Memory mapped file java NIO

Question

I understand how to create a memory mapped file, but my question is let's say that in the following line:

FileChannel roChannel = new RandomAccessFile(file, "r").getChannel();
ByteBuffer roBuf = roChannel.map(FileChannel.MapMode.READ_ONLY, 0, SIZE);

Where i set SIZE to be 2MB for example, does this means that it will only load 2MB of the file or will it read further in the file and update the buffer as i consume bytes from it?

Answer 1

Where i set SIZE to be 2MB for example, does this means that it will only load 2MB of the file or will it read further in the file and update the buffer as i consume bytes from it?

It will only load the portion of the file specified in your buffer initialization. If you want it to read further you'll need to have some sort of read loop. While I would not go as far as saying this is tricky, if one isn't 100% familiar with the java.io and java.nio APIs involved then the chances of stuffing it up are high. (Eg: not flipping the buffer; buffer/file edge case mistakes).

If you are looking for an easy approach to accessing this file in a ByteBuffer, consider using a MappedByteBuffer .

RandomAccessFile raf = new RandomAccessFile(file, "r");
FileChannel fc = raf.getChannel();
MappedByteBuffer buffer = fc.map(FileChannel.MapMode.READ_ONLY, 0, fc.size());

The nice thing about a using an MBB in this context is that it won't necessarily actually load the entire buffer into memory, but rather only the parts you are accessing.

Answer 2

The size of the buffer is the size you pass in. It will not grow or shrink.

The javadoc says:

Maps a region of this channel's file directly into memory.

...

size - The size of the region to be mapped; must be non-negative and no greater than Integer.MAX_VALUE

EDIT:

Depending on what you mean by "updated with new data", the answer is yes.

The view of a file provided by an instance of this class is guaranteed to be consistent with other views of the same file provided by other instances in the same program. The view provided by an instance of this class may or may not, however, be consistent with the views seen by other concurrently-running programs due to caching performed by the underlying operating system and delays induced by network-filesystem protocols. This is true regardless of the language in which these other programs are written, and whether they are running on the same machine or on some other machine. The exact nature of any such inconsistencies are system-dependent and are therefore unspecified.

So, other systems may do caching, but when those caches are flushed or otherwise up-to-date, they will agree with the view presented by the FileChannel .

You can also use explicit calls to the position method and other methods to change what is presented by the view.

Changing the channel's position, whether explicitly or by reading or writing bytes, will change the file position of the originating object, and vice versa. Changing the file's length via the file channel will change the length seen via the originating object, and vice versa. Changing the file's content by writing bytes will change the content seen by the originating object, and vice versa.