Re: The first 10 files

From:
Knute Johnson <nospam@knutejohnson.com>
Newsgroups:
comp.lang.java.programmer
Date:
Sat, 26 Jan 2013 18:37:09 -0800
Message-ID:
<ke23sm$afb$1@dont-email.me>
On 1/26/2013 1:14 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?


import java.io.*;
import java.nio.*;
import java.nio.file.*;

public class FileSystemsTest {
     public static void main(String[] args) throws IOException {
         long start = System.currentTimeMillis();
         Path dir = FileSystems.getDefault().getPath(".");
         int i=10;
         DirectoryStream<Path> stream = Files.newDirectoryStream(dir);
         for (Path path : stream) {
             System.out.println(path.getFileName());
             if (--i <= 0)
                 break;
         }
         long stop = System.currentTimeMillis();
         System.out.println(stop - start);
     }
}

300003 files in the directory, almost 1.7GB of files, Windows XP, Java 7
and it takes 16 ms to run. Somebody else should try this out on their
computer to see if it works as fast.

..
..
..
01/26/2013 05:46 PM 58,890 9998.txt
01/26/2013 05:46 PM 58,890 9999.txt
01/26/2013 06:31 PM 1,316 FileSystemsTest.class
01/26/2013 06:29 PM 636 FileSystemsTest.java
01/26/2013 05:44 PM 650 MakeFiles.java
            30003 File(s) 1,766,702,602 bytes
                2 Dir(s) 49,387,085,824 bytes free

C:\Documents and Settings\Knute Johnson\bigdirectory>java FileSystemsTest
0.txt
1.txt
10.txt
100.txt
1000.txt
10000.txt
10001.txt
10002.txt
10003.txt
10004.txt
16

C:\Documents and Settings\Knute Johnson\bigdirectory>

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Knute Johnson

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