In c++ how do I go about using pointers to get the average of an array?

Question

I'm new to programming and I'm still having trouble with arrays, pointers, and functions. I'd like to know whats wrong with this and how I can fix it. Specifically why the pointer isn't working with the function. Here is the program I'm trying to write: Write a program that DYNAMICALLY creates a pointer to an array large enough to hold a user-defined number of test scores. Once all the scores are entered (in the main function), the array should be passed to a function that RETURNS a DOUBLE for the average score. In the user output, the average score should be formatted with two decimals. Use pointer notation; do not use array notation .

#include <iostream>
#include <iomanip>
#include <memory>
using namespace std;

double getAverage(int, int);

int main()
{ 
    int size = 0;
    cout << "How many scores will you enter? ";
cin >> size;

unique_ptr<int[]> ptr(new int[size]);


cout << endl;
int count = 0;

//gets the test scores

for (count = 0; count < size; count++)
{
    cout << "Enter the score for test " << (count + 1) << ": ";
    cin >> ptr[count];
    cout << endl;
}
//display test scores
cout << "The scores you entered are:";
for (count = 0; count < size; count++)
    cout << " " << ptr[count];
cout << endl;

double avg;
avg = getAverage(ptr, size);
cout << setprecision(2) << fixed << showpoint << endl;

cout << "The average is " << avg << endl;

return 0;
}

double getAverage(int *ptr, int size)
{
double average1;
double total = 0;
for (int count = 0; count < size; count++)
{
    total = total + *(ptr + count);
}
average1 = total / size;

return average1;
}

Answer 1

First of all your function getAverage() has different prototype than that you defined. And secondly you try to pass std::unique_ptr<int []> object into a function that instead expects a int*. But std::unique_ptr<int []> is a different type than int* and not implicitly convertible. So to pass int *use std::unique_ptr::get function. like

#include <iostream>
#include <iomanip>
#include <memory>
using namespace std;

double getAverage(int *, int);

int main()
{ 
    int size = 0;
    cout << "How many scores will you enter? ";
cin >> size;

unique_ptr<int[]> ptr(new int[size]);


cout << endl;
int count = 0;

//gets the test scores

for (count = 0; count < size; count++)
{
    cout << "Enter the score for test " << (count + 1) << ": ";
    cin >> ptr[count];
    cout << endl;
}
//display test scores
cout << "The scores you entered are:";
for (count = 0; count < size; count++)
    cout << " " << ptr[count];
cout << endl;

double avg;
avg = getAverage(ptr.get(), size);
cout << setprecision(2) << fixed << showpoint << endl;

cout << "The average is " << avg << endl;

return 0;
}

double getAverage(int *ptr, int size)
{
double average1;
double total = 0;
for (int count = 0; count < size; count++)
{
    total = total + *(ptr + count);
}
average1 = total / size;

return average1;
} 

Answer 2

In the assignment there is written

Use pointer notation; do not use array notation

This means that you should not use subscripting.

In main the variable ptr is declared as having the type std::unique_ptr<int[]>.

unique_ptr<int[]> ptr(new int[size]);

You are trying to pass it to the function getAverage

avg = getAverage(ptr, size);

that has the corresponding parameter of the type int.

double getAverage(int, int);

Though then you defined the function as having the parameter of the type int * in any case there is no explicit conversion from the type std::unique_ptr<int[]> to the type int *. You should use method get of the smart pointer.

Also the function parameter should be declared like const int * because the array is not changed in the function.

The program can look the following way

#include <iostream>
#include <iomanip>
#include <memory>

double getAverage(const int *a, size_t n)
{
    double total = 0.0;

    for (const int *p = a; p != a + n; ++p) total += *p;

    return n == 0 ? total : total / n;
}

int main()
{
    size_t n = 0;

    std::cout << "How many scores will you enter? ";
    std::cin >> n;

    std::unique_ptr<int[]> ptr(new int[n]);

    std::cout << std::endl;

    size_t i = 0;
    for ( int *current = ptr.get(); current != ptr.get() + n; ++current )
    {
        std::cout << "Enter the score for test " << ++i << ": ";
        std::cin >> *current;
    }

    std::cout << std::endl;

    std::cout << "The scores you entered are:";
    for (const int *current = ptr.get(); current != ptr.get() + n; ++current)
    {
        std::cout << " " << *current;
    }
    std::cout << std::endl;

    double avg = getAverage( ptr.get(), n );

    std::cout << std::setprecision(2) << std::fixed << std::showpoint;

    std::cout << "\nThe average is " << avg << std::endl;

    return 0;
}

The program output might look like

How many scores will you enter? 10

Enter the score for test 1: 1
Enter the score for test 2: 2
Enter the score for test 3: 3
Enter the score for test 4: 4
Enter the score for test 5: 5
Enter the score for test 6: 6
Enter the score for test 7: 7
Enter the score for test 8: 8
Enter the score for test 9: 9
Enter the score for test 10: 10

The scores you entered are: 1 2 3 4 5 6 7 8 9 10

The average is 5.50

Answer 3

You almost had it. Just need to change this chunk:

using namespace std;

double getAverage(int*, int); // <--here

int main()
{
    int size = 0;
    cout << "How many scores will you enter? ";
    cin >> size;
    int *ptr = new int[size]; //<--and here (but you'll need to delete it later)
    //unique_ptr<int[]> ptr(new int[size]);

I don't recommend mixing smart pointers and standard pointers until you have a better understanding of standard pointers. edit: I mean mixing them in terms of the same effective pointer.