基本上,对于这个数学测验,我必须编程的最后一件事是,我必须询问用户是否愿意回答更多的问题,如果愿意,则重新运行Main方法中的所有内容。如果不是,则打印再见。说“不”很容易,但是如果他们说“是”,我不确定如何告诉它重新运行main方法。下面是我的main方法中的代码。
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int digit = 0;
String result1 = getUserChoice("");
digit = getNumberofDigit1(digit);
int numberOfProblems = amountOfProblems();
for (int i = 0; i < numberOfProblems; i++) {
int number1 = getRandomNumber1(digit);
int number2 = getRandomNumber2(digit);
System.out.println("Enter your answer to the following problem: \n" +
number1 + result1 + number2);
int correctAnswer = getCorrectAnswer(number1, result1, number2);
int userAnswer = getUserAnswer();
CheckandDisplayResult(correctAnswer, userAnswer);
}
System.out.println("Would you like to solve more probelms(Y/N)? ");
String moreProblems = in.next();
if ("Y".equals(moreProblems)){
digit = 0;
result1 = getUserChoice("");
digit = getNumberofDigit1(digit);
numberOfProblems = amountOfProblems();
for (int i = 0; i < numberOfProblems; i++) {
int number1 = getRandomNumber1(digit);
int number2 = getRandomNumber2(digit);
System.out.println("Enter your answer to the following problem: \n" +
number1 + result1 + number2);
int correctAnswer = getCorrectAnswer(number1, result1, number2);
int userAnswer = getUserAnswer();
CheckandDisplayResult(correctAnswer, userAnswer);
}
System.out.println("Would you like to solve more probelms(Y/N)? ");
moreProblems = in.next();
if ("Y".equals(moreProblems)){
}
System.out.println("Thank you for taking this quiz, Goodbye!");
}
现在我尝试了一些类似的东西,
if "Y".equals(moreProblems)){复制并传递main方法}
但这有一个需要无限循环的错误,因为你必须在每个if of yes中有更多的问题语句,这意味着它永远不会结束编码,你将永远保持复制和粘贴。
发布于 2013-10-27 12:27:26
你可以把所有你想要“重新运行”的代码放在一个while循环中:
boolean run = true;
while (run) {
// Here your code
// Here input if user want to re-run
if (getUserChoice("").equals("NO"))
run = false;
}
发布于 2013-10-27 12:30:31
除了其他人的建议之外,这是我更喜欢的方法:
while(true) {
//do all your stuff
if(/*some exit condition*/) { break; }
}
发布于 2013-10-27 12:26:33
您可以做的是将main()
中的所有内容移动到另一个称为interact()
的静态方法中。然后在main()
中,只要用户想要与您的程序交互,就有调用interact()
的逻辑。换句话说,将数学测验放入一种方法中,并将测验呈现到main()
中。如果需要,您的程序将更易于阅读和进一步修改。
https://stackoverflow.com/questions/19614599
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