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Java Programming Notes # 1796
Preface
New features in SDK Version 1.4.0
The recently released JavaTM 2 SDK, Standard Edition Version 1.4 contains a
large number of new features, including the concept
of IO channels. The first lesson in this miniseries, entitled
FileChannel Objects
in Java, Background Information, introduced you to the concept of channels
from a read/write IO viewpoint.
Memory-mapped IO
The previous lesson, entitled
Introduction
to Memory-Mapped IO in Java, introduced
you to the basics of memory-mapped IO and showed you how to perform memory-mapped IO for data of type byte.
In this lesson, I will show you how to do memory-mapped IO for data
records containing mixed types of data. I will also show you how to
manipulate memory maps using view buffers such as FloatBuffer.
Viewing tip
You may find it useful to open another copy of this lesson in a separate
browser window. That will make it easier for you to scroll back and
forth among the different listings and figures while you are reading about
them.
Supplementary material
I recommend that you also study the other lessons in my extensive collection
of online Java tutorials. You will find those lessons published at
Gamelan.com.
However, as of the date of this writing, Gamelan doesn’t maintain a consolidated
index of my Java tutorial lessons, and sometimes they are difficult to
locate there. You will find a consolidated index at
www.DickBaldwin.com.
What is a FileChannel?
Sun describes an object of the FileChannel class simply as “A
channel for reading, writing, mapping, and manipulating a file.”
Discussion
and Sample Code
Once you map a file into memory, changes that you make to the memory map are
automatically propagated to the file.
Methods of the ByteBuffer class apply
Because the memory map is an object instantiated from a subclass of the
ByteBuffer class, you can use the methods of the ByteBuffer class to
manipulate the data in the memory map, and hence to manipulate the data in the file.
The ByteBuffer class provides numerous methods that allow you to treat
the contents of the buffer as different data types. This includes methods
such as:
- The method named asFloatBuffer, which returns a
float view of the data in the ByteBuffer object. This
float view is provided via an object of the FloatBuffer class,
and allows you to manipulate the data in the ByteBuffer object as
type float. - The methods named getFloat and putFloat, which allow you to
manipulate the data in the ByteBuffer object as though it were type float rather
than type byte.
Will discuss sample program in fragments
I will illustrate memory-mapped IO for data of mixed types using the sample program named
Channel05. As usual, I will discuss this program in fragments.
You will find a complete listing of the program in Listing 11 near the end of the lesson.
Description of the program
This program,
which was tested using JDK 1.4.0 under Win2000, illustrates memory-mapped IO for data of mixed types.
The program begins by creating and populating a disk file named junk.txt
with data of mixed types.
Then the program creates a memory map of the file and modifies the map.
This, in turn, modifies the contents of the file.
Then the program reads and displays the modified file, demonstrating that
it was modified when the contents of the map were modified.
The code to this point illustrates the use of methods such as putDouble,
getDouble, putInt, and getInt to manipulate memory-map
data of mixed types.
Following this, the program modifies the map (and hence the disk file) to contain only data of type
float. This portion of the program illustrates the use of methods such as
asFloatBuffer to get a float view of the buffer, and to manipulate
the data in the map (and the disk file) as type float.
Then the program reads and displays the modified file, the entire
contents of which have now been converted to data of type float.
The main method
For simplicity, this
program consists of a main method and a couple of static convenience
methods for displaying data.
The first fragment in Listing 1 shows the beginning of the main method.
public static void main(
String[] args){
ByteBuffer buf =
ByteBuffer.allocate(16);
buf.putDouble(1.0/3);
buf.putInt(16000*16000);
buf.putInt(-32000*32000);
showMixedData(buf,
"Display raw buffer contents");
Listing 1
|
Create, populate, and display
The code in Listing 1 creates, populates, and displays a ByteBuffer
object with one double value and two int values. If you have
been studying this series of lessons from the beginning (Understanding
the Buffer class in Java), this code should be familiar to you.
(The method named showMixedData is a convenience method of my own
design, whose purpose is to display the contents of the ByteBuffer
object. This method, which can be viewed in Listing 11 near the end of the
lesson, is very similar to methods that I have explained in previous lessons.
Therefore, I won’t discuss it in detail here.)
The output
The output produced by the code in Listing 1 is shown in Figure 1.
Display raw buffer contents 0.3333333333333333 256000000 -1024000000 Figure 1 |
As you can see, the output consists of one double value and two int
values.
The code in Listing 2 creates a file named junk.txt and writes the
contents of the ByteBuffer object into the file.
