Question

基本上，这种掷骰子游戏几乎可以正常运行。在Java中运行它。我现在只有两个问题：

如果玩家在第二局获胜，则不会注册为胜利并结束游戏。我并没有花太多时间来解决此问题，因为我只是不知道如何阻止它。
我在比赛，胜利和失败计数器方面遇到了麻烦。如何添加这些内容？

之前，我一直在将代码与许多掷骰游戏进行比较，但没有很多帮助，并且很少有示例可以说明问题。同样，我在代码中采用的方法主要由我的讲师强制执行，并尽我所能执行。如果我的代码中有任何令人发指的罪过，那是我的错，但需要很多复杂的东西。

import java.util.Scanner;
import java.util.Random;

//ISSUES: Roll 2 won't initiate a win if it equals the point. Win, loss, and game counters.


public class Craps_MCCH {
public static void main(String[] args){

    Scanner keyboard = new Scanner(System.in);
    Random rand = new Random();

    System.out.println("Hey, over there. Yeah, you. What's your name?");
    String name = keyboard.nextLine();
    System.out.println("Are you serious? " + name + "? We don't use real names here.");
    System.out.println("This is a serious operation. Give me your codename.");
    String code = keyboard.nextLine();
    System.out.println("Oh. " + code +". Okay. Sure.");
    System.out.println();
    System.out.println("So, want to play Craps?");
    System.out.println("y/n");

    char letter = '0';
    letter = keyboard.nextLine().charAt(0);

    while(letter == 'y' || letter == 'Y'){
        System.out.println("Press Enter to roll.");
        keyboard.nextLine();

        int die1 = rand.nextInt(6)+1;
        int die2 = rand.nextInt(6)+1;




        System.out.println("For roll 1, you rolled a " + die1 + " and a " + die2);
        int outcome = die1 + die2;

        if(outcome == 7 || outcome ==11){
            System.out.println("Wow, won already! No prize for new gize.");
        }
        else if(outcome == 2){
            System.out.println("We call that snake-eyes. You lose. Do better");
        }
        else if(outcome == 12){
            System.out.println("Choo-choo it's the boxcar children! Also you lose.");
        }
        else {
            System.out.println(outcome + " is now your point. Roll it again to win.");

            int die3 = rand.nextInt(6)+1;
            int die4 = rand.nextInt(6)+1;
            int outcome2 = (die3+die4);


            System.out.println();
            System.out.println("Let's roll. Press Enter.");
            keyboard.nextLine();
            int rcount = 3;

            System.out.println("Roll 2: \n" + die1 + " and " + die2);
            while(outcome2 != 7 || outcome2 != 11){
                if(outcome == outcome2){
                    System.out.println("Huzzah! You've won!");
                    //win = win+1;
                    break;
                }
                else{
                    System.out.println("Point is " + outcome);
                    die3 = rand.nextInt(6)+1;
                    die4 = rand.nextInt(6)+1;
                    outcome2 = die3+die4;
                    System.out.println("Roll " + rcount++ + ":");
                    System.out.println("You rolled " + outcome2);
                    System.out.println("Press Enter.");
                    keyboard.nextLine();
                }

            }
            if(outcome2 == 7 || outcome2 == 11){
            System.out.println("After all that... no dice.");
            //loss = loss+1;
        }

        }


        //System.out.println("Current wins: " + win);
        //System.out.println("Current losses: " + loss);
        System.out.println();
        System.out.println("Wins: ");
        System.out.println("Losses: ");
        System.out.println("Games played: ");
        System.out.println();
        System.out.println("Would you like to play again?");
        System.out.println("y/n");

        letter = keyboard.nextLine().charAt(0);


    }


}

}

我希望比赛在第二局结束，但最终还是获胜，但我不知道该如何实现。此外，还需要计算我的赢，输和比赛。

我们鼓励我们尽可能地让我们的教授开心，因为他是一个愚蠢的家伙，不喜欢运行相同的无聊的旧程序，所以我尝试用他的打印语句来招待他，如果看起来很抱歉愚蠢的人，但只是希望它对他很有趣，并赋予它一些个性。

Answer 1

要创建计数器，您需要使变量可用于整个方法。在键盘和兰特之后立即初始化游戏，赢和输计数器：

    Scanner keyboard = new Scanner(System.in);
    Random rand = new Random();
    int gameCount = 0;
    int winCount = 0;
    //int lossCount

您不需要损失计数。这只是gameCount-winCount。

与其说：

    winCount = winCount + 1;

您可以说：

    winCount++; //Which means the same thing, but looks professional

如果在循环中定义它们，则它们仅在该循环块中可用，并且每次循环重复时都会重置。

关于您的游戏为何未获得胜利的原因，仅使用break关键字打破最内层的循环。这让您进入第一时间循环。

此外，没有理由将字母分配给“ 0”。只需声明一下即可：

    char letter; //You don't need to assign it to anything.

Answer 2

我不知道如何玩掷骰子，但要重新考虑while循环。

掷骰子游戏，游戏计数器和得分问题

2 个答案: