Problems reading a huge file of 12 MB (java.lang.OutOfMemoryError)

Question

i need to open a file of 12 Megabytes, but actually i'm doing it creating a buffer of 12834566-byte, because the size of the file is 12MB and i am developing this app for Android mobile systems.

Then, i supose i have to read with blocks of 1024 Kbytes instead of one block of 12 Mbytes, with a for, but i don't know how to do it, i need a little help with it.

This is my actual code:

File f = new File(getCacheDir()+"/berlin.mp3");
        if (!f.exists()) try {
          InputStream is = getAssets().open("berlin.mp3");
          int size = is.available();
          byte[] buffer = new byte[size];
          is.read(buffer);
          is.close();
          FileOutputStream fos = new FileOutputStream(f);
          fos.write(buffer);
          fos.close();
        } catch (Exception e) { throw new RuntimeException(e); }

Please, can someone tell me what i have to changue in this code to read blocks of 1024 Kbytes instead of one block of 12 Mbytes?

THanks!

You can't create a byte[] progressively. You have to feed the data to wherever the data is going as you read it. — Peter Lawrey
– Peter Lawrey, Commented Dec 12, 2011 at 16:53

Peter Lawrey · Accepted Answer · 2011-12-12 17:06:43Z

4

Try copying 1 KB at a time.

File f = new File(getCacheDir()+"/berlin.mp3");
if (!f.exists()) try {
     byte[] buffer = new byte[1024];
     InputStream is = getAssets().open("berlin.mp3");
     FileOutputStream fos = new FileOutputStream(f);
     int len;
     while((len = is.read(buffer)) > 0) 
        fos.write(buffer, 0, len);
} catch (Exception e) { 
     throw new RuntimeException(e); 
} finally {
     IOUtils.close(is); // utility to close the stream properly.
     IOUtils.close(fos);
}

Does Android support symbolic or hand links like UNIX? If it does, this would be faster/more efficient.

edited Dec 12, 2011 at 17:06

answered Dec 12, 2011 at 16:54

Peter Lawrey

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2 Comments

Marek Potociar Over a year ago

You are not closing streams properly.

Nikhil Agrawal Over a year ago

@peter Larey great answer sir. Solved my problem. +1 for you sir.

Marek Potociar · Accepted Answer · 2011-12-12 17:02:10Z

File f = new File(getCacheDir()+"/berlin.mp3");
InputStream is = null;
FileOutputStream fos = null;
if (!f.exists()) try {
    is = getAssets().open("berlin.mp3");
    fos = new FileOutputStream(f);
    byte[] buffer = new byte[1024];
    while (is.read(buffer) > 0) {
        fos.write(buffer);
    }
} catch (Exception e) { 
    throw new RuntimeException(e); 
} finally { 
    // proper stream closing
    if (is != null) {
        try { is.close(); } catch (Exception ignored) {} finally {
           if (fos != null) {
               try { fos.close(); } catch (Exception ignored2) {}
           }
        }
    }
}

Joop Eggen · Accepted Answer · 2011-12-12 16:55:43Z

0

        import org.apache.commons.fileupload.util.Streams;

        InputStream in = getAssets().open("berlin.mp3");
        OutputStream out = new FileOutputStream(f);
        Streams.copy(in, out, true);

answered Dec 12, 2011 at 16:55

Joop Eggen

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