C + + Can I use arrays here to shorten my code?

Question

This is my first post on here so please don't kill me for my noobishness.

I recently made a program for fun to put in a ton of numbers and have it put out the mean, not very useful but I thought I would see if I could. I would love it if someone could explain to me how I could improve my code using arrays instead of lots of variables, but still achieve the same thing, maybe even more efficiently.

My code looks like this:

#include "stdafx.h"
#include <iostream>

using namespace std;

int main() {

    int q1;
    int q2;
    int q3;
    int q4;
    int q5;
    int q6;
    int q7;
    int q8;
    int q9;
    int q10;
    int q11;
    int q12;

    int f;
    //Used for the total of all values

    int t;
    //Used for the total to be divided

    int a;
    //Used for dividing the numbers.
    cout << "We will be finding a mean. Enter the amount of numbers that will be        entered, the maximum is 12: ";

    cin >> a;

    cout << "Now enter what numbers you want to find the mean for, because the maximum is 12, if you have less than 12, enter 0 for the rest: ";
    cin >> q1;
    cin >> q2;
    cin >> q3;
    cin >> q4;
    cin >> q5;
    cin >> q6;
    cin >> q7;
    cin >> q8;
    cin >> q9;
    cin >> q10;
    cin >> q11;
    cin >> q12;

    f = q1 + q2 + q3 + q4 + q5 + q6 + q7 + q8 + q9 + q10 + q11 + q12;

    cout << f / a << '\n';


    system("pause");
}

Any advice is very appreciated! This was made in Visual Studio just in case you needed to know.

Answer 1

Of course arrays can make your life easier!

Here's how you could have accomplished the same task as above, with arrays:

#include "stdafx.h"
#include <iostream>

using namespace std;

int main() {

    int totalNums;
    cout << "We will be finding a mean.\n";
    cout << "You can only enter up to 12 numbers;

     // Declare an array to hold 12 int's
    int nums[12];

    // i will count how many numbers have been entered
    // sum will hold the total of all numbers
    int i, sum = 0;
    
    for(i = 0; i < 12; i++) {
        cout << "Enter the next number: ";
        cin >> nums[i];
        sum += nums[i];
    }    

    cout << "The mean is: " << (sum / totalNums) << '\n';


    //Try to avoid using system!
    system("pause");
}

---

But, why use an array?

There's no need to keep any of the numbers after you add them to the total, so why use an array?

You can accomplish the same task without an array and with only one variable for the numbers!

#include "stdafx.h"
#include <iostream>

using namespace std;

int main() {

    int totalNums;
    cout << "We will be finding a mean.\n";
    cout << "Enter the amount of numbers that will be entered: ";
    cin >> totalNums;

    // i will count how many numbers have been entered
    // sum will hold the total of all numbers
    // currentNum will hold the last number entered
    int i, sum = 0, currentNum = 0;
    
    for(i = 0; i < totalNums; i++) {
        cout << "Enter the next number: ";
        cin >> currentNum;
        sum += currentNum;
    }    

    cout << "The mean is: " << 1.0 * sum / totalNums << '\n';


    //Try to avoid using system!
    system("pause");
}

Answer 2

Arrays can be considered as series of variables each of which has ids. integers between 0 and (number of elements) - 1 (both inclusive) are available ids.

Using that with loop, your code can be like this (sorry, I hate stdafx.h):

#include <cstdlib>
#include <iostream>

using namespace std;

int main() {
    int q[12];

    int f;
    //Used for the total of all values

    int t;
    //Used for the total to be divided

    int a;
    //Used for dividing the numbers.
    cout << "We will be finding a mean. Enter the amount of numbers that will be        entered, the maximum is 12: ";

    cin >> a;

    cout << "Now enter what numbers you want to find the mean for, because the maximum is 12, if you have less than 12, enter 0 for the rest: ";
    for (int i = 0; i < 12; i++) {
        cin >> q[i];
    }

    f = 0;
    for (int i = 0; i < 12; i++) {
        f += q[i];
    }

    cout << f / a << '\n';

    system("pause");
}

You may use the numbers read in the future, but currently the numbers aren't used except for calculating the sum, so you can omit the array and do addition while reading. Also I deleted the variable t, which is unused and stopped using using namespace std;, which is not considered as good.

