This code counts the number of digits in an integer.

let digitCount number = (int) (log10 ((float)number)) + 1;

For instance, it tells us 123456789 has 9 digits.

let input = 123456789;
printfn "%i has %i digits" input (digitCount input);


A weakness of the code is that it uses two casts. Can we use math to count the number of digits in an integer without having to cast twice? The reason to cast to a float is to make the type compatible with a the log10 function. The reason to cast back in to an integer is to represent a "count" of digits instead of a "measure" of digits.

A second, related weakness of the code is the number of parentheses, which makes the code more difficult to read.

  • 1
    \$\begingroup\$ Using log10 seems like overkill. You can do this recursively by dividing (well, integer division) the number by 10 repeatedly. \$\endgroup\$
    – Dair
    Jan 18, 2018 at 6:50
  • \$\begingroup\$ You can also cast as a string and then count the length. \$\endgroup\$ Jan 18, 2018 at 18:39

1 Answer 1


Your code contains a bug. What happens if number is 0? log10 0 should return -Infinity, so we end up with the smallest int plus one. Therefore, we need to check whether input is 0 first:

let digitCount number = if number == 0 then 1 else (int) (log10 ((float)number)) + 1;

At that point we're not really succinct anymore, so there's no harm in using a recursive function instead:

let rec digitCount number = if number < 10 then 1 else 1 + digitCount (number / 10)

However, that's not tail recursive, so we would want to rewrite it using a loop that's tail recursive. That's left as an exercise, though.

That being said, one can optimize counting digit a lot.

Andrei Alexandrescu held a talk about optimizing digits_base_10 in C++ several years ago. The original talk is offline, unfortunately, but the slides are still there.


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