how to store input in a dynamic array?

Question

So I am trying to make a program that stores user input into a dynamic array but I cant do it right. When I try to put a number let us say 1, and I want to try again then I want to view history, the only thing that shows up is the last number I have input. And sometimes there is a large number sowing up like 1214098101909279242 like that.

Here is my code:

#include<iostream>
    using namespace std;
    int main(){
        const int size = 20;
        int *num = new int[size];
        char answer;
        while(true){
        cout<<"ENTER NUMBER: \n";
        cin>>*num;
        cout<<"TRY AGAIN? ";
        cin>>answer;
            switch(answer){
                case 'y':
                    num[size+1];
                    system("cls");
                    break;
                default:
                    cout<<"INPUT HISTORY: \n";
                    for(int i=0;i<=size;i++){
                        cout<<num[i];
                    }
            }
        }
        return 0;
    }

Answer 1

I have two major issues with your code:

1. There is no need for dynamic allocation in this case, since size is a const value (and can be defined as constexpr, which is better).
2. instead of using the standard containers (such as std::array (for constant size array) or std::vector (for dynamically expanding array)) you use a plain pointer array, which is ill-advised, and in this case can cause memory leak (for example, if you move this code into an utility function, since you never delete the memory of num array!), you should use std::unique_ptr instead! (and better read about the RAII idiom).
3. Spacing and indentation - your code should be readable to other coders as well.

About your code, the original code have the following line:

cin >> *num

which means put the value in stdin into the FIRST position of num. In both C/C++, arrays are just pointers to the first address of memory that was allocated for the array. In order to access the other items, you need to use the operator[] or use pointer arithmetic with the operator* (for example:

cin >> *(num + count)

will get the item at position count). Also, as pointed above, there is no safety measure for disallowing writing to an invalid memory that you can count on. Therefore, you should use the std containers which enforce index safety.

Just for reference, here is the code I would have wrote to get the same functionality:

int main() {
    constexpr size_t SIZE = 20;
    std::array<int, SIZE> arr;
    arr.fill(0);

    int last_filled_position = 0;
    bool cont = true;

    while (cont)
    {
        int val;
        char answer;

        cout << "ENTER NUMBER:" << std::endl;

        cin >> val;

        cout<<"TRY AGAIN? Yes/Print/Stop";

        cin >> answer;

        arr.at(last_filled_position++) = val;

        switch (answer)
        {
            case 'y':
                system("cls");
                break;
            case 'p':
                cout << "INPUT HISTORY: " << std::endl;

                for (int val: arr) // With more complex types, you should use const auto&
                {
                    cout << val << ", ";
                }

                cout << endl;
                break;
            case 's':
                cont = false;
                cout << "STOPPING";
                break;
        }
    }

    return 0;
}

Answer 2

So I have change the code a little bit. Is my code valid for some reason? because it works fine now. I can input numbers and when i view all the numbers, it works fine.

#include<iostream>
using namespace std;
int main(){
    const int size = 20;
    int *num = new int[size];
    char answer;
    int count = 1;
    while(true){
    cout<<"ENTER NUMBER: \n";
    cin>>num[count-1];
    cout<<"TRY AGAIN? ";
    cin>>answer;
        switch(answer){
            case 'y':
                count++;
                system("cls");
                break;
            default:

                cout<<"INPUT HISTORY: \n";
                for(int i=0;i<=count-1;i++){
                    cout<<num[i]<<endl;
                }
                count++;
        }
    }
    return 0;
}