Renaming file after it has been opened as a RandomAccessFile and mapped as a MappedByteBuffer

Question

I am trying to rename a file that I used as a RandomAccessFile before.

When I try to rename the file I get an error on the renameTo call. When I use the Windows application Process Monitor I see that there was no rename call.

How is it possible that I cannot rename a file that was opened as a RandomAccessFile?

The following code will replicate the problem that I am having:

File f = new File("testfile.txt");

FileChannel fc = new RandomAccessFile(f, "rw").getChannel();
MappedByteBuffer mem = fc.map(FileChannel.MapMode.READ_WRITE, 0, 8);
mem.position(0);
fc.close();

File oldfile = new File("testfile.txt");
File newName = new File("testfile2.txt");
Boolean success = oldfile.renameTo(newName);
success = f.renameTo(newName);

Is it because the file is still open ( I see you did one close)? On Microsoft's windows you can not rename a file that is open. — ctrl-alt-delor, Dec 09 '14 at 20:35
I have just noticed that you renamed it twice, is that significant. (“Testfile Testfile …”) — ctrl-alt-delor, Dec 09 '14 at 20:38

score 1 · Accepted Answer · edited May 23 '17 at 12:27

1

The file is still open. You have to unmap the file from memory before you can rename it. You can find the solution here: How to unmap a file from memory mapped using FileChannel in java?

For example (This method can be dangeraus):

public static void unmap(MappedByteBuffer buffer)
{
   sun.misc.Cleaner cleaner = ((DirectBuffer) buffer).cleaner();
   cleaner.clean();
}

edited May 23 '17 at 12:27

Community

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answered Dec 09 '14 at 20:41

Laurence

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Renaming file after it has been opened as a RandomAccessFile and mapped as a MappedByteBuffer

1 Answers1