Using a byte array

Question

The following code is giving a compilation error of storing char into byte. How do store the bytes read by the fin.read() function into a byte array and print it?

import java.io.FileInputStream;

class IO {
    public static void main(String args[]) {
        int i;
        int j = 0;
        FileInputStream fin = new FileInputStream("file.txt");
        byte[] b = new byte[100];
        do {
            i = fin.read();
            j++;
            b[j] = (char) i;
        } while (i != -1);
        System.out.println(b);
    }
}

Output:

possible loss of precision
found   : char
required: byte
            b[j] =(char) i;
                  ^
1 error

How do I make it work? Read the file in to byte array and then display?

Answer 1

You can read directly into a byte array using InputStream.read(byte [])

Answer 2

Read a file and print it:

    File file = new File("C:\\MyFile.txt");
    FileInputStream fis = null;
    BufferedInputStream bis = null;
    DataInputStream dis = null;

    try {
      fis = new FileInputStream(file);

      // Here BufferedInputStream is added for fast reading.
      bis = new BufferedInputStream(fis);
      dis = new DataInputStream(bis);

      // dis.available() returns 0 if the file does not have more lines.
      while (dis.available() != 0) {

      // this statement reads the line from the file and print it to
        // the console.
        System.out.println(dis.readLine());
      }

      // dispose all the resources after using them, you need to move this to the finally block and check for null!!
      fis.close();
      bis.close();
      dis.close();

    } catch (FileNotFoundException e) {
      e.printStackTrace();
    } catch (IOException e) {
      e.printStackTrace();
    }
  }

Answer 3

Try using the method FileUtils.readFileToByteArray(file) found in Apache Commons IO. It's more straightforward than writing the code by hand.

Answer 4

For your question the answer is straightforward. You are converting the int to char but you want to store it as byte. So you have to convert it to byte during conversation. i.e

import java.io.*;
class IO{
public static void main(String args[]){
    int i;
    int j=0;
    FileInputStream fin = new FileInputStream("file.txt");
    byte[] b = new byte[100];
    do{
        i= fin.read();
        j++;
        b[j] =(byte) i;
    }while( i != -1);
    System.out.println(b);
}

}

Also you do not need to read it as int. directly read as char or as byte.

Answer 5

Change the following line b[j] = (char) i; to b[j] = (byte) i;.

It will give you another compiler error. But the first problem will be solved.

The first one does not work because Java uses Unicode for char type so it has two bytes. You are casting the int value to char but then trying to assign it to a byte variable and for doing that you have to cast explicitly to byte, this is neither C nor C++ anymore.

You could also consider using more straightforward method fin.read(b) which would read up to 100 bytes in your case and return -1 when the EOF is reached. This way you would not have to cast i explicitly to byte.

For example like this:

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

class IO {
    public static void main(String args[]) {
        byte[] b = new byte[100];
        try (FileInputStream fin = new FileInputStream("file.txt");) {
            while (true) {
                int i = fin.read(b);
                if (i < 0) {
                    break;
                }
                if (i < b.length) {
                    b = Arrays.copyOf(b, i);
                }
                System.out.println(Arrays.toString(b));
            }
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}