Math / Count number of digits with Logarithm
Suppose that a number $n$ has $d$ digits, then:
$$
10^{d-1} \leq n \lt 10^{d}
$$
Because $10^d$ is the smallest integer with $d + 1$ digits.
Now, take logs base 10 of this relation:
$$
log_{10}{(10^{d-1})} \leq log_{10}{n} \lt log_{10}{(10^d)}
$$
This becomes:
$$
d-1 \leq log_{10}{n} \lt d
$$
If you now take the integer part of $log_{10}{n}$, throwing away everything to the right of the decimal point, you will get $\lfloor log_{10}{n} \rfloor = d-1$. Thus, $d = \lfloor log_{10}{n} \rfloor + 1$.
For example, $234$ is a $3$ digits number, take the logs base 10:
$$
log_{10}{(234)} \approx 2.369215
$$
If we take the integer part and throw away everything to the right of the decimal point, it’s $2$.
The calculation can be done programmatically like this:
#include <math>
int digits = log10(n) + 1;
Source: Proof: How many digits does a number have? - StackExchange Mathematics◹