While Loop Error

[Peter Kolozvary]

I'm trying to create a Dice Game for class and am getting an error when copying the way to enter the while loops straight from the book.

Here is the code and the error:

import java.util.Scanner; // Needed for the Scanner class
import java.util.Random; // Needed for the Random class
// Name:Peter Kolozvary
// Date:    September 16 2008
// Course:ITCS-1215-001 (Fall 2008) 
// Submission: Assignment 2 - Dice Game
 

 
    public class Dice
   {
       public static void main(String[] args)
      {
         int guess;  // Input for the users guess
int dice1;  // The first die
int dice2;  // The second die
char repeat;// To hold 'y' or 'no'
String playAgain;   // To hold input



// Create a Scanner object to read input.
Scanner keyboard = new Scanner(System.in);

do
{
      
      //  Get the users guess
         System.out.print("Please select a number between 2 & 12. ");
         guess = keyboard.nextInt();

// Create a random class object.
Random randomNumbers = new Random();

// Get two random dice rolls.
dice1 = randomNumbers.nextInt(6) + 1;
dice2 = randomNumbers.nextInt(6) + 1;

// Display the first roll.
System.out.println(" Here is the first roll.");
System.out.println(dice1 + " & " + dice2);


// Ask the user if they want to play again.
System.out.println("Would you like to play again?");
System.out.println("Y = yes, N = no ");
playAgain = keyboard.nextLine();   //  Read a line.
repeat = input.charAt(0); //  Get the first char.

} while (repeat == 'Y' || repeat == 'y');
      
      }
   }    

Dice.java:48: cannot find symbol
symbol  : variable input
location: class Dice
repeat = input.charAt(); //  Get the first char.
                                 ^
1 error
 
 ----jGRASP wedge2: exit code for process is 1.
 ----jGRASP: operation complete.    

[Rene Larsen]

Where in the code do you initialize the 'input' variable??
[ September 07, 2008: Message edited by: Rene Larsen ]

[Rob Spoor]

And more importantly, where do you declare it?

[Peter Kolozvary]

Thanks for the help. I changed the input to playAgain instead but I'm now getting another error.

Okay, I'm obviously a beginner. Here is my way of thinking. Maybe you guys and gals can tell me where I'm off. I get most of the code up till this part and am confused with the repeat line below. Why would the error say string is out of range?

// Ask the user if they want to play again.
System.out.println("Would you like to play again?");
System.out.println("Y = yes, N = no ");
playAgain = keyboard.nextLine(); // Read a line.
repeat = playAgain.charAt(0); // Get the first char.

} while (repeat == 'Y' || repeat == 'y');

The system.out seems simple. It prints the message to the user and asks if they want to play again and gives them options.

Next the line starting with playAgain takes what the user types and assigns it to the playAgain string. Does that sound correct?

The line with the repeat is where I'm confused. It sounds like the repeat is assigning the first character of playAgain as what is used to determine if we're going to play again or not.

The last while statement just means to repeat if the user enters a Y or y.

[Rene Larsen]

You need to have a Scanner for each input - or change your code to have one

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

as the input 'controller', and then call

instream.readLine()

for each input.
[ September 07, 2008: Message edited by: Rene Larsen ]

[Campbell Ritchie]

I think you could have got away with changing the line with input in to

playAgain = keyboard.nextLine();

Look through the String class and you find methods allowing you to test whether a String starts with something, and to change a case to lower case or upper case. That might make your while testing easier.
I would suggest for a simple application like that you will find Scanners much easier to use than BufferedReaders.

[Rene Larsen]

Originally posted by Campbell Ritchie:
I think you could have got away with changing the line with input in to

playAgain = keyboard.nextLine();

Look through the String class and you find methods allowing you to test whether a String starts with something, and to change a case to lower case or upper case. That might make your while testing easier.
I would suggest for a simple application like that you will find Scanners much easier to use than BufferedReaders.

The problem is that when the Scanner is used in

guess = keyboard.nextInt();

then the call to

playAgain = keyboard.nextLine();

will not be made (the run of the program don't stop at the line, for the user to type to the input stream) - therefore my suggestion of the use of one Scanner for each input - OR the use of one BufferedReader.
[ September 07, 2008: Message edited by: Rene Larsen ]

[Campbell Ritchie]

Yes, I see what you mean, that Scanner sticks to the line until nextLine is called. That can be confusing, I agree. Do you think keyboard.next() would work?

Or putting the input in the while like this?

do
{
. . .
} while((playAgain = input.nextLine()) != null && playAgain.toLowerCase().startsWith("y"));

[Rene Larsen]

You only get one shot with a Scanner, and when you put nextLine() inside the while loop - then you need to type ALL input in that loop in one line ("Please select a number between 2 & 12. ", "Would you like to play again?", etc.)

I don't think that this solution is good enough - it will work, but the input will be confusing.

[Campbell Ritchie]

You only put the bit about whether you want to go back in the () after while . . .

Of course, you might need a nextLine() call immediately before that.

[Rene Larsen]

You still need to type ALL 'answers' in only one input.

In the Javadoc for the Scanner class, are the following written:

A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace. The resulting tokens may then be converted into values of different types using the various next methods.

[Campbell Ritchie]

I see what you mean. But a line end counts as whitespace and we usually use the enter key after input.

This works all right:

    private Scanner inScan;
. . .
    public void go()
    {
        int i, j;
        String input;
        do
        {
            out.print("Enter two ints: ");
            i = inScan.nextInt();
            j = inScan.nextInt();
            out.printf("%d * %d = %d%nPlease enter yes to continue: ", i, j, i * j); 
        }while((input = inScan.next()) != null && input.toLowerCase().startsWith("y"));
    }//end go

You can enter 1 2 yes or

1
2
yes
but 1 2 yes prints "please enter yes" after that, then "enter two ints."

[Campbell Ritchie]

And, by the way [pedantic mode]that's not called a while loop, it's a do loop (or a do-while loop).[/pedantic mode]