使用try-catch和while循环捕获错误

Question

我试图确保用户输入是整数，但是当我使用下面的代码时，我只会得到print语句的无限循环。关于如何改进的任何建议？

boolean valid = false;
System.out.println("What block are you gathering? (use minecraft block ids)");
while(valid == false){
    try{
        block = input.nextInt();
        valid = true;
    }
    catch(InputMismatchException exception){
        System.out.println("What block are you gathering? (use minecraft block ids)");
        valid = false;
    }
}

Answer 1

nextInt()不会消耗无效的输入，因此它将反复尝试读取相同的无效值。要解决此问题，您需要通过调用next()或nextLine()接受任何值来显式使用它。

顺便说一句，为使您的代码更整洁并避免昂贵的操作（如创建异常），您应使用hasNextInt()。

这是组织代码的方法

System.out.println("What block are you gathering? (use minecraft block ids)");
while(!input.hasNextInt()){
    input.nextLine();// consume invalid values until end of line, 
                     // use next() if you want to consume them one by one.
    System.out.println("That is not integer. Please try again");
}
//here we are sure that next element will be int, so lets read it
block = input.nextInt();