Question

运行时它会正确捕获异常，但是当我选择一个案例时会导致无限循环。我已经尝试了几个小时寻找解决方案，没有任何帮助。

int whatYouWant = 1;
boolean contin = true;
while (whatYouWant != 0) {
    while (contin) {
        try {
            Scanner scanner = new Scanner(System.in);
            System.out.println("Which program would you like to run?" + " Input 0 to exit the program.");
            whatYouWant = scanner.nextInt();
            contin = false;
            scanner.close();
        } catch (Exception e) {
            System.out.println("Try again.");
            contin = true;
        }
    }
    AllCode a = new AllCode();
    switch (whatYouWant) {
    case 1:
        // method call that calls a method with another try catch block
        break;
    case 2:
        // another method call
        break;
    case 3:
        // another method call and so on and so forth
        break;
    case 0:
        whatYouWant = 0;
        break;
    default:
        System.out.println("Please try again.");
        break;
    }
}

以下是交换机调用的方法示例：

void multiplytheNumbers() {// Using the parameter to

    boolean run = true;
    while (run) {
        try {
            Scanner sc = new Scanner(System.in);
            System.out.println("Enter the 1st number:");
            int num1 = sc.nextInt(); /// getting user input from the user.
            System.out.println("Enter the 2nd number:");
            int num2 = sc.nextInt(); /// getting more user input
            int product = num1 * num2; /// using the multiplication operator
            int sum = num1 + num2; /// using then addition operator
            int diff = num1 - num1; /// using the subtraction operator
            int quotient = num1 / num1; /// using the quotient operator
            System.out.println(sum);
            System.out.println(diff);
            System.out.println(product);
            System.out.println(quotient);
            num1 = num1++; // using the increment operator to increase num1
                            // by 1
            num2 = num2--; // using the decrement operator to decrease num2
                            // by 1
            run = false;
            sc.close();
        } catch (Exception e) {
            System.out.println("Invalid Input. Try again.");
        }
    }
}

Answer 1

所以这是因为如果Scanner.nextInt()与int不匹配，它会抛出异常，但是如果你调用Scanner.next()，你会看到你之前输入的相同字符串。因此，每次进入循环时，它都会尝试解析与Integer相同的字符串输入，并且每次都会抛出异常。

您可以通过以下代码注意到此行为：

Scanner scanner = new Scanner(System.in);
while (contin) {
    try {
        System.out.println("Which program would you like to run?" + " Input 0 to exit the program.");
        whatYouWant = scanner.nextInt();
        contin = false;
    } catch (Exception e) {
        System.out.println(scanner.next());
        System.out.println("Try again.");
        contin = true;
    }
}
scanner.close();

您会注意到，在调用scanner.next()时，它仍将扫描相同的字符串。

要获得快速解决方案，您应该尝试Integer.parseInt(scanner.next());代替scanner.nextInt();

Answer 2

我正在defualt中阅读default而不是switch。

带有try块的无限循环 - java

2 个答案: