CODEWITHSUNDEEP
c++
Adam Pawelski
Adam Pawelski

Reputation: 1

Is there a way to not include a negative number in an average, when entering a negative number is how you terminate the program?

Sorry about last time for those who saw my previous thread. It was riddled with careless errors and typos. This is my assignment:

"Write a program that will enable the user to enter a series of non-negative numbers via an input statement. At the end of the input process, the program will display: the number of odd numbers and their average; the number of even numbers and their average; the total number of numbers entered. Enable the input process to stop by entering a negative value. Make sure that the user is advised of this ending condition."

And here is my code:

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    int number, total1=0, total2=0, count1=0, count2=0;
    do
    {
        cout << "Please enter a number. The program will add up the odd and even ones separately, and average them: ";
        cin >> number;

        if(number % 2 == 0)
        {
            count1++;
            total1+=number;
        }
        else if (number >= 0)
        {
            count2++;
            total2+=number;
        }
    }
    while (number>=0);
        int avg1 = total1/count1;
        int avg2 = total2/count2;
        cout << "The average of your odd numbers are: " << avg1 << endl;
        cout << "The average of your even numbers are " << avg2 << endl;
}

It seems to be working fine, but when I enter a negative number to terminate the program, it includes it with the rest of the averaged numbers. Any advice to get around this? I know it's possible, but the idea escapes me.

Upvotes: 0

Views: 5077

Answers (3)

Mark Garcia
Mark Garcia

Reputation: 17708

After here:

cout << "Please enter a number. The program will add up the odd and even ones seperately, and average them: ";
cin >> number;

Immediately check if the number is negative

if(number < 0) break;

Now you wouldn't need to use your do-while loop in checking if the number is negative. Thus, you can use an infinite loop:

while(true) {
   cout << "Please enter a number. The program will add up the odd and even ones seperately, and average them: ";
   cin >> number;

   if(number < 0) break;

   // The rest of the code...
}

ADDITIONAL: There is something wrong in your code. You aren't showing the user how much the number of even and odd numbers are, and the total number of numbers entered.

ANOTHER ADDITIONAL: You should use more meaningful variable names:

int totalNumEntered = 0, sumEven = 0, sumOdd = 0, numEven = 0, numOdd = 0;

Of course I am not limiting you to these names. You can also use other similar names.

FOR THE INTEGER DIVISION PROBLEM: You must cast your expression values to the proper type (in this case, it is float). You should also change the averages variables' types to float:

float avg1 = float(total1) / float(count1);
float avg2 = float(total2) / float(count2);

Upvotes: 1

adzm
adzm

Reputation: 4136

Immediately after cin >> number, check for < 0, and break if so. Try to step through the program line by line to get a feel for the flow of execution. Have fun learning, and good luck!

Upvotes: 0

Kerrek SB
Kerrek SB

Reputation: 477388

Your main loop should be like this:

#include <iostream>

for (int n; std::cout << "Enter a number: " && std::cin >> n && n >= 0; )
{
    // process n
}

Or, if you want to emit a diagnostic:

for (int n; ; )
{
    std::cout << "Enter a number: ";

    if (!(std::cin >> n)) { std::cout << "Goodbye!\n"; break; }

    if (n < 0) { std::cout << "Non-positve number!\n"; break; }

    // process n
}

Upvotes: 2

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