CODEWITHSUNDEEP
java
Matt
Matt

Reputation: 11

problems quitting java program with 'Q'

I wrote a program to determine a prime number and am required to use 'Q' as the quit function to get out of the program. Below is the code i have written. I am wondering how I would modify method of quitting the program to make it exit correctly. 

import java.util.Scanner;

public class Practice
{
   public static void main(String[]args)
    {
     Scanner keyboard = new Scanner(System.in);
     int number, i=2;
     String quit;
     boolean prime = true;



     System.out.println("Please enter a number to determine if it is prime, to quit     please enter Q.");
     number = keyboard.nextInt();

     quit = keyboard.nextLine();


     for (i=0; i<number;i++)
        {
           if ((number%2==0) || (number==1) || (number%3==0))
           {
              prime = false;
              }
           }
     if (prime==false)
        {
        System.out.println(number+" is not a prime number.");
        }
        else if(prime==true)
        System.out.println(number+" is a prime number.");
    if (quit.charAt(0)=='Q')

       { System.exit(0);}

and my output upon entering 'Q' is:

   ----jGRASP exec: java Practice

  Please enter a number to determine if it is prime, to quit please enter Q.
  Q
  Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Scanner.java:909)
at java.util.Scanner.next(Scanner.java:1530)
at java.util.Scanner.nextInt(Scanner.java:2160)
at java.util.Scanner.nextInt(Scanner.java:2119)
at Practice.main(Practice.java:15)

  ----jGRASP wedge2: exit code for process is 1.
   ----jGRASP: operation complete.








  }

}

Upvotes: 0

Views: 2417

Answers (4)

michaelsnowden
michaelsnowden

Reputation: 6192

Try this instead:

 String input;
 int number = 1;
 int i=2;

Then do this:

 System.out.println("Please enter a number to determine if it is prime, "
                  + "\nto quit, please enter Q.");
 input = keyboard.nextLine();
 if (input.charAt(0)=='Q') 
 { 
    System.exit(0);
 }
 else 
 {
      try  
      {  
          number = Integer.parseInt(input);
      }  
      catch(NumberFormatException nfe)  
      {  
          System.out.println("Please enter a number or Q!");
      }  
 }

Upvotes: 0

tycoon177
tycoon177

Reputation: 41

You should read in the numbers as Strings with keyboard.nextLine(); and check first if the it contains 'Q'. If it does, quit, else do, number = Integer.parseInt(yourVariableHere); and go from there

Upvotes: 0

Jason C
Jason C

Reputation: 40315

You have:

 number = keyboard.nextInt();
 quit = keyboard.nextLine();

Think for a moment about what actually happens here. 'Q' is not a number and so nextInt(), as documented, throws an InputMismatchException. Your use of nextLine() doesn't make much sense following that.

You have a couple of options:

  • Use Scanner.hasNextInt() to determine if the next token truly is a well-formed integer. If not, read it as a string and check it.
  • Use Scanner.next() always, and check if it is "Q" before parsing it with Integer.parseInt() (being prepared to handle errors there as well).

The Scanner.next* functions don't implicitly skip invalid values. The Scanner is only doing what you tell it to do, and if you tell it to grab an integer but the input is not an integer, it rightfully complains. It is up to you to use the Scanner appropriately and check your possible input cases.

Upvotes: 0

goSun
goSun

Reputation: 29

When the input through the console is given as "Q", the value of type String is getting assigned to a variable of type int, which is incorrect.

number = keyboard.nextInt();

The type of in the input needs to be checked, before it can be assigned to a int variable.

Upvotes: 1

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