dice game

[Stephen Foy]

Hi,

Could someone give me some feedback on how to improve this, and of silly things ive done?

import java.util.*;
 
class dice {
 
int[] diceRound = new int[3]; // Holds the random numbers
int[] values = new int[3];  // Holds the highest values
 
/** highestValue
*   Takes the values array, and adds together to return the score of the game.
* @return: returns sum of highest values */
 
public int highestValue() {
int sum = 0;
for (int i = 0;i < values.length;i++) {
sum = ((sum + values[i]));
}
return sum;
}
 
/** diceNumber
* Generates a random number
*@return: returns the randomNumber */
 
public int diceNumber() {
int realNum = (int) Math.floor(Math.random() * 6) + 1;
return realNum;
}
 
/** rollDice
*This methid plays the game, it rolls the dice once with 3 random numbers, gives user option
*to keep the highest or roll again, maximum 2 trys, then does this all the way down to one dice
*
*Assigns highest values to array "values".
*/
 

public void rollDice() {
 
dice generate = new dice();
 
int largest = Integer.MIN_VALUE;
boolean UserRollsAgain = false;
int length = diceRound.length;
int moveRolls = 1;
int counter = 0;
String answer = " ";
 

while (moveRolls <= 3) {// Allows user to play the game 3 times
 
UserRollsAgain = false;
 
if (moveRolls == 2) {
length = diceRound.length - 1; largest = 0;// If second roll then, two dice must be used
}
else if (moveRolls == 3) {
length = diceRound.length - 2; largest = 0; // If third roll then, one dice.
}
 

// Loops program while counter < 2 and user rolls again
 
while (!UserRollsAgain) {
 
for (int i = 0;i < length;i++) {// Rolls dice and determines highest value
diceRound[i] = generate.diceNumber();// and assigns it to the array values.
System.out.print(diceRound[i]+" | ");
 
if (diceRound[i] > largest) {
largest = diceRound[i];
}
values[i] = largest;
}
 
// If statements, determines whether uses can roll again or not.
 
if (counter < 2) {
System.out.print("\nKeep "+largest+" or roll again? y or n: ");
Scanner in = new Scanner(System.in);
answer = in.nextLine();
}
 
if (counter == 2) {
System.out.println ("\nYour Number is "+largest+" You Have No more rolls\n");
UserRollsAgain = true;
moveRolls++;
counter = 0;
}
 
else if (answer.length() == 1 && answer.charAt(0)=='y') {
counter++;
largest = 0;
UserRollsAgain = false;
}
 
else if (answer.length() == 1 && answer.charAt(0)=='n') {
System.out.println ("\nYour Number is "+largest+"\n");
UserRollsAgain = true;
moveRolls++;
}
 
else { System.out.println ("Error"); UserRollsAgain = true; counter = 2;}
}
 
}
System.out.println(" ");
 
} // rollDice ends;
 
/** main method
*runs the game, and displays highest value.
*gives user option to play game again
*/
 

public static void main(String[] args) {
dice game = new dice();
boolean loopGame = false;
 
while (!loopGame) {
 
game.rollDice();
 
System.out.println("Your Score is: "+game.highestValue()+"\n");
 
System.out.print("Would you like to play again y or n: ");
Scanner in = new Scanner(System.in);
String answer = in.nextLine();
 

if (answer.length() == 1 && answer.charAt(0)=='y') {
loopGame = false;
}
else if (answer.length() == 1 && answer.charAt(0)=='n') {
loopGame = true;
}
 
else { loopGame = true; }
 
}
}
}

[Stan James]

A neat exercise is to write out the simplest description of the game you can think of, then try to make the code look like that. I don't know if I have the rules quite right but it might look something like:

main
   playing = true
   while ( playing )
      playOneGame();
      playing = promptUserPlayAgain();
   end while
 
private playOneGame
   highestThrow = 0
   for i = 1 to 3
      newThrow = getBestOfThreeThrows()
      highestThrow = max( highestThrow, newThrow )
      display highestThrow 
      if i < 3
         if !promptUserThrowAgain()
            break for loop
   next 

If the rules look right, see if you can make the core of your program have almost exactly that many lines and say almost exactly those rules to the reader. The other methods like getBestOfThreeThrows() should be easy to build out of the code you already have.