How do I catch InputMismatchException for doubles a, b, and c in this code?

Question

So I made a calculator that solves the quadratic formula, but I want the program to throw an exception when the variables a, b, or c aren't valid double numbers. However, I can't figure out how to get the variables into the equations that I want them to be in, so here is the code.

I don't know what background to put, I'm really new to java programming and I couldn't find answers to my specific problem anywhere else.

public static void main(String[] args) {

    Scanner input = new Scanner (System.in); //scanner

    short repeat = 1;
    while (repeat == 1) {

        System.out.println("Enter your equation by entering a, b, and c."); //introduction
        System.out.println("Press enter evey time you enter a number.");

        try {
            double a = input.nextDouble();
        }
        catch (InputMismatchException e) {
            System.out.println("That's not a valid number.");
        }

        double b = input.nextDouble();
        double c = input.nextDouble();

        double answer1 = ( (-b) + Math.sqrt( Math.pow(b, 2) - (4*a*c) ) ) / (2*a); //answers
        double answer2 = ( (-b) - Math.sqrt( Math.pow(b, 2) - (4*a*c) ) ) / (2*a);

        System.out.println("Your answers are: " + answer1 + " and " + answer2);

        System.out.println("Would you like to calculate more numbers? 1 for yes, 0 for no: ");
        repeat = input.nextShort();

    }

input.close();

}

I wanted the try/catch thing to work and for it to output a value that can be used in the equations, but the equations don't register the variable a. How would I go about doing the try/catch or any other method of displaying an error message?

If the next word is not a `double` it won't be read so every other attempt will fail. You can use `input.nextLine()` to read the whole line. — Peter Lawrey, Apr 03 '19 at 17:07
What do you define as an invalid double number? You will automatically throw that exception if you enter something that is not a number. Furthermore, you could also use ifs to manually throw it if you don't like the number. Ex) `if (a < 1) throw new InputMismatchException();`. Also, you may want to consider putting the two lines after the `catch` into your `try` block as well. — JamieT, Apr 03 '19 at 17:09

score 0 · Answer 1 · answered Apr 03 '19 at 17:16

According to Java Docs Scanner throws InputMismatchException if the next value is not a valid double value, so you should not take care of it manually.

But if you want to display a message when an exception occurs, you should catch it with try-catch block and do what you want to do.

the equations don't register the variable a

This happens because a is part of the "try" block scope. Once you exit this block, the variable is gone.

What you probably want to do is:

 double a; //declaring variable a (main() method scope)
 try {
     a = input.nextDouble(); //assigning a new value
 }
 catch (InputMismatchException e) {
     System.out.println("That's not a valid number.");
     throw e; //notice, the exception is still thrown after the message is printed
 }

So now a is in the main method's scope and you can use it after the try-catch block.

score 0 · Accepted Answer · answered Apr 03 '19 at 17:32

public static void main(String[] args) {
    Scanner input = new Scanner (System.in); //scanner

    short repeat = 1;
    while (repeat == 1) {


        System.out.println("Enter your equation by entering a, b, and c."); //introduction
        System.out.println("Press enter evey time you enter a number.");

        try {
            double a = Double.parseDouble(input.nextLine());
            double b = Double.parseDouble(input.nextLine());
            double c = Double.parseDouble(input.nextLine());

            double answer1 = ( (-b) + Math.sqrt( Math.pow(b, 2) - (4*a*c) ) ) / (2*a); //answers
            double answer2 = ( (-b) - Math.sqrt( Math.pow(b, 2) - (4*a*c) ) ) / (2*a);

            System.out.println("Your answers are: " + answer1 + " and " + answer2);
        }
        catch (NumberFormatException e) {
            System.out.println("That's not a valid number.");
        }



        System.out.println("Would you like to calculate more numbers? 1 for yes, 0 for no: ");
        repeat = Short.parseShort(input.nextLine());

    }
    input.close();

}

You have initalize the repeat variable to 1 and not to 0, since the program will not enter the while loop otherways.
You need to put your calculations for answer1 and answer2 in the try-catch-block too, because if the user enters an invalid value, the result shouldn't get calculated.
You should use only one Scanner instance, so you mustn't instance it in the loop.

Thanks, I was using an older version of my code which I forgot to update so all of the errors that I made are gone except for the try/catch system. I understand that you can just put everything into the block! — Christopher Tan, Apr 03 '19 at 17:41
I have a question, why use double a = Double.parseDouble(input.nextLine()); instead of double a = input.nextDouble(); ? Just wondering! — Christopher Tan, Apr 03 '19 at 17:51
With `input.nextDouble()` it depends on your region if , or . is used as decimal point. (but you can change the `Locale`-object of the scanner or the comma delimiter), and `parseDouble()` uses . as decimal point always. You can use `.nextLine()` as well as `.nextDouble()`/`.nextShort()`, but you mustn't mix them. — DasElias, Apr 03 '19 at 18:18

How do I catch InputMismatchException for doubles a, b, and c in this code?

2 Answers2