FileOutputStream fos =
new FileOutputStream(
"junk.txt");
FileChannel fChan =
fos.getChannel();
buf.position(0);
System.out.println(
"Bytes written: " +
fChan.write(buf));
System.out.println("File size: "
+ fChan.size() + "n");
Listing 2
|
All of this code should also be familiar to you if you have been studying the
lessons in this miniseries from the beginning.
The output
The code in Listing 2 produces the output shown in Figure 2.
Bytes written: 16 File size: 16 Figure 2 |
A file size of sixteen bytes is what we would expect for one double
value, (which requires eight bytes), and two int values, (which
require four bytes each).
Close streams, etc.
The code in Listing 3 closes streams and channels, and turns objects over to
the garbage collector.
fos.close();
fChan.close();
fos = null;
fChan = null;
buf = null;
Listing 3
|
Following the execution of the code in Listing 3, the original data is
available to the program only by way of the disk file that now contains it.
Get FileChannel object
The code in Listing 4 gets a read/write FileChannel object,
connected to the file named junk.txt, which contain the data.
The FileChannel object will be used to map the
contents of the file into a memory map.
FileChannel rwCh =
new RandomAccessFile(
"junk.txt","rw").
getChannel();
Listing 4
|
(By now, you are probably getting tired of me telling you that you should
already understand the code with no further discussion on my part. From
this point on, if I present code and don’t discuss it, I will be assuming that
you already understand the code with no further discussion on my part.)
Map the file to memory and display the map
The code in Listing 5:
- Maps the contents of the file named junk.txt into a read/write
memory map. - Closes the channel used to perform the mapping operation.
- Displays the contents of the memory map.
long fileSize = rwCh.size();
ByteBuffer mapBuf = rwCh.map(
FileChannel.MapMode.READ_WRITE,
0, fileSize);
rwCh.close();
showMixedData(mapBuf,
"Map contents");
Listing 5
|
The output
The output produced by the code in Listing 5 is shown in Figure 3.
Map contents 0.3333333333333333 256000000 -1024000000 Figure 3 |
Figure 3 shows the contents of the memory map. Happily, the output values in Figure 3 match the original values originally
written to the file, confirming that no data modification or corruption has
taken place up to this point in the program.
Modify the map and the file
The code in Listing 6 invokes the relative putInt method on the map object’s
reference to change the value of the first int value in the map.
(This will also change the value in the first int value stored in the
disk file.)
//Modify one value in the middle
// of the map buffer
mapBuf.position(8);
mapBuf.putInt(127);
showMixedData(mapBuf,
"Modified map contents");
Listing 6
|
The code in Listing 6 causes the int value to be changed from 256000000
to 127.
The output
This is illustrated by Figure 4, which shows the output produced by the code
in Listing 6. Figure 4 confirms that the first int value in the
memory map has been modified as described above.
Modified map contents 0.3333333333333333 127 -1024000000 Figure 4 |
Read and display the modified contents of the disk
file
The code in Listing 7 reads the contents of
the disk file into a new ByteBuffer object and displays that data.
//Get new channel for read only
FileChannel newInCh =
new RandomAccessFile(
"junk.txt","r").
getChannel();
//Allocate a new ByteBuffer
ByteBuffer newBuf =
ByteBuffer.allocate(
(int)fileSize);
//Read file data into the new
// buffer, close the channel, and
// display the data.
System.out.println(
"Bytes read = "
+ newInCh.read(newBuf));
newInCh.close();
showMixedData(newBuf,
"Modified file contents");
Listing 7
|
The output
The output produced by the code in Listing 7 is shown in Figure 5.
Bytes read = 16 Modified file contents 0.3333333333333333 127 -1024000000 Figure 5 |
Figure 5 confirms that the change made to the first int value in the
memory map was automatically propagated to the associated disk file.
Recap
A disk file containing one double value and two int values was
mapped to a ByteBuffer object. The putInt method of the
ByteBuffer class was used to modify the first of the two int values.
This change was automatically propagated to the associated disk file.
The disk file was read and displayed, confirming that the change to the
memory map
was automatically propagated to the disk file.
Memory-mapped IO for single-type data
That completes the illustration of the use of memory-mapped IO for mixed data
types. The remainder of this lesson will concentrate on the use of
memory-mapped IO for data of a single type other than type byte.
Change to type float
The code in Listing 8 gets a FloatBuffer view of the ByteBuffer
object containing the memory map of the data in the disk file.
mapBuf.position(0);
FloatBuffer fBuf =
mapBuf.asFloatBuffer();
Listing 8
|
Manipulate the map data as type float
Having gotten a FloatBuffer view of the map, the program can
manipulate the contents of the map (and hence the contents of the disk file)
as type float.
