Converting ASCII representation of positive numbers to integer

A number can be represented as ASCII characters (char) or integers (int). Thus, the question is how we can convert from char to int. For example, we want to convert '123' to 123 or '45' to 45.

To solve this problem, we need to revisit the ASCII table for a little bit. Moreover, we are interested in the characters '0' to '9' in the table. The special thing about those characters is that they stand consecutively in the table. That means their decimal values are consecutive positive integers. For example, '0' is 48; '1' is 49; and '2' is 50. Thus, if we have '1' - '0', we'll get an integer value of 1 because 49 - 48 = 1. That's how we're gonna convert the number from char to int. Here is the algorithm in C++:

#include<iostream>
#include<math.h>
using namespace std;

int digitsToInt(char arrayOfChars[], int size)
{
  int digitPos = size - 1;
  int result = 0;
  for (int i = 0; i < size; i++)
  {
    result += ((int)(arrayOfChars[i] - '0')) * pow(10, digitPos);
    digitPos--;
  }
  return result;
}

Explanation: our function takes an array of characters containing the char representation of the number and the size of the array.

1. digitPos keeps track of which digit we're converting. It is important to do so because we must convert the digit into appropriate value in integer (remember that ASCII digits are between '0' and '9'). If the number passed in is '102', then the first digit '1' represents 100 in integer. The algorithm must know that and multiply '1' with 100 after converting it into int 1.
2. result is the final number in int. Our strategy here is to convert each digit into its appropriate int value and then add all of them up. That will give us the final result in integer. For example, if we need to convert '12', we will convert '1' to 1 first and multiply it by 10. Next, we convert '2' to 2 and multiply it by 1 (2 is the last digit). Therefore, the final result is 10 + 2 = 12.
3. We loops through the array and convert each digit into the appropriate int value and add that value into the result. As explained above, by subtracting the digit by '0' and casting the result into int we'll get the correct int number corresponding to that digit. However, that's not all. Notice how we use the function pow(10, digitPos) to get the correct value for each digit. This method works because our numbers are base 10 numbers. Thus, by multiplying a number by 10 to the power of its position, we'll get the correct value for that number. Don't worry if you are confused, we'll do an example later.
4. Lastly, we decrement digitPos by one because we will move to process the next digit which has one order less than the current number.
5. Return the final result which is the sum of all the int values of our digits.

Example: we'll convert the number '123' into integer.

1. Our digitPos is 2 because we process the first digit '1' first and it is in the hundredth position. So, '1' - '0' gives int 1. Then, 1 * 10 ^ 2 = 100. Add that into our result. We have 100 after the first iteration. Moreover, digitPos decreases by 1 because we are about to process the tenth digit.
2. Second iteration, digitPos is 1. And we have '2' - '0' = int 2. Multiplying 2 by ten to the power of its position, we have 2 * 10 ^ 1 = 20. Add that value to the result and we end up with the sum of 120. Finally, we decrement digitPos by 1.
3. Last iteration, digitPos is 0. And, '3' - '0' = int 3. Multiplying 3 by ten to the power of its position, we have 3 * 10 ^ 0 = 3. Add 3 into the result and we have int 123. Loop ends and our algorithm returns the result of 123.

Thanks for reading and as always feel free to leave any comment or suggestion :)