Reading a binary file into ArrayList in Java

Question

I want to read binary file in java. I have 153(1.bin, 2.bin...153.bin) bin file. I must read that. I thought that I must use ArrayList for buffering. But I could not do that. How can I do this ?

After some research, I found that way in this title(Reading a binary input stream into a single byte array in Java). There is code like below in this title.

    StringBuilder sb = new StringBuilder();
    String fileName  = "/path/10.bin";
    byte[] buffer;

    try {

        buffer = Files.readAllBytes(Paths.get(fileName));

        for(byte b : buffer){

            sb.append(String.format("%02X ", b));
        }
        System.out.println(sb.toString());

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

Is it true way for my question or do I must use Arraylist for buffering ? If I use single byte array for buffering, do I must clear the buffer for the other binary files.

Edit : 153 unit means 153 file(1.bin,2.bin ... 153.bin)

FileNotFoundException will never be thrown by Files.readAllBytes(); what you'll get instead is a NoSuchFileException if the file does not exist. See the javadoc for FileSystemException as well. — fge
– fge, Commented Feb 24, 2015 at 10:05
And you want to buffer all of that into memory?? Is that reasonable? What do you want to do with this content once it is in memory? — fge
– fge, Commented Feb 24, 2015 at 11:43

fge · Accepted Answer · 2015-02-24 10:11:44Z

1

Your question is unclear. For one, you don't tell what those "units" are, how long they are etc. Second, all your code does is dump the contents of the file in hexadecimal.

What I suggest you do here is map the file into memory and use a class to wrap that around, and make it implement Closeable.

See FileChannel.open() and FileChannel.map(). Please note however that it is unsafe to map more than 1 GiB in memory. This is not a "real" mmap().

answered Feb 24, 2015 at 10:11

fge

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Raildex · Accepted Answer · 2015-02-24 10:46:20Z

1

I'm not sure what you mean by "units". Byte data is read something like that:

File f = new File ("File.txt");
FileInputStream fis = new FileInputStream (f);
byte[] bytes = new byte[(int) f.length ()];
fis.read (bytes, 0, (int) f.length () );

Make sure your File is not too big.

answered Feb 24, 2015 at 10:46

Raildex

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score 1 · Accepted Answer · 2015-02-24 10:48:53Z

1

I'm assuming a unit is one byte. Is this correct?

An ArrayList is not appropriate for byte buffering. It is a wrapper class for an array that implements the List interface (by which most of it's power is defined.)

http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html

If you simply want to read bytes in from a file, you could use FileInputStream.

http://docs.oracle.com/javase/tutorial/essential/io/bytestreams.html

Here's a simple example that reads input from a file containing "hello world!"

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;

public class Input {
public static void main(String[] args) throws FileNotFoundException, IOException {
    try (FileInputStream in = new FileInputStream("test.txt")) {
        int c;
        while ( (c = in.read()) != -1 )
            System.out.print((byte)c + " ");
    }
}
}

Output in bytes: 104 101 108 108 111 32 119 111 114 108 100 33 13 10

edited Feb 24, 2015 at 10:48

answered Feb 24, 2015 at 10:33

user3834947

1 Comment

bzkrtmurat Over a year ago

I understand this way, but I want to do buffering with this datas.

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