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POGetting content of a collection in a thread-safe manner
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  1. POGetting content of a collection in a thread-safe manner
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    copied!<p>I would like to <strong>read</strong> the content of a java Collection in a multi-threaded way. There have been a lot of questions here with the same context, but none on the specific read point.</p> <p>I have a collection of integer. I just want several threads to iterate over it, each thread pulling one integer at a time. I want to make sure all the collection is iterated, and that no integer is pulled twice by two different threads. </p> <p>Frankly, I do not know what works. I know that Iterators are not thread-safe, but when it comes to read only I do not know. I did some test to try and get thread faults, but did not reach 100% certainty:</p> <pre class="lang-java prettyprint-override"><code>int imax = 500; Collection&lt;Integer&gt; li = new ArrayList&lt;Integer&gt;(imax); for (int i = 0; i &lt; imax; i++) { li.add(i); } final Iterator&lt;Integer&gt; it = li.iterator(); Thread[] threads = new Thread[20]; for (int i = 0; i &lt; threads.length; i++) { threads[i] = new Thread("Thread " + i) { @Override public void run() { while(it.hasNext()) { System.out.println(it.next()); } } }; } for (int ithread = 0; ithread &lt; threads.length; ++ithread) { threads[ithread].setPriority(Thread.NORM_PRIORITY); threads[ithread].start(); } try { for (int ithread = 0; ithread &lt; threads.length; ++ithread) threads[ithread].join(); } catch (InterruptedException ie) { throw new RuntimeException(ie); } </code></pre> <p>EDIT: In the real use-case, each of this integer is used to start an intensive work, such as finding whether it is prime. </p> <p>The above example pulls the list of integer without duplicates or misses, but I do not know whether it is by chance.</p> <p>Using a HashSet instead of an ArrayList works as well, but again, it might be by chance.</p> <p>How do you do in practice if you have a general collection (not necessarily a list) and need to pull its content in multithreaded fashion?</p>
 

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