The code in Listing 9 sets the position property of the FloatBuffer
object to zero, and then uses the relative put method of the
FloatBuffer class to store four
float values in the map. (This causes the same four float values to
be stored in the associated disk file.)
fBuf.position(0);
fBuf.put((float)1.0/6);
fBuf.put((float)2.0/6);
fBuf.put((float)3.0/6);
fBuf.put((float)4.0/6);
showFloatData(fBuf,
"Modified map contents");
Listing 9
|
The output
Having stored the four float values in the memory map, the code in
Listing 9 invokes the showFloatData method to display the contents of the
map as float data. This produces the four float values
shown in Figure 6.
Modified map contents 0.16666667 0.33333334 0.5 0.6666667 Figure 6 |
Display the file contents
All that now remains is to display the current contents of the file to confirm that
the file now contains four float values matching the four float
values in the memory map.
This is accomplished by the code in Listing 10.
//Read and display the modified
// file data
//Get new channel for read only
FileChannel floatInCh =
new RandomAccessFile(
"junk.txt","r").
getChannel();
//Allocate a new ByteBuffer
ByteBuffer anotherBuf =
ByteBuffer.allocate(
(int)fileSize);
//Read file data into the new
// buffer, close the channel, and
// display the data.
System.out.println(
"Bytes read = "
+ floatInCh.read(anotherBuf));
floatInCh.close();
anotherBuf.position(0);
FloatBuffer fileBuf =
anotherBuf.asFloatBuffer();
showFloatData(fileBuf,
"Modified file contents");
}catch(Exception e){
System.out.println(e);}
Listing 10
|
The final output
The contents of the disk file, as displayed by the code in Listing 10, are
shown in Figure 7.
Bytes read = 16 Modified file contents 0.16666667 0.33333334 0.5 0.6666667 Figure 7 |
As you can see, the sixteen bytes in the disk file no longer represent one
double value followed by two int values. Now the sixteen bytes
represent four float values.
A sequence of bytes
The contents of a disk file are, after all, simply a sequence of bytes.
We can cause those bytes to represent anything that we want them to represent.
The methods of the FileChannel class, and the methods of the Buffer
class and its subclasses make it relatively easy for us to cause the sequence of
bytes in a disk file to represent any combination of any of the primitive types,
(other than boolean).
Those methods also make it possible for us to manipulate primitive data in a
memory map and to have all changes made to the memory map automatically
propagated to the associated disk file.
That’s it for now
By now you should understand a quite a lot about creating and manipulating memory maps of disk files for any combination of primitive
types other than boolean.
Run the Program
If you haven’t already done so, I encourage you to copy the code from Listing 11 into your text editor, compile it, and execute it. Experiment
with it, making changes, and observing the results of your changes.
Remember, however, that you must be running Java version 1.4.0 or later
to compile and execute this program.
Summary
In this lesson, I have taught you how to do memory-mapped IO for data
records containing mixed types of data. I have also taught you how to
manipulate memory maps using view buffers such as FloatBuffer.
What’s Next?
As time goes on, I plan to publish additional lessons that will help
you learn to use other new IO features including:
- File locks
- Character-set encoders and decoders
- Pattern matching on files using regular expressions
- Socket channels for both clients and servers
- Non-blocking reads
- Non-blocking servers
Complete Program Listing
A complete listing of the program discussed in this lesson is shown in Listing
11 below.
/* File Channel05.java
Copyright 2002, R.G.Baldwin
Illustrates use of FileChannel objects
for mapped-memory IO for data of mixed
types.
First creates and populates a disk
file with data of mixed types.
Then creates a memory map of the
existing file, and modifies the map,
which modifies the contents of the
file.
Then reads and displays the modified
file.
Then modifies the map to contain only
data of type float.
Then reads and modifies the modified
file.
Tested using JDK 1.4.0 under Win2000
The output is:
Display raw buffer contents
0.3333333333333333
256000000
-1024000000
Bytes written: 16
File size: 16
Map contents
0.3333333333333333
256000000
-1024000000
Modified map contents
0.3333333333333333
127
-1024000000
Bytes read = 16
Modified file contents
0.3333333333333333
127
-1024000000
Modified map contents
0.16666667
0.33333334
0.5
0.6666667
Bytes read = 16
Modified file contents
0.16666667
0.33333334
0.5
0.6666667
**************************************/
import java.io.*;
import java.nio.channels.*;
import java.nio.*;
class Channel05{
public static void main(
String[] args){
//Create and populate a ByteBuffer
// object with data of mixed types
ByteBuffer buf =
ByteBuffer.allocate(16);
buf.putDouble(1.0/3);
buf.putInt(16000*16000);
buf.putInt(-32000*32000);
showMixedData(buf,
"Display raw buffer contents");
//Write buffer to disk file
try{
FileOutputStream fos =
new FileOutputStream(
"junk.txt");
FileChannel fChan =
fos.getChannel();
buf.position(0);
System.out.println(
"Bytes written: " +
fChan.write(buf));
System.out.println("File size: "
+ fChan.size() + "n");
//Close stream and channel and
// make objects eligible for
// garbage collection
fos.close();
fChan.close();
fos = null;
fChan = null;
buf = null;
//Get a new FileChannel object
// for reading and writing the
// existing file
FileChannel rwCh =
new RandomAccessFile(
"junk.txt","rw").
