.out.println((int) (16383.999999999999)); // -&gt; 16383 System.out.println((int) (16384.999999999999)); // -&gt; 16385 !!! System.out.println((int) (16385.999999999999)); // -&gt; 16386 !!! </code></pre> <p>...So now I'm looking for an explanation for that behavior ???</p>
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    1. COThe IEEE standard sometimes produces unexpected results. Write these numbers in binary as closely as possible, keeping in mind the bit-width restrictions, consider how IEEE states they should be rounded, and you'll see why it happens.
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    2. COThanks for all the good answers! Unfortunately I can only accept one...
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