How to convert an array of numbers from char to int?

Question

C++ / C#

How to convert an array of char to int numbers in a class that works with "long" numbers?

class Long
{
private:
  char arr[100];
public:
  Long();
  void input();
  void show();
  Long operator+(const Long& l1);
};

void Long::show()
{
  for (int i = 0; i < 5; i++)
  {
    cout << (arr1[i]);
  }
}

Long Long::operator+(const Long& l1)
{
  Long l2;
  int k = 0;
  for (int i = 4; i >= 0; i--)
  {
    if (i == 4)
    {
      if ((arr[i] + l1.arr[i]) >= 10)
      {
        l2.arr[i] = (arr[i] + l1.arr[i]) % 10;
        k = 1;
      }
      else if ((arr[i] + l1.arr[i]) < 10)
      {
        l2.arr[i] = (arr[i] + l1.arr[i]);
      }
    }
    if (i != 4)
    {
      if (k == 1)
      {
        l2.arr[i] = (arr[i] + l1.arr[i] + k) % 10;
        if ((arr[i] + l1.arr[i] + k) >= 10)
        {
          k = 1;
        }
        else if ((arr[i] + l1.arr[i] + k) < 10)
        {
          k = 0;
          continue;
        }
      }
      if ((arr[i + 1] + l1.arr[i + 1] + k) < 10)
      {
        if (k == 0)
        {
          l2.arr[i] = (arr[i] + l1.arr[i]) % 10;
          if ((arr[i] + l1.arr[i] + k) >= 10)
          {
            k = 1;
          }
          else if ((arr[i] + l1.arr[i] + k) < 10)
          {
            k = 0;
            continue;
          }
        }
        else if (k == 1)
        {
          l2.arr[i] = (arr[i] + l1.arr[i] + k);
          if ((arr[i] + l1.arr[i] + k) >= 10)
          {
            k = 1;
          }
          else if ((arr[i] + l1.arr[i] + k) < 10)
          {
            k = 0;
            continue;
          }
        }
      }
    }
  }
  return l2;
}

I understand that the code is written very clumsily, but I would like the sum of two numbers to be displayed in the show method. And in general will it work with char?

Answer 1

Show seems to work fine for numbers that are 5 digits long. You don't need to cast to int for this.

Long a, b;
a.input();
b.input();
Long sum = a + b;
sum.show();

Your code can be greatly simplified:
* Formula l2.arr[i] = (arr[i] + l1.arr[i] + k) % 10; works for the first digit too (i == 4 for you), since at the beginning k = 0.
* The value of k can also be calculated without if: k = (arr[i] + l1.arr[i] + k) / 10 ;