Question

Positive and negative:

Return these four results using C++ reference parameter

Write a function, named sums(), that has two input parameters; an array of floats; and an integer,

n, which is the number of values stored in the array. Compute the sum of the positive values in the array

and the sum of the negative values. Also count the number of positives and negative numbers in each category. Write a main program that reads no more than 10 real numbers and stores them in an array. Stop reading numbers when a 0 is entered. Call the sums() function and print the answers it returns. Also compute and print the average values of the positive and negative sets.

Answer 1

#include <iostream>

using namespace std;

void sums(float array[], int n, float &positiveSum, float &negativeSum, int &positiveCount, int &negativeCount);

int main() {
    float arr[10];
    float pos_sum, neg_sum;
    int pos_count, neg_count, size = 0;

    for (int i = 0; i < 10; ++i) {
        cout << "Enter a number(0 to stop): ";
        cin >> arr[i];
        if (arr[i] == 0) { break; }
        size++;
    }

    sums(arr, size, pos_sum, neg_sum, pos_count, neg_count);
    cout << "Sum of positive numbers is " << pos_sum << endl;
    cout << "Sum of negative numbers is " << neg_sum << endl;
    cout << "Count of positive numbers is " << pos_count << endl;
    cout << "Count of negative numbers is " << neg_count << endl;
    cout << "Average of positive numbers is " << pos_sum/pos_count << endl;
    cout << "Average of negative numbers is " << neg_sum/neg_count << endl;
    return 0;
}

void sums(float array[], int n, float &positiveSum, float &negativeSum, int &positiveCount, int &negativeCount) {
    positiveSum = 0;
    positiveCount = 0;
    negativeCount = 0;
    negativeSum = 0;
    for (int i = 0; i < n; ++i) {
        if (array[i] > 0) {
            positiveCount++;
            positiveSum += array[i];
        } else {
            negativeCount++;
            negativeSum += array[i];
        }
    }
}