不明白 Java Scanner 对象在这个例子中是如何工作的（hasNextInt，next...？）

Question

In my program I need to take an input from a user, but it must be an integer. If the user doesn't input an integer, I need the program to reprompt them to enter an integer, and keep doing so until they enter one. I found an example while loop code snippet online which works perfectly, but I am having problems understanding why and how it works, and I'd really like to understand better:

Scanner reader = new Scanner(System.in);      
int guess=0;             
System.out.println("Guess the number"); 

while (!reader.hasNextInt()) {      //I get that this is a "not have" boolean
    System.out.println("That's not a number. Please enter a valid number");
    reader.next();
}

guess= reader.nextInt();
System.out.println(guess);

My questions are:

1. Does the !hasNext condition in the while loop actually trigger the scanner object to ask the user for input and then check that input, or does it only check anything that has already been inputted?
2. Why is it necessary to have the reader.next(); line after the while loop's main statement and what is this doing exactly? I know it's necessary as the program doesn't work when I take it out. But is it prompting for input? If yes what happens to this input?
3. When guess= reader.nextInt(); runs, why doesn't it re-prompt the user for input at this point?

Sorry that these are probably really basic questions, I'm new to coding and Java and just can't get my head around what's happening internally in this particular example, though have no problem doing other basic stuff with the Scanner.

Answer 1

1. hasNextInt() only checks whether the next input is an integer. It won't consume any input at all. The ! is just negate theu outcome of this function.

2. As hasNextInt() won't consume, then we need to use next() to consumeo the user input to let hasNextInt() to check with user's next input value. As you won't need it at all, then no need to assign it to any variables.

3. Scanner won't display the prompt at all. The prompt is printed by System.out.println() .

   For the line nextInt() , it is used to consume the next user input. Since hasNextInt() must be true when it execute this line, there must be one integer input waiting for consumption, so this method can return immediately with that user input.

Answer 2

It's not particularly intuitive, for sure, so don't feel bad for not getting it immediately.

It seems to me that the problem you're having with your understanding is that that methods such as nextInt may or may not prompt for user input, depending if anything is already in the Scanner.

---

Here's the sequence of events:

All your code executes until you hit !reader.hasNextInt() . There's no input so it "blocks" (waits) until there is some input from the user.

If the user enters 'A', that's not an integer so we enter the body of the while loop. We then print the error message.

Now, hasNextInt doesn't "consume" (process) the user input when it's checking whether or not it's an integer, so we still have that invalid user input of 'A' sitting in our scanner. We call reader.next() to effectively discard that value.

Now we're back to !reader.hasNextInt() . The scanner is empty once again, so we prompt for user input. If they enter another non-integer, that process will simply keep repeating.

Say this time we do have a valid user input - they've entered the number 2. This passes the check so our while loop ends and we continue along.

We've now got some input in our scanner, but we've not consumed it. We're sure it's an integer because of the while loop condition. We can now consume the input with reader.nextInt() and assign it to our variable.

Answer 3

 Try it :)
 public static void main(String[] args) {
        int option;
        if(args!=null&&args.length>0){
        option = Integer.parseInt(args[0]);
        }
        else{
        System.out.println("Enter:\n 1 to run x \n 2 to run y \n");
        Scanner keyboard = new Scanner(System.in);
         option = keyboard.nextInt();
        }
        switch (option) {
            case 1:
                xx
                break;
            case 2:
                yy
                break;
            default:
                break;
        }
    }