What's the problem with this code?

Question

Hi When I run the following code I am getting NumberFormatException can anybody help me out in debugging code.

import java.io.*;
public class Case1 {
   public static void main(String args[])
   {  
       char ch='y';int i=0;
       BufferedReader bf=new  BufferedReader(new InputStreamReader(System.in));
       System.out.println("ch before while:::"+ch);
       while(ch=='y'||ch=='Y'){
           try{

       System.out.println("Enter the option");
         i=Integer.parseInt(bf.readLine());
       System.out.println("after"+i);

  switch {
       case 1 ystem.out.println("1");   break;
       case 2 ystem.out.println("1"); break;
       }
       System.out.println("do u want to continue(Y/y");
       ch=(char)bf.read();
       System.out.println("ch after execution:::"+ch);


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

}}

First, please format your code. Second, at which line does the exception occur? — Bobby, Sep 28 '10 at 09:28
If the code doesn't compile, an IDE (Eclipse, Netbeans, IntelliJ IDEA, etc.) should explain what line is wrong in your code and why. — Benoit Courtine, Sep 28 '10 at 09:34
check this link http://forums.sun.com/thread.jspa?forumID=54&threadID=762831 — Emil, Sep 28 '10 at 10:08

score 2 · Answer 1 · edited May 23 '17 at 12:11

2

This problem i have faced in my daytoday work. The bf.readLine() gives you the empty string("") or character values [A-Z].So do a precondition check like

    // To allow only Integer to be parsed.

      String  rawText = br.readLine().trim();
      if (    isNumeric (rawText)              // returns false for non numeric
           && rawText.matches("-?\\d+?")           // Only Integers.
         )

Updates :

// isNumeric implementation Due credit to CraigTP

Please refer for brilliant logic

How to check if a String is numeric in Java

public static boolean isNumeric(String str) 
{ 
  NumberFormat formatter = NumberFormat.getInstance(); 
  ParsePosition pos = new ParsePosition(0); 
  formatter.parse(str, pos); 
  return str.length() == pos.getIndex(); 
}

edited May 23 '17 at 12:11

Community

1
1

answered Sep 28 '10 at 09:37

Dead Programmer

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How will you implement the `isNumeric` method? – dogbane Sep 28 '10 at 09:58
1

this `isNumeric` method won't work in this case because it allows doubles too. For example, the string "5.5" will pass the `isNumeric` check, but throw a `NumberFormatException` on `Integer.parseInt`. – dogbane Sep 28 '10 at 10:19

score 1 · Answer 2 · answered Sep 28 '10 at 09:33

1

System.out.println("Enter the option");
         i=Integer.parseInt(bf.readLine());

Problem is here.

You are reading some non numeric input and trying to parse it into int. Thats the exceptional case.

answered Sep 28 '10 at 09:33

jmj

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score 0 · Answer 3 · answered Sep 28 '10 at 09:32

0

Maybe because readline string is like '123\n'.

answered Sep 28 '10 at 09:32

foret

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No, because readLine() returns the line without line-termination-string: http://download.oracle.com/javase/6/docs/api/java/io/BufferedReader.html#readLine%28%29 – Mnementh Sep 28 '10 at 10:37

Emil · Answer 4 · 2010-09-28T10:38:14.690

I replaced the inputstream with a Scanner.

public class Case1 {


   public static void main(String args[])
   {  
       char ch='y';int i=0;
      Scanner s=new Scanner(System.in);
       System.out.println("ch before while:::"+ch);
       while(ch=='y'||ch=='Y'){


       System.out.println("Enter the option");
       System.out.flush(); 
       i=s.nextInt();

       System.out.println("after"+i);

  switch(i) {
       case 1:System.out.println("1");   break;
       case 2:System.out.println("1"); break;
       }
       System.out.println("do u want to continue(Y/N)");
       ch=(char)s.next().charAt(0);
       System.out.println("ch after execution:::"+ch);


   }


}}

score 0 · Answer 5 · answered Sep 28 '10 at 12:21

We can only use Explicit Type Conversion for data types which are type compatible with each other. But as you are trying to do type conversion on

ch=(char)bf.read();

you are actually trying to cast an Integer as char(since return type of bf.read() is int). But Integer and char are not compatible, hence the error.

What's the problem with this code?

5 Answers5

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