Question

我目前正在处理Ackerman功能问题，我们必须为用户输入编写故障安全代码。因此，如果用户输入通常会导致程序崩溃，那么它只会发送消息。我能够找出异常，如果整数值太大但我无法弄清楚如何检查用户输入是否为整数。我尝试过尝试使用“InputMismatchException”捕获块，但是代码开始混乱并且出错或只是不起作用。

public static void main(String[] args) {

//creates a scanner variable to hold the users answer
Scanner answer = new Scanner(System.in);


//asks the user for m value and assigns it to variable m
System.out.println("What number is m?");
int m = answer.nextInt();




//asks the user for n value and assigns it to variable n
System.out.println("What number is n?");
int n = answer.nextInt();


try{
//creates an object of the acker method
AckerFunction ackerObject = new AckerFunction();
//calls the method
System.out.println(ackerObject.acker(m, n));
}catch(StackOverflowError e){
    System.out.println("An error occured, try again!");
}



}

}

Answer 1

你必须把

int n = answer.nextInt();

在try块中

。然后你可以捕获java.util.InputMismatchException

这对我有用：

public static void main(String[] args) {

    //creates a scanner variable to hold the users answer
    Scanner answer = new Scanner(System.in);

    int m;
    int n;
    try{
        //asks the user for m value and assigns it to variable m
        System.out.println("What number is m?");
        m = answer.nextInt();
        //asks the user for n value and assigns it to variable n
        System.out.println("What number is n?");
        n = answer.nextInt();
    }catch(InputMismatchException e){
        System.out.println("An error occured, try again!");
    }
}

Ackermans的功能尝试捕获问题

1 个答案: