Beginner · loops + integer math

Count the Digits

Read a positive integer and print how many digits it has. Introduces the most useful number trick in beginner problems: chopping digits off with / 10 and % 10.

At a glanceCount the digits of N by dividing by 10 in a while-loop until N reaches 0 — the loop shape every later digit problem reuses.

Problem

Input: one positive integer N (1 ≤ N ≤ 10⁹).

Output: the number of digits in N.

Samples:

Input	Output
`12345`	`5`
`7`	`1`

Your solution

main.cpp — your solution

Custom input for "Run my input"

Judging runs in your browser against hidden tests. A runaway loop is cut off after 3 seconds — still, make sure every loop has an exit. Your code is saved in this browser as you type.

Stuck? Hints

Hint 1 — the chopping pattern

n % 10 gives the last digit; n / 10 removes it (integer division! see why 7/2 = 3). So 12345 becomes 1234 becomes 123… until it hits 0. Count how many chops that took.

Hint 2 — choosing the loop

You don't know in advance how many iterations you need — that's a while job: while (n > 0) { count++; n /= 10; }. Compare with the for vs while guide: unknown iteration count → while.

Solution (try it first!)

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    int count = 0;
    while (n > 0) {
        count++;
        n /= 10;
    }
    cout << count << endl;
    return 0;
}

This chopping pattern is half of the next problem too — Reverse a Number uses the same two moves in a different order. That's how algorithmic thinking works: patterns recombine.

Deep dive

Why this works

Integer division by 10 peels off the last digit; that pair / 10 and % 10 is the entire basis of digit manipulation, and it shows up in half the problems on this site.

The cost

O(d) where d is the digit count — at most 10 for an int, so effectively constant.

Where it goes wrong

while (n > 0) returns 0 digits for n = 0. Zero needs its own answer.
log10(n) + 1 looks clever and breaks on 0 and on floating-point rounding at exact powers of ten.

Next step

Add the digits you are peeling: Sum of Digits.

Two things to remember

/ 10 removes the last digit; % 10 reads it.
Zero is the input that breaks every digit loop.

🐞 This version fails — open it and try

while (n >= 10) stops one digit early

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    int count = 0;
    while (n >= 10) {
        n /= 10;
        count++;
    }
    cout << count << endl;
    return 0;
}