CODEWITHSUNDEEP
javaexception
EricGrahamMacEachern
EricGrahamMacEachern

Reputation: 37

program finishes after exception is handled in Java

This is for homework.

I am trying to create a program that takes the average of ten input numbers. When the user enters a character that is not a number, the NumberFormatException exception is caught. The program was finishing after the exception was caught, so I changed it to use recursion to call the method again after the exception is caught, but now it prints multiple averages, some of which are not correct.

How do I change the program so that it continues to ask for input after an exception has been caught instead of finishing? I do not want the program to finish after catching the exception. 

import java.util.Scanner;

import java.util.Scanner;

public class PrintAverage {

    int average;

    public static void main(String args[]) {
        System.out.println("You are going to enter ten numbers to find their average.");
        getInput();
    }

    private static void getInput() {
        String input;
        int sum = 0;
        int[] arrayOfIntegers = new int[10];
        double average = 0;
        Scanner scanner = new Scanner(System.in);

        try {
            for (int i = 0; i < arrayOfIntegers.length; i++) {
                System.out.println("Enter the next number.");
                input = scanner.nextLine();
                arrayOfIntegers[i] = Integer.parseInt(input);
            }
        } catch (NumberFormatException exception) {
            System.out.println("The last entry was not a valid number.");
            getInput();
        }

        for (int k = 0; k < arrayOfIntegers.length; k++) {
            sum = sum + arrayOfIntegers[k];
        }
        average = sum / arrayOfIntegers.length;
        System.out.println("The average is " + average);

    }
}

Upvotes: 1

Views: 54

Answers (2)

Uncle Iroh
Uncle Iroh

Reputation: 6045

Localize the exception. Change this:

try {
        for (int i = 0; i < arrayOfIntegers.length; i++) {
            System.out.println("Enter the next number.");
            input = scanner.nextLine();
            arrayOfIntegers[i] = Integer.parseInt(input);
        }
    } catch (NumberFormatException exception) {
        System.out.println("The last entry was not a valid number.");
        getInput();
    }

to this:

    for (int i = 0; i < arrayOfIntegers.length; i++) {
        System.out.println("Enter the next number.");
        input = scanner.nextLine();
        try {
           arrayOfIntegers[i] = Integer.parseInt(input);
        } catch (NumberFormatException exception) {
            System.out.println("The last entry was not a valid number.");
            i--; //so you don't lose one of the 10.
        }
    }

Upvotes: 1

pyb
pyb

Reputation: 5269

Your try/catch is not specific enough, so it catches the exception after asking all the numbers you want.

  1. Identify the line that is throwing the NumberFormatException
  2. Move your try/catch around it
  3. Store the input in a temporary variable that can be null (primitive int don't allow that)
  4. Keep asking for a value as long as it's null (= as long as it's not been set to a legit number)

Upvotes: 3

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