Java input/output problems

Question

I'm very new with Java, so I wanted to create simple program, which would ask me to input some random value, and then to print it. Problem is when I input number 1, output is 51, input 3-output 51, input 77-output 55. What is wrong with this? Code looks like this:

public static void main(String[] args) throws IOException
{
    System.out.print("Input:");
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    int val=br.read();
    System.out.print("Output:");
    System.out.println(val);
}

Might want to use "scanner" for simple reading :)

Dylan Meeus
– Dylan Meeus

2013-11-10 21:57:13 +00:00
Commented Nov 10, 2013 at 21:57 — Dylan Meeus
– Dylan Meeus, Commented Nov 10, 2013 at 21:57

Stefan Haustein · Accepted Answer · 2013-11-10 22:04:16Z

3

You just read a single char and print the unicode :)

try something like

String s=br.readLine();
System.out.print("Output:");
System.out.println("Input " + s);
int val = Integer.parseInt(s):
System.out.println("As integer: " + val);

If you just want to read a single character:

System.out.print("Input:");
Reader r = new new InputStreamReader(System.in);
int val = r.read();
System.out.print("Output:");
System.out.println((char) val);

If you want to read a single digit:

System.out.print("Input:");
Reader r = new new InputStreamReader(System.in);
int val = r.read() - '0';
System.out.print("Output:");
if (val < 0 || val > 9) {
  System.out.println("error, digit expected");
} else {
  System.out.println(val);
}

edited Nov 10, 2013 at 22:04

answered Nov 10, 2013 at 21:55

Stefan Haustein

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Comments

Sandeep Singh · Accepted Answer · 2016-10-24 06:13:37Z

1

you can also use BufferedReader as

public static void main(String[] args) {
BufferedReader reader = new BufferedReader(new     InputStreamReader(System.in));
String input = reader.readLine();
input number = Integer.parseInt(input);
System.out.println(input);
}

answered Oct 24, 2016 at 6:13

Sandeep Singh

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