what is the problem in the following code?

**rotaryfreak** · Jul 14 '10, 04:41 PM

um, i have no idea why its giving you the answer it is. i tried your code out for myself inputting 12 and the program would echo 49.

i have always been shown to grab user input using the Scanner class and creating a scanner object

Code:

import java.util.Scanner;

public class ToRead {

	public static void main(String[] args) {
		Scanner kb = new Scanner(System.in);
		int x;
		System.out.println("Enter a digit");
		try {
			x = kb.nextInt();
			System.out.println("Entered Value:" + x);
		} catch (Exception e) {
			System.out.println("You had entred wrong value");
		}
	}
}

now, when i try to input 12, it echos the inputted number, 12.

You should check out the scanner class, it's quite useful and has lots of features for grabbing user input from keyboard, etc. etc.

Java Scanner Class

hope this helps :)

**newlearner** · Jul 15 '10, 01:04 PM

Code:

import java.io.BufferedReader;
import java.io.InputStreamReader;


public class SysRead {
	public static void main(String[] args) { 
		String x; 
		System.out.println("Enter a digit"); 
		try { 
			BufferedReader x1 = new BufferedReader(new InputStreamReader(System.in));
			System.out.println("Entered Value:" + Integer.parseInt(x1.readLine())); 
		} catch (Exception e) { 
			System.out.println("You had entred wrong value"); 
		} 
	} 
}

Try this out.... hope ul understand the difference

**Oralloy** · Jul 15 '10, 02:05 PM

BTW, 50 is the (decimal) ASCII value for '2'. Check your input once again.

What your program is doing is reading the raw byte, hoisting that value to an integer, and then string converting the integer to generate your output.

Your code likely would have worked if you'd declared 'x' as a char.

what is the problem in the following code?

what is the problem in the following code?

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