Re: The first 10 files

From:
=?ISO-8859-15?Q?Arne_Vajh=F8j?= <arne@vajhoej.dk>
Newsgroups:
comp.lang.java.programmer
Date:
Sat, 26 Jan 2013 13:26:54 -0500
Message-ID:
<51041ff8$0$284$14726298@news.sunsite.dk>
On 1/26/2013 7:24 AM, Arved Sandstrom wrote:

On 01/26/2013 05:14 AM, Wojtek wrote:

Using:

int max = 10;
int count = 0;

for (File thisFile : aDir.listFiles())
{
  doSomething(thisFile);

  if ( ++count >= max )
    break;
}

gives me the first ten files in aDir. But if aDir contains 30K files,
then the listFiles() will run for a long time as it builds an array for
the 30K files.

Is there a way to have Java only get the first "max" files?


One way of doing it, which you can find by Googling but should occur to
you if you read the File Javadocs carefully, is below.

         Integer limit = Integer.parseInt(args[0]);
         File testDir = new File(".");
         File[] files = testDir.listFiles(new MyFileFilter(limit));

     static class MyFileFilter implements FileFilter {

         int maxFiles;

         public MyFileFilter(int maxFiles) {
             this.maxFiles = maxFiles;
         }

         @Override
         public boolean accept(File pathname) {
             return maxFiles-- > 0;
         }
     }
}


If the problems is as described by OP then that must be the
correct solution.

"will run for a long time as it builds an array for the 30K files"

does not happen with this solution.

But I am a bit skeptical about whether a String[] with 30K elements
is really the bottleneck.

If the real bottleneck is the OS calls to get next file, then
a filter like this will not help.

Arne

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