Question

Sample Code is :-

import java.io.*;

public class WriteInt{
    public static void main(String [] args)
    {
    WriteInt obj = new WriteInt();
    obj.write();
    }

    public void write(){
    File file = null;
    FileOutputStream out = null;
    int [] arr = {6};
    try{
    file= new File("CheckSize.txt");
    out = new FileOutputStream(file);
    for(int i =0; i<arr.length;i++)
    {
        System.out.println("Trying to write to file:-"+ file);
        out.write(arr[i]);
    }

    }
    catch(IOException ioex){
    ioex.printStackTrace();
    } 
    finally{
    if(out != null)
    {
        System.out.println("Closing the stream");
        try{
        out.close();
        }   
        catch(IOException ioex){
        ioex.printStackTrace();
        }
    }

    else{
        System.out.println("Stream not open");
        }   
    }
    }

    }

Since I am using Byte-Oriented Stream to write data to a file; My Question is that will the data be written to file in 4 steps (1 byte) in each step. Considering int to be of 4 bytes. Please correct me if I am wrong.

Answer 1

out.write(arr[i]) will write only the lowest byte of int. The best solution is to use java.io.DataOutputStream which has writeInt(int) method.

    DataOutputStream out = new DataOutputStream(new FileOutputStream("file"));
    out.writeInt(arr[i]);