Creating error message for invalid input with Java calculator

Question

I have to create a calculator in Java. Everything works ok so far, except I can't quite figure out how to produce an error message when somebody enters a value that is not an integer when it asks for a number. I tried a "try, catch" statement but it still throws me an error in the IDE which is :

Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Unknown Source)
    at java.util.Scanner.next(Unknown Source)
    at java.util.Scanner.nextInt(Unknown Source)
    at java.util.Scanner.nextInt(Unknown Source)

here is the code for the rest of my program:

import java.util.InputMismatchException;
import java.util.Scanner;
public class Calculator {

static int num1, num2;
static int memory;
static String operation;
static String menu = "\nChoose an operation:\n+ Add\n- Subtract\n* Multiply\n/ Divide\n^ Exponent\n~ Square Root\n Exit";
static boolean run = true;

public static void menu (String menu){
    System.out.println (menu);
};

public static void add (int num1, int num2) {
    System.out.println (num1 +num2);}

public static void subtract (int num1, int num2) {
    System.out.println (num1 - num2);           
};

public static void multiply (int num1, int num2) {
    System.out.println (num1 * num2);           
};

public static void divide (int num1, int num2) {
    if (num2 !=0) {
        System.out.println (num1/num2);}
    else{
        System.out.println ("Cannot divide by 0");
    };
};

public static void exp (int num1, int num2) {
    System.out.println (Math.pow(num1,num2));           
};

public static void sqrt (int num1) {
    System.out.println (Math.sqrt(num1));           
};

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

    do{
        menu(menu);
        operation = scanner.next();

        try {
        switch (operation) {
            case "+":
                System.out.println ("Enter first number:");
                num1 = scanner.nextInt();
                System.out.println ("Please enter a valid number");
                System.out.println ("Enter second number:");
                num2 = scanner.nextInt();
                add (num1,num2);    
                break;
            case "-":
                System.out.println ("Enter first number:");
                num1 = scanner.nextInt();
                System.out.println ("Enter second number:");
                num2 = scanner.nextInt();
                subtract (num1, num2);
                break;
            case "*":
                System.out.println ("Enter first number:");
                num1 = scanner.nextInt();   
                System.out.println ("Enter second number:");
                num2 = scanner.nextInt();
                multiply (num1, num2);
                break;
            case "/":
                System.out.println ("Enter first number:");
                num1 = scanner.nextInt();
                System.out.println ("Enter second number:");
                num2 = scanner.nextInt();
                divide (num1, num2);
                break;
            case "^":
                System.out.println ("Enter first number:");
                num1 = scanner.nextInt();
                System.out.println ("Enter exponent:");
                num2 = scanner.nextInt();
                exp (num1, num2);
                break;
            case "~":
                System.out.println ("Enter number:");
                num1 = scanner.nextInt();
                sqrt(num1);
                break;
            case "Exit":
                System.out.println("You have exited the calculator");
                System.exit(0);
                run = false;
                break;
            default:
                System.out.println("Invalid");
            }
        }catch (InputMismatchException e) {

        }
    }while(run == true);



    scanner.close();
        }

};

thanks!

ok update: i got the try catch to throw the error, but when i run it with the error, it runs the default case of the switch statement, and then runs again (so you see the menu twice) and I only need it to show up once

at com.catalyst.training.calculator.Calculator.main(Calculator.java:70). where is tje line 70? — David Herrero
– David Herrero, Commented Sep 24, 2015 at 15:23
So just to clarify where do you wana see the error message. Also don't use try catch for this instead make use of a while loop and keep on listening for an int. — Vivin
– Vivin, Commented Sep 24, 2015 at 15:47
When somebody is prompted to enter a number, and enters something other than an Int I want an error message and then have them prompted to enter a valid number — Bryan Neill
– Bryan Neill, Commented Sep 24, 2015 at 15:50
Dont use try catch for this. Instead use a while loop to keep on listening till you get a valid input. — Vivin
– Vivin, Commented Sep 24, 2015 at 15:52
If I use a do while loop can I still use the InputMismatchException in the while? — Bryan Neill
– Bryan Neill, Commented Sep 24, 2015 at 15:54

Vivin · Accepted Answer · 2015-09-24 16:05:33Z

1

case "+":
                System.out.println ("Enter first number:");
                while(!scanner.hasNextInt())
                {
                    System.err.println("Enter valid int");
                    scanner.next();
                }

                num1 = scanner.nextInt();
                System.out.println ("Enter second number:");
                while(!scanner.hasNextInt())
                {
                    System.err.println("Enter valid int");
                    scanner.next();
                }

                num2 = scanner.nextInt();
                add(num1, num2);

                break;

Use something like this instead of a try catch block. This way it will listen till you get a valid num1. Similarly do for the other number too.

edited Sep 24, 2015 at 16:05

answered Sep 24, 2015 at 15:54

Vivin

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3 Comments

Bryan Neill Over a year ago

Sorry if this is a dumb question, but when I enter that into my case it won't run the rest of the code and prompt for a number

Vivin Over a year ago

Yea you will have to modify your code a bit. Dont just copy paste lol

Vivin Over a year ago

I have updatd the answer

ergonaut · Accepted Answer · 2015-09-24 15:24:16Z

1

Wrap the entire method with a try catch block. Your try/catch block is localized to a single area, it will only throw exceptions in that section.

answered Sep 24, 2015 at 15:24

ergonaut

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Comments

SlopeOak · Accepted Answer · 2015-09-24 15:25:09Z

0

You're using the Scanner.nextInt() method. When this sees an input that cannot be cast to an int, it will throw an InputMismatchException.

http://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextInt()

You can verify that the next input is of type int by first checking "hasNextInt()".

answered Sep 24, 2015 at 15:25

SlopeOak

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1 Comment

Bryan Neill Over a year ago

right, so how do I throw an error message if it is not an int

Rabbit Guy · Accepted Answer · 2015-09-24 15:32:30Z

0

It appears that your error is thrown when trying to multiply (case "*")... It is outside of a try/catch and therefore you will get this error. Either try/catch for each input or wrap the whole thing into a single swoop.

answered Sep 24, 2015 at 15:32

Rabbit Guy

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