How can I find odds and evens using functions and then find the sum of the total amount of odds and the sum of the total amount of evens?

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Hello

So I'm trying to create program which will ask the user to enter 10 integers and then check if the numbers that they enter are odds or evens by using functions like int numOfOdd , int numOfEven or something similar to that. After that I want to find the sum of the total amount of odds and the sum of the total amount of evens and then print everything out with a function maybe like this one void outputResults ( int numOfOdd, int oddTotal, int numOfEven, int evenTotal ); If there's anyone can help with this, I thank you in advance. Also, I said that I want to check the odds and evens using functions but down here you can see that I used loops as I found making functions is kind of confusing, but if someone can show me how to do it with function, that would be awesome too.

Down here is my code

What I have tried:

C++

<pre>
#include<iostream>

#include<string>

using namespace std;

int main()
{
    int numbers[10], even = 0, odd = 0, i;
    cout << "Please enter 10 integers ";
    for (i = 0; i < 10; i++)
        cin >> numbers[i];
    for (i = 0; i < 10; i++)
    {
        if (numbers[i] % 2 == 0)
            even++;
        else
            odd++;
    }
    cout << "The amount of even numbers: " << even;
    cout << "The amount of odd numbers: " << odd;
    cout << endl;
    return 0;
}

Posted 15-Dec-21 5:18am

Tiến Dũng Đỗ

Updated 15-Dec-21 20:35pm

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Comments

Richard MacCutchan 15-Dec-21 11:44am

In the loop where you check for even or odd numbers you can add each number to the respective sums at the same time.

3 solutions

Solution 2

You are almost there. Try this :

C++

#include<iostream>

#include<string>

using namespace std;

inline bool isEven( int value )
{
    return (value  % 2 == 0);
}


int main()
{
    const int numberCount = 10;
    int sumEven = 0;
    int sumOdd = 0;
    int numbers[numberCount], even = 0, odd = 0, i;

    cout << "Please enter " << numberCount << " integers ";
    for (i = 0; i < numberCount; i++)
        cin >> numbers[i];

    for (i = 0; i < numberCount; i++)
    {
        if( isEven( numbers[i] )
        {
            even++;
            sumEven += numbers[i];   // update the sum of even values
        }
        else
        {
            odd++;
            sumOdd += numbers[i];    // update the sum of odd values
        }
    }

    cout << "The count of even numbers: " << even << "their sum is " << sumEven;
    cout << "The count of odd numbers : " << odd  << "their sum is " << sumOdd;
    cout << endl;
    return 0;
}

If you want, and I usually do, you can put all of your input values in a file and then run the program with input redirected from that file. This way you don't have to enter the values every time. You can do this by typing this at the command line : MyProgram << InputData.txt. With the appropriate names for your program and the input file. The input file should have one value per line. This will make things much easier I think.

Posted 15-Dec-21 6:40am

Rick York

Updated 15-Dec-21 18:00pm

v2

Comments

CPallini 15-Dec-21 16:14pm

Hey Rick, have a second look at your isEven function.
Also, have a second look at your the summing up code.

Tiến Dũng Đỗ 15-Dec-21 21:45pm

Yes sir, there're 4 errors in there

Rick York 15-Dec-21 23:57pm

You are progressing if you can see the errors - or did the compiler tell you? :)

Rick York 15-Dec-21 23:58pm

It's a puzzle - let's see if he can fix it. :)

Tiến Dũng Đỗ 16-Dec-21 0:57am

Yes sir, I did fix them :) But for the summing up code, I tried both your way and this: sumOfEven = sumOfEven+i; it will only print out 10 no matter what

CPallini 16-Dec-21 2:02am

:-D

Solution 3

Let's focus on odd values.
'Traditional' implementation

C++

<iostream>
using namespace std;

enum { N = 10 };

bool odd(int i);
int count_of_odd(int a[], int size);
int sum_of_odd(int a[], int size)


int main()
{
  int a[N];
  cout << "please enter " << N << " integers\n";

  for (int n=0; n<N; ++n)
  {
    cin >> a[n];
    if ( odd(a[n] ))
      cout << a[n] << " is odd\n";
  }

  cout << "count of odd integers " << count_of_odd(a, N) << "\n";
  cout << "sum of odd integers " << sum_of_odd(a, N) << "\n";
}

bool odd(int i) {return (i & 1) == 1;}

int count_of_odd(int a[], int size)
{
  int count = 0;
  for (int n = 0; n<size; ++n)
    count += odd(a[n]) ? 1 : 0;

  return count;
}

int sum_of_odd(int a[], int size)
{
  int sum = 0;
  for (int n = 0; n<size; ++n)
    sum += odd(a[n]) ? a[n] : 0;

  return sum;
}

'Less-traditional' implementation

C++

#include <iostream>
#include <array>
#include <algorithm>
#include <numeric>

using namespace std;

enum { N = 10 };
int main()
{

  auto odd = [] (int i) { return ((i & 1) == 1); };
  auto add_odd =  [odd] (int sum, int i) { return (sum + (odd(i) ? i : 0)); };

  array<int, N> a;
  cout << "please enter " << N << " integers\n";

  for (auto & x : a)
  {
    cin >> x;
    if ( odd(x) )
      cout << x << " is odd\n";
  }

  cout << "count of odd integers " << count_if( a.begin(), a.end(), odd) << "\n";
  cout << "sum of odd integers " << accumulate( a.begin(), a.end(), 0, add_odd ) << "\n";

}

Choose your poison...

Posted 15-Dec-21 20:35pm

CPallini

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OriginalGriff · Accepted Answer · 2021-12-15T06:09:00

Create a function called numOfEven which accepts and array and which returns an integer.
In the function, count the even numbers - you have code for that - and return it.
Do the same for a numOfOdd function.
Then call them, save the results, and sum the results to a total varaibale
Print the results.

Simple.