Unexpected output of Exception Handling

Question

Can you please tell me what is wrong with my code. It is supposed to ask for input until both inputs entered are integers

public static void main(String[] args) {
    int a, b;
    Scanner input = new Scanner(System.in);
    int i = 0;

    while (i == 0) {
        try {

            System.out.print("Enter first number");
            a = input.nextInt();
            System.out.println("The number you entered is " + a);
            System.out.println("Enter second number");
            b = input.nextInt();
            System.out.println("The number you entered is " + b);

            i++;
        } catch (InputMismatchException e) {
            e.printStackTrace();
        }
    }
    System.out.println("End");
}

Think about implementing a method that requests a single integer (and it should ask again if the input was invalid). Then call this method twice. This is much easier, than requesting and checking two valid individual numbers. Also the user doesn't need to re-enter the first number, if the second one was invalid. — Tom
– Tom, Commented Feb 21, 2015 at 23:30

Reimeus · Accepted Answer · 2015-02-21 21:50:47Z

2

If an invalid integer is entered that content will remain in the input buffer and be passed through to the exception block ad infinitum. These invalid characters need to be consumed using next or nextLine. hasNextInt should be used to check that an integer is actually available before calling nextInt

edited Feb 21, 2015 at 21:50

answered Feb 21, 2015 at 21:35

Reimeus

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Comments

Amitesh Rai · Accepted Answer · 2015-02-21 22:53:09Z

-1

Your Looping condition is :

int i = 0;
while (i == 0) {  // Looping against i==0
    try {
        System.out.print("Enter first number");
        a = input.nextInt();
        System.out.println("The number you entered is " + a);
        System.out.println("Enter second number");
        b = input.nextInt();
        System.out.println("The number you entered is " + b);
        i++; // Increments the value of i to 1, which makes the loop to exit after first iteration
    } catch (InputMismatchException e) {
        e.printStackTrace();
    }
}

Refactor your code as :

boolean areBothIntegers = true;
        while (areBothIntegers) {
            try {
                System.out.print("Enter first number");
                a = input.nextInt();
                System.out.println("The number you entered is " + a);
                System.out.println("Enter second number");
                b = input.nextInt();
                System.out.println("The number you entered is " + b);
            } catch (InputMismatchException e) {
                e.printStackTrace();
                areBothIntegers = false;
            }
        }

Sample Output :

Enter first number : 10

The number you entered is 10

Enter second number : 21

The number you entered is 21

Enter first number : 21

The number you entered is 21

Enter second number : 34

The number you entered is 34

Enter first number : 21.5 // Not an int value, so throws the exception and exits

java.util.InputMismatchException
    at java.util.Scanner.throwFor(Scanner.java:909)
    at java.util.Scanner.next(Scanner.java:1530)
    at java.util.Scanner.nextInt(Scanner.java:2160)
    at java.util.Scanner.nextInt(Scanner.java:2119)
    at jdbc.statements.ADS.main(ADS.java:15)

edited Feb 21, 2015 at 22:53

answered Feb 21, 2015 at 22:14

Amitesh Rai

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10 Comments

Amitesh Rai Over a year ago

@Tom Yes it does..after then only i posted this.

Tom Over a year ago

And this code asks the user for input until he entered two valid numbers?

Tom Over a year ago

It is supposed to ask for input until both inputs entered are integers That is what OP tries to accomplish. Your loop asks for more input as long the user enters numbers. This is the direct opposite.

enu Over a year ago

It actually asks for input as long as both inputs are valid, however I got the point .. respect!

enu Over a year ago

@Armitesh Your view is right but your code does not reflect your view

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