Number of \ (backslash) characters in a string

Question

I'm trying to built a code that counts the number of \ in a string. My code can work with any other symbols (for example, if I need to count the number of 'a's, that works). I already tried to modify the catcode of the backslash symbol to 12, but with no success.

Any ideas?

My (incomplete) solution:

\newcounter{alternatives}
\renewcommand\thealternatives{\alph{alternatives}}
\makeatletter
\def\endtab{\vspace{-4 mm}}
\def\endtabi{\endtab}

\newcount\testcount
\def\gobblechar{\let\char= }
    \newcount\slashcount
\def\countunlessnil
{%
    \let\a=0
    \let\\=\a
    \ifx\char\a
%       \slashcount=
        \advance\slashcount by 1
    \fi
    \ifx\char\nil
        \let\next=\relax%
    \else%
        \let\next=\auxcountchar%
    \fi
    \let\\=\\
    \next
}%
\def\auxcountchar{%
  \afterassignment\countunlessnil\gobblechar%
}
\def\countchar#1{\edef\xx{#1} \expandafter\auxcountchar\xx\nil}
\def\analyze#1{%
  \countchar{#1}%
}
\newlength{\alternativessize}
\newlength{\abcde}
\settowidth{\abcde}{a)~b)~c)~d)~e)}
\WithSuffix\DeclareRobustCommand\tab+{\@tabbb}
\def\@tabbb#1 \endtab
{
    \tabiiii#1
    \endtab\\
        \analyze{dfgreggegewg\a}
    \the\testcount
%   \setcounter{alternatives}{0}
    \settowidth{\alternativessize}{#1}
    \setlength{\alternativessize}{\linewidth - \alternativessize}
    \setlength{\alternativessize}{\alternativessize - \abcde}
    \the\value{alternatives}\\[4 mm]
    \ifthenelse{\equal{\value{alternatives}}{5}}
    {
    \setlength{\alternativessize}{0.17\alternativessize}
    }
    {
        \setlength{\alternativessize}{0.27\alternativessize}
    }
    \setcounter{alternatives}{0}
%   \setlength{\alternativessize}{0.18\alternativessize}
    \begin{minipage}[t]{\linewidth}
            \tabiii#1
    \end{minipage}\vspace{-2 mm}\\
    \endtab\\
\setcounter{alternatives}{0}
}
\def\tabiii#1\\
{
    \def\tempa{#1}
    \ifx\tempa\endtabi
        \endtab

    \else
        \if\relax\detokenize{#1}\relax
        \else
            \stepcounter{alternatives}\thealternatives)~~\@ifnextchar {_}{}#1
            \hspace{\alternativessize}
        \fi
        \expandafter\tabiii
    \fi
}
\def\tabiiii#1\\
{
    \def\tempa{#1}
    \ifx\tempa\endtabi
        \endtab
    \else
        \if\relax\detokenize{#1}\relax
        \else
            \stepcounter{alternatives}\thealternatives)~~\@ifnextchar {_}{}#1
            \hspace{\alternativessize}
        \fi
        \expandafter\tabiii
    \fi
}
\makeatother

When tabiii is called, it detokenize the sequence: aaaaa\\ bbbbb\\ ccccc\\ ddddd\\

and so on, putting a marker a), b), c) in front of each line. I need to count the number of \, OR get the count of the \thealternative. Reason: I'm trying to make a code that cant put 4 or 5 alternatives in the same line, but I need to know the string size BEFORE TeX read it. This is why I'm calling tabiii and tabiiii: tabiiii reads the sentence to calculate the size of the spaces, and tabiii prints the content on the PDF file. Any ideas?

Answer 1

Here is a solution involving xstring (in order to count the occurrences) and catcode changes:

\documentclass{article}
\usepackage{xstring}
\begingroup
\catcode`\|0
|catcode`|\12
|makeatletter
|gdef|mymacro{|begingroup|catcode`|\12 |mymacro@i}
|gdef|mymacro@i#1{|endgroup|StrCount{#1}\}
|endgroup
\begin{document}
\mymacro{\ifnum`\\=0 \textbf{0}\else\textif{non zero}\fi}
\end{document}

Answer 2

Here are two ways to get the number of \\ tokens in an argument. In both cases the counter \l_cnobs_count_int will contain that number (three for the example).

First way (with token lists)

\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand{\countbstl}{m}
 {
  \cnobs_main_one:n { #1 }
 }

\int_new:N \l_cnobs_count_int

\cs_new_protected:Npn \cnobs_main_one:n #1
 {
  \int_zero:N \l_cnobs_count_int
  \tl_map_inline:nn { #1 } 
   {
    \tl_if_eq:nnT { ##1 } { \\ } { \int_incr:N \l_cnobs_count_int }
   }
  \int_show:N \l_cnobs_count_int
 }
\ExplSyntaxOff

\countbstl{aaa\\ bbb \\ ccc \\}

\stop

Second way (with regular expressions)

\documentclass{article}
\usepackage{xparse,l3regex}
\ExplSyntaxOn
\NewDocumentCommand{\countbsregex}{m}
 {
  \cnobs_main_two:n { #1 }
 }

\int_new:N \l_cnobs_count_int

\cs_new_protected:Npn \cnobs_main_two:n #1
 {
  \int_zero:N \l_cnobs_count_int
  \regex_count:nnN { \c{ \\ } } { #1 } \l_cnobs_count_int
  \int_show:N \l_cnobs_count_int
 }

\ExplSyntaxOff

\countbsregex{aaa\\ bbb \\ ccc \\}

\stop

One might easily modify this to count the number of control sequences; by changing the \regex_count:nnN line into

  \regex_count:nnN { \c{ .* } } { #1 } \l_cnobs_count_int

an input such as

\countbsregex{\textbf{aaa}\\ bbb \\ ccc \\}

will store 4 in the counter. A further change

  \regex_count:nnN { \c{ .* } | \cA. } { #1 } \l_cnobs_count_int

will count also active characters, so

\countbsregex{\textbf{aaa}\\ bbb~bbb \\ ccc \\}

stores 5 in the counter.

---

It's hard to say how to use this in your very complicated (and unexplained) code.