Highlighting token groups inside an expression

Question

What I want to do is replace {...} (token group) by \{...\} (curly braces) at any level of nesting inside an expression. I'm only able to do it for level = 0. There should be an extra curly brace around the second 'o' in the example given. Any suggestion?

\documentclass{article}
\ExplSyntaxOn
\cs_new:Nn
__erw_unbrace_scan:n
{ %\group_begin:
  __erw_scan:n#1
  %\group_end:
}
\cs_new:Nn
__erw_aux:n
{  \tl_if_single_token:nTF{#1}{#1}
  { {__erw_unbrace_scan:n{#1}}}}
\cs_new:Npn
__erw_scan:w #1 #2 #3 \q_recursion_stop
{\quark_if_recursion_tail_stop:n{#2}
  __erw_aux:n{#1}
  __erw_scan:w {#2} #3 \q_recursion_stop}
\cs_new:Nn
__erw_scan:n
{__erw_scan:w #1 {dummy}\q_recursion_tail\q_recursion_stop}
\ExplSyntaxOff
\begin{document}
\ExplSyntaxOn
\tl_if_single_token:nTF{x}{T}{F}\par
\tl_if_single_token:nTF{{x}}{T}{F}\par
__erw_scan:n{f{o{o}}bar} % expected: f{o{o}}bar
\ExplSyntaxOff
\end{document}

plante · Accepted Answer · 2021-09-05T16:58:13.787

6

If a TeX solution is allowed (this is fully expandable)

\makeatletter
\def\erw#1{\erwAuxi#1\@gobbletwo{}}
\def\erwAuxi#1#{#1\erwAuxii}
\def\erwAuxii#1{\{\erw{#1}\}\erwAuxi}
\makeatother
\erw{f{oo}bar{baz}{} {a{bc{def}}}}

uses the #{ trick.

edited Sep 05 '21 at 16:58

answered Sep 05 '21 at 06:58

plante

1,349

1

Beautiful! +1 :-) – Przemysław Scherwentke Sep 05 '21 at 07:17

score 3 · Answer 2 · answered Sep 05 '21 at 08:04

This is not fully expandable, but does the job and the same code can be adjusted for other balanced tokens.

\documentclass{article}
\ExplSyntaxOn
\NewDocumentCommand{\showbraces}{m}
 {
  \erwann_showbraces:n { #1 }
 }
\tl_new:N \l__erwann_showbraces_tl
\cs_new_protected:Nn \erwann_showbraces:n
 {
  \tl_set:Nn \l__erwann_showbraces_tl { #1 }
  % start the recursion
  __erwann_showbraces:
 }
\cs_new_protected:Nn __erwann_showbraces:
 {
  \regex_match:nVTF { {.} } \l__erwann_showbraces_tl
   {% there is a match, modify
    \regex_replace_all:nnN { {(.?)} } { \c{{}\1\c{}} } \l__erwann_showbraces_tl
    % reapply
    __erwann_showbraces:
   }
   {% no match, deliver the result
    \tl_use:N \l__erwann_showbraces_tl
   }
 }
\prg_generate_conditional_variant:Nnn \regex_match:nn { nV } { T, F, TF }
\ExplSyntaxOff
\begin{document}
\showbraces{f {oo}bar{baz}{} {a{bc{def}}}}
\end{document}

The reason I got started toying with recursion is because regex is not expandable as you reminded, but for the narrow task at hand your answer suffices. Do you know why my solution does not work? — Erwann, Sep 05 '21 at 11:57

score 2 · Answer 3 · answered Sep 05 '21 at 14:11

Without Expl3, here is a token cycle approach.

While brace conversion could easily mess with macro arguments, the MWE shows that for macros that do not rely on braced arguments, they can be part of the converted stream.

\documentclass{article}
\usepackage{tokcycle}
\stripgroupingtrue
\Groupdirective{\addcytoks{\{}\processtoks{#1}\addcytoks{\}}}
\newcommand\groupscan[1]{\tokcyclexpress{#1}\the\cytoks}
\begin{document}
\groupscan{f{o{o}}bar}
\groupscan{f{o{{\today}}}bar}
\end{document}

score 1 · Answer 4 · answered Sep 05 '21 at 11:31

If braces are stripped out in the normal course of processing, then letting them be stripped out and inserting text braces just beforehand would keep everything balanced and not require recursion, wouldn't it?

MWE

\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
%=====
\tl_new:N \l_mytemp_tl
%=====
\NewDocumentCommand { \scanbraces } { m } {
\tl_set:Nn \l_mytemp_tl { #1 } % no expansion
%\tl_show:N \l_mytemp_tl 

\regex_replace_all:nnN 
            { (\c{[A-Za-z]*}) } %any control sequence with Latin letters
            { \c{token_to_str:N} \1  } %convert to string
            \l_mytemp_tl    

\regex_replace_all:nnN 
            { (\cB.) } %match on opening group
            { \1\c{\{}  } %insert a text brace
            \l_mytemp_tl

\regex_replace_all:nnN 
        { (\cE.) } %match on closing group
        { \1\c{\}}  } %insert a text brace
         \l_mytemp_tl

\tl_use:N \l_mytemp_tl


}
\ExplSyntaxOff
\newcommand\ftest[1]{\textit{#1}}
\newcommand\qq[2][]{\begin{quotation}#1#2\end{quotation}}
\begin{document}
For a command defined as \verb|\newcommand\ftest[1]{\textit{#1}}|
then keying in
\qq[\ttfamily]{\scanbraces{\ftest{o{o}}bar} }
typesets
\qq{\ftest{o{o}}bar}
\end{document}

@egreg

Highlighting token groups inside an expression

4 Answers4