Re: The first 10 files

From:
Robert Klemme <shortcutter@googlemail.com>
Newsgroups:
comp.lang.java.programmer
Date:
Sun, 27 Jan 2013 13:55:18 +0100
Message-ID:
<amkmdqFlumnU1@mid.individual.net>
On 27.01.2013 03:43, Arne Vajh=F8j wrote:

On 1/26/2013 9:35 PM, Arne Vajh=F8j wrote:

On 1/26/2013 9:02 PM, Arved Sandstrom wrote:

If OP happens to be on Java 7, then I will suggest using:

java.nio.file.Files.newDirectoryStream(dir)

It is a straight forward way of getting the first N files.

And it is is as likely as the exception hack to not to read
all filenames from the OS.


import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.Iterator;

public class ListFilesWithLimit {
     public static void main(String[] args) throws IOException {
         Iterator<Path> dir =
Files.newDirectoryStream(Paths.get("/work")).iterator();
         int n = 0;
         while(dir.hasNext() && n < 10) {
             System.out.println(dir.next());
         }
     }
}


For earlier Java versions we could emulate that with a second thread.

package file;

import java.io.File;
import java.io.FileFilter;
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.SynchronousQueue;
import java.util.concurrent.TimeUnit;

public final class ListFileTestThreaded2 {

   private static final class CountFilterThread extends Thread
implements FileFilter {

     private final File dir;
     private final int maxFiles;
     private final BlockingQueue<List<File>> queue;
     private List<File> filesSeen = new ArrayList<File>();

     public CountFilterThread(File dir, int maxFiles,
BlockingQueue<List<File>> queue) {
       this.dir = dir;
       this.maxFiles = maxFiles;
       this.queue = queue;
     }

     @Override
     public void run() {
       try {
         dir.listFiles(this);

         if (filesSeen != null) {
           send();
         }
       } catch (InterruptedException e) {
         e.printStackTrace();
       }
     }

     private void send() throws InterruptedException {
       queue.put(filesSeen);
       filesSeen = null;
     }

     @Override
     public boolean accept(final File f) {
       try {
         if (filesSeen != null) {
           filesSeen.add(f);

           if (filesSeen.size() == maxFiles) {
             send();
             assert filesSeen == null;
           }
         }

         return false;
       } catch (InterruptedException e) {
         throw new IllegalStateException(e);
       }
     }
   }

   private static final int[] LIMITS = { 10, 100, 1000, 10000,
Integer.MAX_VALUE };

   public static void main(String[] args) throws InterruptedException {
     for (final String s : args) {
       System.out.println("Testing: " + s);
       final File dir = new File(s);

       if (dir.isDirectory()) {
         for (final int limit : LIMITS) {
           final SynchronousQueue<List<File>> queue = new
SynchronousQueue<List<File>>();
           final CountFilterThread cf = new CountFilterThread(dir,
limit, queue);
           cf.setDaemon(true);
           final long t1 = System.nanoTime();
           cf.start();
           final List<File> entries = queue.take();
           final long delta = System.nanoTime() - t1;
           System.out.printf("It took %20dus to retrieve %20d files,
%20.5fus/file.\n",
               TimeUnit.NANOSECONDS.toMicros(delta), entries.size(),
(double) TimeUnit.NANOSECONDS.toMicros(delta)
                   / entries.size());
         }
       } else {
         System.out.println("Not a directory.");
       }
     }

     System.out.println("done");
   }

}

https://gist.github.com/4648256

It's not guaranteed though that this will be faster. And it's
definitively not simpler than the straight forward approach. :-)

Cheers

    robert

--
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/

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