Count Number of Digits in a Number in C++

  1. Use std::to_string and std::string::size Functions to Count Number of Digits in a Number in C++
  2. Use std::string::erase and std::remove_if Methods to Count to Count Number of Digits in a Number in C++

This article will demonstrate multiple methods about how to count the number of digits in a number C++.

Use std::to_string and std::string::size Functions to Count Number of Digits in a Number in C++

The most straightforward way to count the number of digits in a number is to convert it to the std::string object and then call a built-in function of the std::string to retrieve a count number. In this case, we implemented a separate templated function countDigits that takes a single argument, which is assumed to be an integer type and returns the size as an integer. Note that the following example outputs incorrect numbers for negative integers since it counts the sign symbol in too.

#include <iostream>
#include <vector>
#include <string>

using std::cout; using std::cerr;
using std::endl; using std::string;
using std::to_string;

template<typename T>
size_t countDigits(T n)
{
    string tmp;

    tmp = to_string(n);
    return tmp.size();
}

int main() {
    int num1 = 1234567;
    int num2 = -1234567;

    cout << "number of digits in " << num1 << " = " << countDigits(num1) << endl;
    cout << "number of digits in " << num2 << " = " << countDigits(num2) << endl;

    exit(EXIT_SUCCESS);
}

Output:

number of digits in 1234567 = 7
number of digits in -1234567 = 8

To remedy the flaw in the previous implementation of function countDigits, we will add a single if statement to evaluate if the given number is negative and return one less of the string size. Notice that if the number is greater than 0, the original value of string size is returned, as implemented in the following example code.

#include <iostream>
#include <vector>
#include <string>

using std::cout; using std::cerr;
using std::endl; using std::string;
using std::to_string;

template<typename T>
size_t countDigits(T n)
{
    string tmp;

    tmp = to_string(n);
    if (n < 0)
        return tmp.size() - 1;
    return tmp.size();
}

int main() {
    int num1 = 1234567;
    int num2 = -1234567;

    cout << "number of digits in " << num1 << " = " << countDigits(num1) << endl;
    cout << "number of digits in " << num2 << " = " << countDigits(num2) << endl;

    exit(EXIT_SUCCESS);
}

Output:

number of digits in 1234567 = 7
number of digits in -1234567 = 7

Use std::string::erase and std::remove_if Methods to Count to Count Number of Digits in a Number in C++

The previous example provides a fully sufficient solution for the above problem, but one can over-engineer the countDigits using the std::string::erase and std::remove_if functions combination to remove any non-number symbols. Note also that this method may be the stepping stone to implement a function that can work with floating-point values. The method is called erase-remove idiom in C++, and it can be a more efficient solution (more details can be found here). Mind though, the following sample code is not compatible with floating-point values.

#include <iostream>
#include <vector>
#include <string>

using std::cout; using std::cerr;
using std::endl; using std::string;
using std::to_string;

template<typename T>
size_t countDigits(T n)
{
    string tmp;

    tmp = to_string(n);
    tmp.erase(std::remove_if(tmp.begin(), tmp.end(), ispunct), tmp.end());
    return tmp.size();
}

int main() {
    int num1 = 1234567;
    int num2 = -1234567;

    cout << "number of digits in " << num1 << " = " << countDigits(num1) << endl;
    cout << "number of digits in " << num2 << " = " << countDigits(num2) << endl;

    exit(EXIT_SUCCESS);
}

Output:

number of digits in 1234567 = 7
number of digits in -1234567 = 7
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