Please Help in Buffered Reader getting error.....

Question

Getting error
java.lang.NumberFormatException: For input string: ""
     at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
     at java.lang.Integer.parseInt(Integer.java:468)
     at java.lang.Integer.parseInt(Integer.java:497)
     at charread.main(charread.java:18)
    import java.io.*;
    class charread
        InputStreamReader ir=new InputStreamReader(System.in);
        BufferedReader br=new BufferedReader(ir);
        public void main()throws IOException
            InputStreamReader ir=new InputStreamReader(System.in);
            BufferedReader br=new BufferedReader(ir);
            System.out.println("enter a character");
            char p=(char)(br.read());
            System.out.println("entered character = "+p);
            System.out.flush();
            System.out.println("enter a number");
            String s=br.readLine();
            int n=Integer.parseInt(s);
            System.out.println("enter a number = "+n);
        Please Help.....

Answer 1

JoachimSauer wrote:
First you read a single character, then you read the rest of the line.
You probably entered a character and pressed enter.
read() returned the character, readLine() returned the rest of the line, which is an empty String. And empty String can't be converted to an int.
It's probably the easiest approach to just replace read() with a readLine() and take the first character of that String (after verifying that there is a character and optionally that there's only one character in that String).Alternatively you can call readLine() after you read the char.
However this is merely a quick fix and I am concerned about other aspects of your program.
InputStreamReader ir=new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(ir);The alternative to wrapping a BufferedReader around an InputStream is to use a Scanner, contrast the above example with the one below:
Scanner sin = new Scanner(System.in);
System.out.flush();Calling flush in this situation in unnecessary.
int n=Integer.parseInt(s);Place the following code in a try/catch block every single time you parse a String from a user into an Integer in case an Exception occurs, e.g. user enters a word or letter.
Integer n = null;
while(n == null) {
  try {
    String s = br.readLine();
    n = Integer.parseInt(s);
  } catch(NumberFormatException e) {
    System.out.println("Please enter an Integer");
}Finally onto Class naming conventions, Class names begin with a capital letter, variables and methods begin with a lower case letter.
class charreadAvoid naming methods "void", especially with a return value of null. Or one might think, well in my case anyway that this is the entry point into your application.
public void main()Mel