Reading Integer user input in DataInputStream in java?

Question

I am trying to get input from user using DataInputStream. But this displays some junk integer value instead of the given value.

Here is the code:

import java.io.*;
public class Sequence {
    public static void main(String[] args) throws IOException {
    DataInputStream dis = new DataInputStream(System.in);
    String str="Enter your Age :";
    System.out.print(str);
    int i=dis.readInt();
    System.out.println((int)i);
    }
}

And the output is

Enter your Age :12

825363722

Why am I getting this junk value and how to correct the error?

Answer 1

The problem is that readInt does not behave as you might expect. It is not reading a string and convert the string to a number; it reads the input as *bytes:

Reads four input bytes and returns an int value. Let a-d be the first through fourth bytes read. The value returned is:

(((a & 0xff) << 24) | ((b & 0xff) << 16) |  
((c & 0xff) << 8) | (d & 0xff))

This method is suitable for reading bytes written by the writeInt method of interface DataOutput.

In this case, if you are in Windows and input 12 then enter, the bytes are:

• 49 - '1'
• 50 - '2'
• 13 - carriage return
• 10 - line feed

Do the math, 49 * 2 ^ 24 + 50 * 2 ^ 16 + 13 * 2 ^ 8 + 10 and you get 825363722.

If you want a simple method to read input, checkout Scanner and see if it is what you need.

Answer 2

In order to get the data from the DataInputStream you have to do the following -

        DataInputStream dis = new DataInputStream(System.in);
        StringBuffer inputLine = new StringBuffer();
        String tmp; 
        while ((tmp = dis.readLine()) != null) {
            inputLine.append(tmp);
            System.out.println(tmp);
        }
        dis.close();

The readInt() method returns the next four bytes of this input stream, interpreted as an int. According to the java docs

However you should have a look at Scanner.