CODEWITHSUNDEEP
javasystem.exit
Sveinn Kallinen
Sveinn Kallinen

Reputation: 23

Two console inputs required to System.exit(0) during else if

So I'm working on a copy of a simple Dice game that was an example from the Maxwell Sanchez YouTube JAVA on Eclipse tutorials. What I started playing around with is simple ways to implement a text based menu of sorts.

What I'm trying to accomplish is a Y or N input method of either restarting the program, or killing it. I'm a total noob, coming here after a tiny bit of Arduino. I'm liking JAVA but there are many things I don't understand.

My problem right now is, everything appears to work so far, except that if you get to the end and type N to quit, It requires 2 inputs of N to actually execute the else if statement. Is that something that is a bug? Or am I just mis-programing what I'm trying to accomplish.

import java.util.*;
public class diceGame 
{
static int money;
static Scanner in = new Scanner(System.in);
static Random random = new Random();
static String userName;
static String tryAgain;

public static void main(String[] args) 
{
    money = 1000;
    System.out.println("Welcome to this simple dice game! " +
            "Please enter your name.");
    String userName = in.nextLine();
    System.out.println("Hey " + userName + ".");
    rollDice();
}
public static void rollDice()
{
    System.out.println("You have " + money + " coins!");
    System.out.println("Please select a number (1-6) to bet on!");
    int betRoll = in.nextInt();
    System.out.println("Please place your bet!");
    int betMoney = in.nextInt();
    while (betMoney > money)
    {
        System.out.println("You don't have enough coins... you only " +
                    "have " + money + "coins.");
        System.out.println("Please place a realistic bet!");
        betMoney = in.nextInt();
    }

    int dice;
    dice = random.nextInt(6)+1;

    if (betRoll == dice)
    {
        System.out.println("You Win!");
        money+=betMoney*6;
        System.out.println("You have " + money + " coins.");
    }

    else
    {
        System.out.println("Snap! You lost your coins!");
        money-=betMoney;
        System.out.println("You have " + money + " coins.");
    }

    if (money <= 0)
    {
        System.out.println("You've lost all yer coins!");
        System.out.println("Play again?" + " Type y or n");

        if (in.next().equalsIgnoreCase("y")) 
        {
            System.out.println("Maybe you'll win this time!");
            money = 1000;
            rollDice();
        } 

        else if (in.next().equalsIgnoreCase("n")) 
        {
            System.out.println("Maybe next time...");
            System.exit(0);
        } 

        else
        { 
            System.out.println("Invalid character");
        }
    }

    else
        {
                rollDice();
        }
}
}

Upvotes: 2

Views: 114

Answers (2)

Thilo
Thilo

Reputation: 262794

if (in.next().equalsIgnoreCase("y")) 
else if (in.next().equalsIgnoreCase("n"))

In this code, you are calling in.next() twice, once for each condition, so it will read two inputs.

You need to separate the reading from the comparison.

String input = in.next();
if (input.equalsIgnoreCase("y")) 
else if (input.equalsIgnoreCase("n"))

Upvotes: 2

Elliott Frisch
Elliott Frisch

Reputation: 201507

Store the input in a variable, and compare it... or you'll have to input twice.

String choice = in.next();
if (choice.equalsIgnoreCase("y")) 
{
    System.out.println("Maybe you'll win this time!");
    money = 1000;
    rollDice();
} 
else if (choice.equalsIgnoreCase("n")) // <-- not in.next()

Every time you call in.next() you read user input.

Upvotes: 2

Related Questions