exception处理无限循环

我的问题简短而甜蜜。 我不明白为什么我的程序在捕获错误时无限循环。 我做了一个新的try-catch语句，但它循环甚至复制，粘贴和修改了之前有效的程序中的相应变量。 以下是声明本身及以下将是整个计划。 感谢您的帮助！

try { input = keyboard.nextInt(); } catch(Exception e) { System.out.println("Error: invalid input"); again = true; } if (input >0 && input <=10) again = false; } 

程序：

  public class Blanco { public static int input; /** * @param args the command line arguments */ public static void main(String[] args) { // TODO code application logic here nameInput(); } /** * * @param name */ public static void nameInput() { System.out.println("What is the name of the cartoon character : "); Scanner keyboard = new Scanner(System.in); CartoonStar star = new CartoonStar(); String name = keyboard.next(); star.setName(name); typeInput(keyboard, star); } public static void typeInput(Scanner keyboard, CartoonStar star) { boolean again = true; while(again){ System.out.println("What is the cartoon character type: 1 = FOX,2 = CHICKEN,3 = RABBIT,4 = MOUSE,5 = DOG,\n" + "6 = CAT,7 = BIRD,8 = FISH,9 = DUCK,10 = RAT"); try { input = keyboard.nextInt(); } catch(Exception e) { System.out.println("Error: invalid input"); again = true; } if (input >0 && input <=10) again = false; } switch (input) { case 1: star.setType(CartoonType.FOX); break; case 2: star.setType(CartoonType.CHICKEN); break; case 3: star.setType(CartoonType.RABBIT); break; case 4: star.setType(CartoonType.MOUSE); break; case 5: star.setType(CartoonType.DOG); break; case 6: star.setType(CartoonType.CAT); break; case 7: star.setType(CartoonType.BIRD); break; case 8: star.setType(CartoonType.FISH); break; case 9: star.setType(CartoonType.DUCK); break; case 10: star.setType(CartoonType.RAT); break; } popularityNumber(keyboard, star); } public static void popularityNumber(Scanner keyboard, CartoonStar star) { System.out.println("What is the cartoon popularity number?"); int popularity = keyboard.nextInt(); star.setPopularityIndex(popularity); System.out.println(star.getName() + star.getType() + star.getPopularityIndex()); } } 

Answer 1

你的程序永远运行，因为调用nextInt而不改变扫描程序的状态会一次又一次地导致exception：如果用户没有输入int ，调用keyboard.nextInt()将不会改变扫描程序正在查看的内容，所以当你在下一次迭代中调用keyboard.nextInt()时，你会得到一个exception。

您需要添加一些代码来读取用户在处理exception后输入的垃圾以解决此问题：

 try { ... } catch(Exception e) { System.out.println("Error: invalid input:" + e.getMessage()); again = true; keyboard.next(); // Ignore whatever is entered } 

注意：在这种情况下你不需要依赖exception：而不是调用nextInt() ，你可以调用hasNextInt() ，并检查扫描器是否正在查看整数。