Problem.

[Stephen Foy]

Been trying to fix this all night. Cant get it to work.

 
while (number != 'q') { 
 
    try { 
 
System.out.print("Enter a number to start with (to quit press q): ");
Scanner in = new Scanner(System.in);
number = in.nextInt();
 
if (number == 'q') { number = (char)letter; };

When pressing q it wont quit.


[ November 04, 2005: Message edited by: ste fing ]

[Jeff Albertson]

import java.util.*;
 
public class Test {
    public static void main(String[] args) {
       Scanner in = new Scanner(System.in);
       boolean done = false;
       System.out.println("Enter ints, or q to quit");
       while (!done) {
           System.out.print(">");
           if (in.hasNextInt()) {
               int number = in.nextInt();
               System.out.println("processing [" + number + "]");
               String line = in.nextLine();
               System.out.println("discarding rest of line [" + line + "]");
           } else if (in.hasNextLine()) {
               String line = in.nextLine();
               if (line.length() > 0 && line.charAt(0)=='q') {
                   System.out.println("Goodbye!");
                   done=true;
               } else {
                   System.out.println("Unexpected input [" + line + "]");
               }
           } else {
               System.out.println("Unexpected end of input.");
               done=true;
           }
       }
    }
}

Here I assumed that if the user entered several ints on one line,
I would read the first and discard the rest. You may want to read
them all.

[Jesper de Jong]

You are reading from the Scanner object using the method nextInt(). Do you know what nextInt() does when the user enters something else than a valid number, such as the letter 'q'? From the API documentation:

Scans the next token of the input as an int. This method will throw InputMismatchException if the next token cannot be translated into a valid int value as described below.

So, an InputMismatchException is thrown when the user enters 'q'. If you don't want this, use a method like nextLine() instead, and check if it contains 'q' before converting it to an integer yourself (which you can do with Integer.parseInt()).

[Justin bob]

Scanner in = new Scanner(System.in);
do{
System.out.print("Enter an integer: ");
String answer = in.nextLine;
}while(answer != "q")
System.out.prinln("Application Terminated");


i believe this would be much easier....

[Jeff Albertson]

Originally posted by Jesper de Jong:
So, an InputMismatchException is thrown when the user enters 'q'. If you don't want this, use a method like nextLine() instead, and check if it contains 'q' before converting it to an integer yourself (which you can do with Integer.parseInt()).

Whilst this is true, one of the annoying things about parsing methods like
Integer.parseInt is that you can't look before you leap -- if the next token
is not an int you have to handle this in a catch block instead of in normal code
flow. On the other hand, Scanner has methods like hasNextInt which is true
iff a call to nextInt would succeed at this point, so that you can encorporate
this test into normal code flow. Try it!

[Joanne Neal]

Originally posted by Justin bob:
Scanner in = new Scanner(System.in);
do{
System.out.print("Enter an integer: ");
String answer = in.nextLine;
}while(answer != "q")
System.out.prinln("Application Terminated");


i believe this would be much easier....

Use equals() to compare Strings

}while(!answer.equals("q"))

[Jeff Albertson]

Yes, that code needs help -- it doesn't even compile...