[Tex/LaTex] Removing a backslash from a character sequence

catcodesdetokenize

For indexing I wanted to write a macro \macroname that removes leading backslashes from macro names but leaves the names of environments untouched. That is

\macroname{\relax}   -->  relax
\macroname{itemize}  -->  itemize

where the --> is meant to be read as "expands to".

From my understanding, the following should work:

\newcommand\removebs[1]{\if#1\char92\else#1\fi}
\newcommand\macroname[1]{%
  \expandafter\removebs\detokenize{#1}}

However, its outcome is

\macroname{\relax}   -->  \relax   % literally
\macroname{itemize}  -->  itemize

By chance, actually, I found out that if I change the catcode of the backslash to "other" during the definition of the \removebs macro, it works:

{
  \catcode`\|=0
  |catcode`|\=12
  |global|def|removebs#1{|if#1\|else#1|fi}
}

\newcommand\macroname[1]{%
  \expandafter\removebs\detokenize{#1}}

Why is that? From TeX by Topic I'd expect that the comparison made by \if is what Philipp Lehman calls "category code agnostic".

Thanks for your answers!

Best Answer

Joseph has given a working solution. I'd like to explain what goes wrong with your code.

First attempt

\newcommand\removebs[1]{\if#1\char92\else#1\fi}
\newcommand\macroname[1]{%
  \expandafter\removebs\detokenize{#1}}

With \macroname{\relax} you get

\expandafter\removebs\detokenize{\relax}

and then (using • to separate tokens and <space> do denote a space token)

\removebs • \ • r • e • l • a • x • <space>

which becomes

\if • \ • \char • 9 • 2 • \else • \ • \fi • r • e • l • a • x • <space>

and the comparison is between a category code 12 backslash and the token \char, which of course leads to false.

The code \char92 are instruction to print character 92 from the current font.

One might correct the code by checking against a real category code 12 backslash:

\makeatletter
\newcommand{\removebs}[1]{\if#1\@backslashchar\else#1\fi}
\makeatother

but the space produced by \detokenize after \relax would remain.

Second attempt

{
  \catcode`\|=0
  |catcode`|\=12
  |global|def|removebs#1{|if#1\|else#1|fi}
}

This works because it just implements the check against a category code 12 backslash, but it's not necessary, as the token is already available as \@backslashchar in the LaTeX kernel.

An alternative way, without global definitions and category changes, can be

\begingroup\lccode`\|=`\\
\lowercase{\endgroup\def\removebs#1{\if#1|\else#1\fi}}

where the only token that is converted to its lowercase equivalent is | (to a backslash).

Proposed code

% \makeatletter
% \newcommand{\removebs}[1]{\if#1\@backslashchar\else#1\fi}
% \makeatother
\begingroup\lccode`\|=`\\
\lowercase{\endgroup\def\removebs#1{\if#1|\else#1\fi}}
\newcommand{\macroname}[1]{\expandafter\removebs\string#1}

Since all you want is to get the argument letters, it doesn't matter if the first letter in \macroname{itemize} has category code 12.

Related Solutions

[Tex/LaTex] How to make a real backslash (escape) character

Another way to do this kind of thing is to use a \lccode trick:

\begingroup\lccode`!=`\\\lowercase{\endgroup\def\@backslashchar{!}}

This avoids any need to change catcodes thanks to the special properties of \lowercase. Compared to the \expandafter\@gobble\string method, it's a bit more flexible as you can choose the catcode of the resulting character inside your macro. Compared to a change of catcodes, you don't have to use \global because of how \lowercase interacts with \endgroup.

Here's the detail of the code. The \begingroup is here to keep the lccode changes local. The

\lccode`!=`\\

means that when lowercasing ! you will get the backslash \ (the choice of ! is arbitrary, you just need a normal character which will not appear elsewhere inside the \lowercase). The important thing is that the lowercase version of ! wille be a normal character since \lowercase doesn't change catcodes.

The code \lowercase{\endgroup\def\@backslashchar{!}} is thus equivalent to \endgroup\def\@backslashchar{\} but with \ not being special (so \ followed by } won't be interpreted as \} but as two separate entities).

At the end, the result is that you have defined \@backslashchar to be a \ with normal catcode. It can thus be used inside a \write command without causing the same problems as \ does.

This lccode trick works with all other special characters. For example if you want a space character (of course, for spaces, the macro \space works fine):

\begingroup\lccode`!=`\ \lowercase{\endgroup\def\@spacechar{!}}

If ever you break the code on two lines, just be careful with the end of lines and put a % after the \:

\begingroup\lccode`!=`\ %
\lowercase{\endgroup\def\@spacechar{!}}

[Tex/LaTex] How to get the category code of a character that is the value of a control sequence

Sadly Knuth in his wisdom gave only few clues in the TeXBook for cases like this.

In order to understand the error we need to firstly understand the definition of \the, which is for the case of catcode \the<codename><8-bit number>, where <codename> stands for either \catcode, \mathcode, \lccode etc...

So clearly in this case catcode TeX expects a number and hence the error generated in the example provided by the OP.

All solutions provided by the other posts revolve around changing the definition one way or another to produce the required character code number and which I am demonstrating here with some different examples:

The example below will produce the right answer in both cases,

% results category 11
\makeatletter
\def\ABC{`@ }
\the\catcode\ABC

% results category 12
\makeatother
\def\ABC{`@ }
\the\catcode\ABC

While comparing two tokens using \ifcat things become a bit more complicated, if you want to compare two active characters you have to say \noexpand - Knuth says so somewhat obscurely in Exercise 20.11!

Consider the following definitions

\catcode`[=13 \catcode`]=13
\def[{*}

The following will result True since we comparing [ with ']' both now being category 13

\ifcat\noexpand[\noexpand]  True \else False \fi

Also \ifcat[* True \else False \fi is True

Since now we have established two facts \the\catcode needs a number and if you use an active character, implicitly or explicitly we can understand why the following will all work!

\def\abc{`A}
\chardef\abc=65

or Joseph's suggestion for comparisons:

\let\abc=A
   \ifcat\noexpand\abc A%
     \TRUE
   \else
     \FALSE
   \fi