How to store each user input into array? - c++

Question

I used a do while loop to ask user to enter integer as long as its not 0 or it will exit the program. I'm stuck on how to store every user input into the dynamically allocated array.
#include <iostream>
using namespace std;
int main() {
int *A;
A= new int();
int n;
do{
cout<<"Enter integer: "<<endl;
cin>>n;
cout<< *A + n << endl;
}while(n!=0);
return 0;
}

Answer 1

You are allocating a single int, not an array of ints.
Also, even if you were allocating an array, a statement like *A + n does not add n to the array. It dereferences A to access the value of the 1st int in the array, and then adds n to that value.
Try something more like this instead:
#include <iostream>
using namespace std;
int main() {
int *A = nullptr;
int count = 0, n;
do{
cout << "Enter integer: " << endl;
cin >> n;
if (n == 0) break;
int *newA = new int[count+1];
for(int i = 0; i < count; ++i) newA[i] = A[i];
newA[count] = n;
delete[] A;
A = newA;
++count;
cout << A[count-1] << endl;
}
while (true);
delete[] A;
return 0;
}
Alternatively, try to avoid reallocating the array on every input, eg:
#include <iostream>
using namespace std;
int main() {
int *A = new int[5];
int count = 0, cap = 5, n;
do{
cout << "Enter integer: " << endl;
cin >> n;
if (n == 0) break;
if (count == cap)
{
int newCap = cap * 1.5;
int *newA = new int[newCap];
for(int i = 0; i < count; ++i) newA[i] = A[i];
delete[] A;
A = newA;
cap = newCap;
}
A[count] = n;
++count;
cout << A[count-1] << endl;
}
while (true);
delete[] A;
return 0;
}
That being said, a better option is to use a std::vector, which will handle the dynamic memory for you, eg:
#include <iostream>
#include <vector>
using namespace std;
int main() {
vector<int> A;
int n;
do{
cout << "Enter integer: " << endl;
cin >> n;
if (n == 0) break;
A.push_back(n);
cout << A.back() << endl;
}
while (true);
return 0;
}