Why does my program ignore zero when counting even numbers?

Question

I'm having an issue when trying to count the number of even integers.

This is the code I'm working with:

int input=0, numeven=0;
Scanner scan = new Scanner(System.in);

input = scan.nextInt();

while (input != 0)
{
    //calculates the total number of even integers
    if (input%2 != 1)
    {
        numeven = numeven+1;
    }
}

I don't know how to set up the while loop: while (input! = 0)

Given the test input 6, 4, -2, 0 it says that I have three even numbers, but the expected outcome is 4 (because 0 is even).

As written here, this looks like an infinite loop if input != 0. scan.nextInt() is out of the while loop. Should it be in the while loop instead? — Auberon, Jan 12 '16 at 18:50
Use a value that's not going to be part of your data as a [sentinel value](https://en.wikipedia.org/wiki/Sentinel_value). -99 is a popular choice for student projects. — Bill the Lizard, Jan 12 '16 at 18:51
How does that code even work? You are only getting one number from user and it is in 4th line. How can you count that you have 3 even numbers if you got only one number from user? Is this your real code? — ctomek, Jan 12 '16 at 18:52
This is just a small part of my code, I end the while loop later on. There's actually a lot of different calculations that I have to do with the inputs. — gnollwarden, Jan 12 '16 at 18:56
I have 3 other if statements that are supposed to calculate different numbers. — gnollwarden, Jan 12 '16 at 18:57
I don't want my while loop to end on a certain value, I want it to look through all the possible numbers that I user can type in. — gnollwarden, Jan 12 '16 at 19:00

score 3 · Accepted Answer · answered Jan 12 '16 at 18:54

3

If you want your loop to work on zero, and treat it as the exit mark too, switch from while to do/while:

do {
    input = scan.nextInt();
    //calculates the total number of even integers
    if (input%2 != 1)
    {
        numeven = numeven+1;
    }
} while (input != 0);

This way your code will process zero along with regular inputs, and stop reading further input upon reaching the end of the loop.

answered Jan 12 '16 at 18:54

Sergey Kalinichenko

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@KoebmandSTO Of course it doesn't. It's implied in the OP's program that he *wants* the input to stop upon reaching zero, so `4` will be ignored by his design. – Sergey Kalinichenko Jan 12 '16 at 19:14
Wanting to count 0s as even numbers imply you want to count them all. Else could just do a numeven++ upon end. – Koebmand STO Jan 12 '16 at 19:16

WalterM · Answer 2 · 2016-01-12T19:07:26.010

You don't want the loop to break when the user enters a 0 or any other integer incase you want to put 0 multiple times.

int numeven=0;
Scanner scan = new Scanner(System.in);

while (true) {
    String input = scan.next();
    try {
        int val = Integer.parseInt(input);
        if (val % 2 == 0)
            numeven++;

    } catch (NumberFormatException e) {
        //enter any input besides an integer and it will break the loop
        break;
    }
}

System.out.println("Total even numbers: " + numeven);

Alternatively this does the same thing. Except it won't consume the last value.

int numeven=0;
Scanner scan = new Scanner(System.in);

while (scan.hasNextInt()) {
    int val = scan.nextInt();
    if (val % 2 == 0)
        numeven++;
}

System.out.println("Total even numbers: " + numeven);

score -1 · Answer 3 · answered Jan 12 '16 at 18:52

-1

Just make the condition of your while loop to be

while( scan.hasNextInt() )

Then it will only loop as long as there are numbers. Inside the loop you can

input = scan.nextInt()

answered Jan 12 '16 at 18:52

ganlaw

1,303
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Why does my program ignore zero when counting even numbers?

3 Answers3