#how do i prevent this scanner.parseInt from throwing NumberFormatException

1 messages · Page 1 of 1 (latest)

final zincBOT
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<@&987246399047479336> please have a look, thanks.

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TJ-Bot
?

You can prevent the NumberFormatException by using Scanner.hasNextInt() to check if the input is an integer before calling Scanner.nextInt(). This way, you can handle non-integer inputs appropriately. Here's an example:

Scanner scanner = new Scanner(System.in);
if (scanner.hasNextInt()) {
    int num = scanner.nextInt();
    // Use num here
} else {
    System.out.println("Invalid input. Please enter an integer.");
}

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eternal ingot
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import java.util.Scanner;

public class Main {
    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);

        int firstNumber;
        String Operator;
        int secondNumber;
        Boolean Continue = true;
        Boolean SkipFirstScannerNext = false;

        System.out.print("\n---------------------------------------------------------------------------------------------------\n\n");
        System.out.println("This is a calculator! You will enter a number, then an operator, and then another number.");
        System.out.print("After you're given the result, you will be given the choice to either continue or stop the program. You will type yes or no to determine which will happen.\n\n");

        while (Continue) {

            System.out.println("Please enter the first number!");
            if (SkipFirstScannerNext) {
                scanner.next();
            }
            SkipFirstScannerNext = true;

            firstNumber = Integer.parseInt(scanner.nextLine());

            System.out.println("Please enter the operator!");
            Operator = scanner.nextLine();

            System.out.println("Please enter the second number!");

            while (1 == 1) {

                secondNumber = scanner.nextInt();

                if (Operator.equals("/") && secondNumber == 0) {

                    System.out.println("You cannot divide by zero! please enter a new second number:");

                } else {

                    break;

                }

            }

            System.out.print("Your result is " + calculateResult(firstNumber, Operator, secondNumber) + "!\n\n");

            System.out.println("Would you like to continue?");
            Continue = doContinue();

        }
    }
final zincBOT
eternal ingot
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had to delete the definitions of the methods cus it was too long

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but the exception is thrown at line 27 i think

final zincBOT
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Mixing any nextXXX method with nextLine from the Scanner class for user input, will not ask you for input again but instead result in an empty line read by nextLine.

To prevent this, when reading user input, always only use nextLine. If you need an int, do

int value = Integer.parseInt(scanner.nextLine());

instead of using nextInt.

Assume the following:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
int age = scanner.nextInt();
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

When executing this code, you will be asked to enter an age, suppose you enter 20.
However, the code will not ask you to actually input a name and the output will be:

Hello , you are 20 years old.

The reason why is that when you hit the enter button, your actual input is

20\n

and not just 20. A call to nextInt will now consume the 20 and leave the newline symbol \n in the internal input buffer of System.in. The call to nextLine will now not lead to a new input, since there is still unread input left in System.in. So it will read the \n, leading to an empty input.

So every user input is not only a number, but a full line. As such, it makes much more sense to also use nextLine(), even if reading just an age. The corrected code which works as intended is:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
// Now nextLine, not nextInt anymore
int age = Integer.parseInt(scanner.nextLine());
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

The nextXXX methods, such as nextInt can be useful when reading multi-input from a single line. For example when you enter 20 John in a single line.

surlyjoe0101_25649 • used /tag id: scanner ·
eternal ingot
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i actually called it a "scanner".parseInt