#include <cstdlib>
#include <iostream>

using std::cin;
using std::cout;

int main() {
    int q;

    int f;
    //Used for the total of all values

    int a;
    //Used for dividing the numbers.
    cout << "We will be finding a mean. Enter the amount of numbers that will be        entered, the maximum is 12: ";

    cin >> a;

    cout << "Now enter what numbers you want to find the mean for, because the maximum is 12, if you have less than 12, enter 0 for the rest: ";
    f = 0;
    for (int i = 0; i < 12; i++) {
        cin >> q;
        f += q;
    }

    cout << f / a << '\n';

    system("pause");
}

Answer 3

You marked this question as C++.

I recommend you do not use "using", and you should prefer vector over array.

Consider the following approach:

#include <iostream>
#include <vector>

int main(int argc, char* argv[]) 
{
   std::cout << "We will be finding a mean." << std::endl
             << "Enter numbers, and press ^d when complete.\n" 
             << std::endl;

   // Declare a vector to hold user entered int's
   std::vector<int> intVec;

   // the vector automatically keeps track of element count

   do {
      std::cout << "number: ";   // prompt
      int t = 0;
      std::cin >> t;             // use std::cin, 
      if(std::cin.eof()) break;  // ^d generates eof()
      intVec.push_back(t);
   }while(1);

   // there are several way to sum a vec, 
   // this works fine:
   int sum = 0;
   for (auto i : intVec)  sum += i;

   std::cout << "\n  sum : " << sum
             << "\ncount : " << intVec.size()
             << "\n mean : " << (sum / intVec.size()) << std::endl;

   return(0);

}    

You can enter single item per line (neatness counts).

You can enter multiple integers separated by white space, but the above will give back a prompt for the integers already entered.

^d - generates an end of file input.
... press and hold 'Control' key and letter 'd' at same time

Note - does not handle error input - try entering a 'number' as 'num' string.

Answer 4

The accepted answer is definitely the most efficient way to transform your code using arrays, but one thing I would add is that in C++ dividing an integer by another integer can only ever result in an integer, and because you're trying to get the mean, it seems like you'd want to have the result in decimals, so you need to do one of two things:

1. Declare sum as a float for the purposes of diving it by totalNums to get the mean.
2. Cast one of the integers to either a float or a double so that the decimals won't get truncated, so the last cout statement would look like this:

cout << "The mean is: " << (double)sum/totalNums << endl;

In C++ the default for precision is 6, but you can change the number of decimal points that are displayed by adding #include <iomanip> and using the setprecision( ) function in the iomanip, which you can just add in the same output line:

cout << setprecision(x) << "The mean is: " << (double)sum/totalNums << endl;

where x is whatever precision you want.

---

If you want to try using dynamic memory

This is definitely not necessary for what you're doing, but it's interesting stuff and good practice!

One more thing is that if you want to be able to have the user enter integers indefinitely, you can dymanically allocate memory during runtime by declaring a array of pointers to integers (so it's an array of address locations instead of an array of integers) and some sentinal value so they can decide when to stop. That code would look like:

#include <iostream>
#include <iomanip>

using namespace std;

main( ) {
    const int ARRAY_SIZE = 200;
    const int SENTINAL = -999;
    int totalNums = 0;
    int sum = 0;

    //declare an array of pointers to integers so
    //the user can enter a large number of integers
    //without using as much memory, because the memory
    //allocated is an array of pointers, and the int
    //aren't allocated until they are needed

    int *arr[ARRAY_SIZE];

    cout << "We will be finding a mean." << endl;
    cout << "Enter integers (up to 200) or enter -999 to stop" << endl;

    //add a conditional into the for loop so that if the
    //user enters the sentinal value they will break out
    //of the loop
    for (int c = 0; c < ARRAY_SIZE; c++) {

        //every time you iterate through the loop, create a new
        //integer by using the new keyword using totalNums as
        //the index
        arr[totalNums] = new int;
        cout << "Enter integer: ";

        //input into the array of pointers by dereferencing it
        //(so it refers to what the pointer is pointer to instead
        //of the pointer)
        cin >> *arr[totalNums];
        if (*arr[totalNums] == SENTINAL)
            break;
        else {
            sum += *arr[totalNums];
            totalNums++;
        }
    }

    cout << setprecision(3) << "The mean is: " << (float)sum / totalNums << endl;
}