CODEWITHSUNDEEP
c++
Arian Hassani
Arian Hassani

Reputation: 99

how to get large numbers as input?

Hi I'm really new to c++ and I wanted to write a code which receives a number from user and sums its digits and keeps doing that until it gets a one-digit number and returns it as a result. But I noticed that when my number is large (like 15 digits long), the wrong number is stored in the variable i declared for storing user input. What do I do?

#include <iostream>
#include <string>
#include <sstream>
using namespace std;

int get_sum(long x) {
    cout << x<<endl;
    if (x < 10) {
        return x;
    }
    else {
        string num_in_str = to_string(x);
        long result=0;
        for (int i = 0; i < num_in_str.size(); i++) {
            int digit = num_in_str[i] - '0';
            result += digit;
        }
        return get_sum(result);

    }
}
int main()
{
    long input;
    cin >> input;
    int final_result = get_sum(input);``
    cout << final_result;

}

Upvotes: 1

Views: 1985

Answers (2)

ivan.ukr
ivan.ukr

Reputation: 3551

Integer data types in C++ can store numbers up to certain limit. In particular, depending on platform and compiler, "long" can be 32 or 64 bit, and store up to 2^31-1 or 2^63-1. If you want to process numbers of an arbitrary precision, I'd suggest to read each input as string, and process it character-by character, like this:

#include <cctype>
#include <iostream>
#include <string>

int main()
{
    std::string s; 
    while (std::getline(std::cin, s)) {
        // parse string
        long sum = 0;
        for (std::size_t i = 0; i < s.length(); ++i) {
            if (std::isdigit(s[i]))
                sum += s[i] - '0';
            else
                break;
        }
        std::cout << sum << std::endl;
        if (sum < 10) break;
    }
    return 0;
}

Upvotes: 1

Arian Hassani
Arian Hassani

Reputation: 99

Oh I found out. I had to use uint64_t data type

Upvotes: 1

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