CODEWITHSUNDEEP
java
mhashim6
mhashim6

Reputation: 527

What does happen inside this catch block?

I found this code online as a password loop game, it's working fine, but my question is: how?

What does happen in this catch block exactly?

I'm curious about this line specifically:

reader.next();

boolean loop = true;
    Scanner reader = new Scanner(System.in);
    System.out.println("PIN: ");

    while (loop) {
        try {
            Integer Code = reader.nextInt();

            if (Code == 8273) {
                System.out.println("Access granted");
                loop = false;
            } else {
                System.out.println("Access denied");
            }
        } catch (Exception e) {
            System.out.println("Please enter a valid PIN!");
            reader.next();
        }
    }

Edit : of course I did deliberately input a Non-integer input to cause the exception.

Edit2 : When I removed that line, the program kept printing Please enter a valid PIN! For ever.

Upvotes: 1

Views: 42

Answers (3)

fge
fge

Reputation: 121820

In fact, what the programmer really wanted here is to capture the next line of input, and verify whether that was a valid integer.

But the code is, admittedly, very confusing. And it relies on the fact that by default, when you "swallow" the next token with anything but .nextLine() with a Scanner, it relies on the current delimiter, which by default matches a newline.

Not good.

Here is a version which is more explicit:

String input;
int code;

while (true) {
    System.out.print("PIN: ");
    input = reader.nextLine();
    try {
        code = Integer.parseInt(input);
    } catch (NumberFormatException ignored) {
        // not an integer!
        System.out.println("Enter a valid PIN!");
        continue;
    }
    if (code == 8273)
        break;
    System.out.println("Access denied");
}

System.out.println("Access granted");

Upvotes: 1

Saad
Saad

Reputation: 305

The catch block simply catches the exception when anything other than an integer is entered. Since Code is an Integer, the input would have to be an integer. After catching the exception and printing the error, the reader moves to the next input until a proper value is entered, and the boolean loop becomes false, which ends the while loop at the end of the if statement once the correct value is entered.

Upvotes: 1

Elliott Frisch
Elliott Frisch

Reputation: 201517

If nextInt throws an exception (because the value entered isn't an int), then the catch block is entered. The last line of which,

reader.next(); // <-- discards invalid token.

Removes the invalid token and then the loop iterates.

Also, don't box the Code1

int code = reader.nextInt();

1Using an Object type and then testing equality with == is a bad idea™. Also, by convention Java variable names start with a lower case letter.

Upvotes: 1

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