Question

Write a program that dynamically allocates an integer array of size of 20 to hold a user-defined number of test scores. Once all the scores are entered, the array should be passed to • a function that calculates the average score, the minimal score, and the maximal score in the array, and returns these scores to the main function (hint: use pass by reference, I gave an example in the class) • a function that searches a specific number. The function returns the index of the number in the array if found, otherwise, returns -1. o for example: array {4,-1,6,10,15, .....}, search 6, the function should return 2 (index). The program should display the above scores in the main function. Use pointer notation rather than array notation whenever possible.

Answer 1

//C++ code

// this code in c++
#include<iostream>
using namespace std;
void calculate(int* arr,int s, int &min, int &max, double &avg);
int search(int* arr,int s, int num);
//Function prototype
int main()
{
   cout << "Enter the size of array: ";
   int size;
   cin >> size;
   int* a = new int[size];
   for (int i = 0; i < size; i++)
   {
       cout << "Enter number " << (i + 1) << ": ";
       cin >> a[i];

   }
   int min=0, max=0;
   double avg=0;
   calculate(a, size, min, max, avg);
   cout << "Max = " << max<<endl;
   cout << "Min = " << min << endl;
   cout << "Avg = " << avg << endl;
   cout << "Number to be searched: ";
   int num;
   cin >> num;
   int index = search(a, size, num);
   if (index < 0)
       cout << num << " not found..." << endl;
   else
       cout << num << " is found at " << index << endl;
   //pause
   system("pause");
}
void calculate(int* arr,int s, int &min, int &max, double &avg)
{
  
   max = *arr;
   min = *arr;
   int sum = 0;
   for (int i = 0; i < s; i++)
   {
       if (min > arr[i])
           min = arr[i];
       if (max < arr[i])
           max = arr[i];
       sum += arr[i];
   }
   avg = (double)sum / s;

  
  
}
int search(int* arr,int s, int num)
{
  
   for (int i = 0; i < s; i++)

   {
       if (arr[i] == num)
           return i;
   }
   return -1;
}

//Output

Enter the size of array: 5 Enter number 1: 45 Enter number 2: 23 Enter number 3: 67 Enter number 4: 43 Enter number 5: 12 (Мax - 67 Min Avg Number to be searched: 43 43 is found at 3 Press any key to continue . 12 38

//If you need any help regarding this solution..... please leave a comment.......... thanks