getChannel();
//Map entire file to memory and
// close the channel
long fileSize = rwCh.size();
ByteBuffer mapBuf = rwCh.map(
FileChannel.MapMode.READ_WRITE,
0, fileSize);
rwCh.close();
showMixedData(mapBuf,
"Map contents");
//Modify one value in the middle
// of the map buffer
mapBuf.position(8);
mapBuf.putInt(127);
showMixedData(mapBuf,
"Modified map contents");
//Read and display the contents
// of the file
//Get new channel for read only
FileChannel newInCh =
new RandomAccessFile(
"junk.txt","r").
getChannel();
//Allocate a new ByteBuffer
ByteBuffer newBuf =
ByteBuffer.allocate(
(int)fileSize);
//Read file data into the new
// buffer, close the channel, and
// display the data.
System.out.println(
"Bytes read = "
+ newInCh.read(newBuf));
newInCh.close();
showMixedData(newBuf,
"Modified file contents");
//Now change the type of data in
// the map and in the file to all
// float
mapBuf.position(0);
FloatBuffer fBuf =
mapBuf.asFloatBuffer();
fBuf.position(0);
fBuf.put((float)1.0/6);
fBuf.put((float)2.0/6);
fBuf.put((float)3.0/6);
fBuf.put((float)4.0/6);
showFloatData(fBuf,
"Modified map contents");
//Read and display the modified
// file data
//Get new channel for read only
FileChannel floatInCh =
new RandomAccessFile(
"junk.txt","r").
getChannel();
//Allocate a new ByteBuffer
ByteBuffer anotherBuf =
ByteBuffer.allocate(
(int)fileSize);
//Read file data into the new
// buffer, close the channel, and
// display the data.
System.out.println(
"Bytes read = "
+ floatInCh.read(anotherBuf));
floatInCh.close();
anotherBuf.position(0);
FloatBuffer fileBuf =
anotherBuf.asFloatBuffer();
showFloatData(fileBuf,
"Modified file contents");
}catch(Exception e){
System.out.println(e);}
}// end main
//---------------------------------//
static void showMixedData(
ByteBuffer buf, String label){
//Displays byte buffer contents
//Save position
int pos = buf.position();
//Set position to zero
buf.position(0);
System.out.println(label);
System.out.println(
buf.getDouble());
System.out.println(buf.getInt());
System.out.println(buf.getInt());
System.out.println();//new line
//Restore position and return
buf.position(pos);
}//end showBufferData
//---------------------------------//
static void showFloatData(
FloatBuffer buf, String label){
//Displays byte buffer contents
//Save position
int pos = buf.position();
//Set position to zero
buf.position(0);
System.out.println(label);
System.out.println(buf.get());
System.out.println(buf.get());
System.out.println(buf.get());
System.out.println(buf.get());
System.out.println();//new line
//Restore position and return
buf.position(pos);
}//end showBufferData
//---------------------------------//
}//end class Channel05 definition
Listing 11
|
Copyright 2002, Richard G. Baldwin. Reproduction in whole or in
part in any form or medium without express written permission from Richard
Baldwin is prohibited.
About the author
Richard Baldwin
is a college professor (at Austin Community College in Austin, TX) and
private consultant whose primary focus is a combination of Java, C#, and
XML. In addition to the many platform and/or language independent benefits
of Java and C# applications, he believes that a combination of Java, C#,
and XML will become the primary driving force in the delivery of structured
information on the Web.
Richard has participated in numerous consulting projects, and he
frequently provides onsite training at the high-tech companies located
in and around Austin, Texas. He is the author of Baldwin’s Programming
Tutorials,
which has gained a worldwide following among experienced and aspiring programmers.
He has also published articles in JavaPro magazine.
Richard holds an MSEE degree from Southern Methodist University and
has many years of experience in the application of computer technology
to real-world problems.
-end-
