counting digts in c++ using log10

#include <iostream>
#include <math.h>

using namespace std;

int main() {
    int n, temp, rem, digits=0, sum=0;
    cout << "Enter a armstrong number: ";
    cin >> n;
    temp = n;
    digits = (int)log10(n) + 1;

    while (n != 0) {
        rem = n % 10;
        sum = sum + pow(rem, digits);
        n = n / 10;
    }

    if (temp == sum) {
        cout << "yes";
    }
    else {
        cout << "not";
    }
}

How does the " digits = (int)log10(n) + 1; " line actually calculates the digits?
can anyone explain?

>Solution :

Math.

Logarithms are basically "exponents in reverse." Log10(100) is 2.0, as 10 to the second power is 100.

Cast to 'int and add one to that are you get 3, which is the number of digits.