在C ++中，我如何使用指针获取数组的平均值？

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.