CODEWITHSUNDEEP
c++arraysfunctionlong-integerdigits
Ulet Natalius
Ulet Natalius

Reputation: 1

Problem in digitizing a number and putting it in an array reversed

I am writing a C++ function which takes a positive number and returns an array containing its digits reversed.

#include <cmath>
#include <vector>

std::vector<int> digitize(unsigned long n) {
  int x = 1;
  for (int i = 1; n >= pow(10,i-1); ++i) { // Supposed to return the correct number of digits
    x = i;
  }
  std::vector<int> result;
  for (x; x > 0; --x) { // Takes the number of digits and begin assembling the array in reverse
    result[x] = n / pow(10,x-1);
  }
  return result;
}

During compiling, it returned warnings, and I'm pretty sure the function hasn't even done what I intended it to do yet. One of the warnings had something to do with the for (x; x > 0; --x) loop (some expression is unused). Shouldn't variable x carry over its value from the previous loop?

I tried modifying the second loop to use i instead of x, but as expected, the value and initialization of variable i didn't carry over from the first loop.

Upvotes: -2

Views: 177

Answers (2)

Pepijn Kramer
Pepijn Kramer

Reputation: 13076

For example you can do something like this (not this answer is not only for you but also for more people starting out with C++)

#include <list>
#include <string>
#include <iostream>

// I just return a string since it is more easy to output the result
auto to_string(std::uint64_t value)
{
    std::list<char> result;
    while( value > 0 )
    {
        // with a list you can insert values in front of all other values
        // push the new `digit` in front of everything else (the `0` is there to make a nice output string
        // the module operator % will get the current last digit of the value
        result.push_front(static_cast<char>(value % 10) + '0');

        // dividing by 10 results in the "removal" of the last digit
        value /= 10ul;
    }
    
    // convert the list of characters to a string (can be done for vector too)
    return std::string{result.begin(), result.end()};
}

int main()
{
    std::cout << to_string(1234) << "\n";
    std::cout << to_string(12340) << "\n";
}

Upvotes: 1

PaulMcKenzie
PaulMcKenzie

Reputation: 35454

I would just rewrite the code to do a simpler approach.

  1. Check for the special case of n being 0.
  2. Write a loop that gets modulo 10 of n and store that value in the vector using push_back.
  3. Divide n by 10
  4. Repeat steps 2) and 3) until n is 0

Example:

#include <vector>
#include <iostream>

std::vector<int> digitize(unsigned long n) 
{
    std::vector<int> answer;
    if ( n == 0 )
        return {0};
    while (n > 0)
    {
        answer.push_back(n % 10);
        n /= 10;
    }
    return answer;
}

int main()
{
    auto v = digitize(1232196);
    for (auto val : v)
        std::cout << val;
}

Output:

6912321

Upvotes: 2

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