Count digits of result of arithmetic operation

Question

Based on J.Wright's answer I wanted to count the characters of an operation. Nevertheless the output of the code below is

16 12321 2

where instead of the bold number I expected to count the number of characters of the previous 12321. What am I doing wrong?

\documentclass{standalone} 
\usepackage{pgfplots}

\usetikzlibrary{calc}

\def\zCountDigits#1{%
  \number\numexpr0\zCountDigitsAux#1\zCountDigitsEnd\relax
}
\def\zCountDigitsAux#1{%
  \ifx\zCountDigitsEnd#1\else+1\expandafter\zCountDigitsAux\fi
}
\def\zCountDigitsEnd{\zCountDigitsEnd}

\begin{document}

\newcommand{\eval}[1]{\pgfmathparse{#1}\pgfmathresult}

% the result of the next is 16, as should
\zCountDigits{0123456789ABCDEF}
% the evaluation gives 12321
\eval{int(111*111)}  
% but the counting the result gives 2, not five
\zCountDigits{\eval{int(111*111)}} 

 \end{document}

score 6 · Accepted Answer · answered Oct 08 '19 at 22:47

The usual problem: \eval is not expandable.

Here's a more straightforward implementation with expl3.

\documentclass{article}
\usepackage{xparse,xfp}

\ExplSyntaxOn

\NewExpandableDocumentCommand{\CountDigits}{m}
 {
  \cp_count_digits:e { #1 }
 }
\cs_new:Nn \cp_count_digits:n { \tl_count:n { #1 } }
\cs_generate_variant:Nn \cp_count_digits:n { e }

\ExplSyntaxOff

\begin{document}

% the result of the next is 16, as it should
\CountDigits{0123456789ABCDEF}
% the evaluation gives 12321
\inteval{111*111}
% the counting of digits in the result gives 5, as it should
\CountDigits{\inteval{111*111}} 

\end{document}

score 5 · Answer 2 · answered Oct 08 '19 at 23:22

@egreg explained you why your approach fails: your macro is not expandable. There are two options: either you switch gears and use the expl3 machinery. There is nothing wrong with that. On the other hand, if you use pgf, tikz and/or pgfplots anyway, it is arguably preferable to build these functions into the pgf ecosystem. For many functions, this has been done already, in particular digitsum is already available. So you could just use it. I actually advise against using a macro but suggest to use the parsing and number printing machinery, but I added a macro, too.

\documentclass{article} 
\usepackage{pgf}
\makeatletter
\newcount\c@Digits
\newcount\c@Powers
\pgfmathdeclarefunction{digitcount}{1}{%
  \begingroup%
  \global\c@Digits=0%
  \expandafter\DigitCount@i#1\@nil%
  \pgfmathparse{int(\the\c@Digits)}%
  \pgfmathsmuggle\pgfmathresult\endgroup}
\def\DigitCount@i#1#2\@nil{%
  \advance\c@Digits by \@ne%
  \ifx\relax#2\relax\else\DigitCount@i#2\@nil\fi%
}
\makeatother
\newcommand{\DigitCount}[1]{\pgfmathparse{digitcount(int(#1))}\pgfmathresult}
\begin{document}
If you really want a marco: \DigitCount{111*111}

Cross--check: \pgfmathparse{int(111*111)}\pgfmathprintnumber\pgfmathresult

I'd suggest parsing within the pgf system: \pgfmathparse{digitcount(int(111*111))}\pgfmathprintnumber\pgfmathresult

Works with strings, too: \pgfmathparse{digitcount("0123456789ABCDEF")}\pgfmathprintnumber\pgfmathresult
\end{document}

Count digits of result of arithmetic operation

2 Answers2