exception thrown for chardata = stdin.readline();

[vb_student]

Hello

i was wondering how this piece of code executed without any problems

import java.io.*;
public class InputToDouble
{
public static void main (String[] args) throws IOException
{
String charData;
double value;

BufferedReader stdin = new BufferedReader (new InputStreamReader(System.in));

System.out.println("Enter a double:");

charData = stdin.readLine();

value = Double.parseDouble( charData ) ;
System.out.println("value: " + value +" twice value: " + 2*value );
}
}


whereas this piece of code

import java.io.*;
/**
*
* @author Moyeen
*/
public class mywelcome {
public static void main (String[] args)
{
String charData;
double value;

BufferedReader stdin = new BufferedReader (new InputStreamReader(System.in));

System.out.println("Enter a double:");

charData = stdin.readLine();

value = Double.parseDouble( charData ) ;
System.out.println("value: " + value +" twice value: " + 2*value );
}
}

gives the error
mywelcome.java [21:1] unreported exception java.io.IOException; must be caught or declared to be thrown

chardata = stdin.readline();

why is the readline causing an exception?

[fartman_900]

I dont know how the IOException is thrown, but all I know is that there could be an error while reading from the keyboard. I checked the Java API and it wasnt much help, but maybe some of the others can explain it better if you need it.

But to fix that error, put a try...catch statement around that section, so it should look like this:

try{
     charData = stdin.readLine();

     value = Double.parseDouble( charData ) ;
}
catch(IOException i){
     //output some kind of message saying that there was a problem 
     //reading from the keyboard
}
catch(NumberFormatException n){     //this is thrown by Double.parseDouble()
    //output a message saying that it was incorrect format for the double
}

Java is very picky about exceptions, if there is a possibility that an exception is thrown, Java either makes you catch and deal with it, or you can just keep throwing it. Its always best to catch the exceptions and make meaningful error messages.

[ComputerJy]

This is how your throw a possible Exception in your code

/**
     * Read a line of text.  A line is considered to be terminated by any one
     * of a line feed ('\n'), a carriage return ('\r'), or a carriage return
     * followed immediately by a linefeed.
     *
     * @return     A String containing the contents of the line, not including
     *             any line-termination characters, or null if the end of the
     *             stream has been reached
     *
     * @exception  IOException  If an I/O error occurs
     */
    public String readLine() throws IOException {
        return readLine(false);
    }

And your code should look like:

public class mywelcome
{
  public static void main (String[] args) {
    String charData ;
    double value = 0 ;
    BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in)) ;
    System.out.println("Enter a double:") ;
    try {
      charData = stdin.readLine() ;
      value = Double.parseDouble(charData) ;
    }
    catch (IOException ex) {
    }
    System.out.println("value: " + value + " twice value: " + 2 * value) ;
  }
}

[ComputerJy]

I checked the Java API and it wasn't much help, but maybe some of the others can explain it better if you need it.

Java API explains better than any other source, a possible exception is thrown (any I/O Exception) since I/O Channels are most likely to fail.

Java is very picky about exceptions, if there is a possibility that an exception is thrown, Java either makes you catch and deal with it, or you can just keep throwing it. Its always best to catch the exceptions and make meaningful error messages.

1- I'd call it cautious not picky, how many times your code crashed because you didn't handle possible exceptions in your code
2- There are a best case scenario for all apps, so it's up to the developer when to handle exceptions