Make Characters Active via Macro in Math Mode

Question

I am trying to make characters active in math mode, but can't seem to get past the most basic of examples. I would like to define a macro which is passed:

1. the character to make active and

2. the macro that should be executed once that macro is encountered.

The MNWE below attempts to make =, + and ! active and color code them. The desired output is that those characters get colored in math mode as. Thus, the input

a=+b $c=+4+d+3!$ e=+f

yields:

where the characters are not altered outside of math mode (the leading a=+b and trailing e=+f).

Notes:

• There is a separate, but related (as it is in the context of active characters) question in the \MyPlus macro to detect a binary addition + versus a unary positive +. Perhaps that should be a separate question?

Related Questions:

• Macro to take a character as argument, make it active, then \def it.
• How do you define macros to toggle a character active and define a macro for it?
• "Activate" active characters in argument passed as macro

Code (non-working):

\documentclass{article}
\usepackage{mathtools}
\usepackage{xparse}
\usepackage{xcolor}

\NewDocumentCommand{\MakeActiveChar}{%
    m% char to make active
    m% code to execute for this character
}{%
    \catcode`\#1=13\relax%
    \def#1{#2}%
}%

\let\OldEqual=
\newcommand*{\MyEqual}[1][green]{\mathrel{\textcolor{#1}{\OldEqual}}}%

\let\OldPlus+
\newcommand*{\MyPlus}[1][blue]{%
    %\catcode`\+=12% reset catcode so we can use it (not needed with \OldPlus?)
    %% Perhaps this if-then construct should be a separate question?
    %% if binary addition operator
        \mathbin{\textcolor{#1}{\OldPlus}}%
    %% if unary positive operator
    %%  \mathrel{\textcolor{red}{\OldPlus}}%
}%

\let\OldFactorial!
\newcommand*{\MyFactorial}{\textcolor{brown}{\OldFactorial}}%

\everymath{%
    \MakeActiveChar{=}{\MyEqual}%
    \MakeActiveChar{+}{\MyPlus}%
    \MakeActiveChar{!}{\MyFactorial}%
}%

\begin{document}
The ``math'' outside here is just to test that characters are active \emph{only} in math mode:

a=+b $c=+4+d+3!$ e=+f 
\end{document}

Answer 1

You have to be careful with the declarations if you load amsmath; in the body of the definition, use \std{<char>} where you need the old meaning.

\documentclass{article}
\usepackage{amsmath}
\usepackage{xcolor,etoolbox}

\makeatletter
\newcommand{\DeclareMathActive}[2]{%
  % #1 is the character, #2 is the definition
  \expandafter\edef\csname keep@#1@code\endcsname{\mathchar\the\mathcode`#1 }
  \begingroup\lccode`~=`#1\relax
  \lowercase{\endgroup\def~}{#2}%
  \AtBeginDocument{\mathcode`#1="8000 }%
}

\newcommand{\std}[1]{\csname keep@#1@code\endcsname}
\patchcmd{\newmcodes@}{\mathcode`\-\relax}{\std@minuscode\relax}{}{\ddt}
\AtBeginDocument{\edef\std@minuscode{\the\mathcode`-}}
\makeatother


\DeclareMathActive{+}{\mathbin{\textcolor{blue}{\std{+}}}}
\DeclareMathActive{!}{\mathclose{\textcolor{brown}{\std{!}}}}
\DeclareMathActive{-}{\mathbin{\textcolor{red}{\hat{\std{-}}}}}

\begin{document}

a=+b $c=+4+d+3!-1$ e=+f

\end{document}

Don't do \let\OldPlus=+.

With Unicode engines (xelatex or lualatex), the patch must be done differently. Here's a version that copes with all engines.

\documentclass{article}
\usepackage{amsmath}
\usepackage{xcolor,etoolbox}
\usepackage{iftex}

\makeatletter
\newcommand{\DeclareMathActive}[2]{%
  % #1 is the character, #2 is the definition
  \expandafter\edef\csname keep@#1@code\endcsname{\mathchar\the\mathcode`#1 }
  \begingroup\lccode`~=`#1\relax
  \lowercase{\endgroup\def~}{#2}%
  \AtBeginDocument{\mathcode`#1="8000 }%
}

\newcommand{\std}[1]{\csname keep@#1@code\endcsname}
\iftutex
  \patchcmd{\newmcodes@}{\Umathcodenum `\-\relax}{\std@minuscode\relax}{}{\ddt}
\else
  \patchcmd{\newmcodes@}{\mathcode`\-\relax}{\std@minuscode\relax}{}{\ddt}
\fi
\AtBeginDocument{\edef\std@minuscode{\the\mathcode`-}}
\makeatother


\DeclareMathActive{+}{\mathbin{\textcolor{blue}{\std{+}}}}
\DeclareMathActive{!}{\mathclose{\textcolor{brown}{\std{!}}}}
\DeclareMathActive{-}{\mathbin{\textcolor{red}{\hat{\std{-}}}}}

\begin{document}

a=+b $c=+4+d+3!-1$ e=+f

\end{document}

Answer 2

You can define \mathdef declarator and then use it:

\input opmac \localcolor

\def\mathdef#1{\mathcode`#1="8000 \bgroup \lccode`~=`#1\lowercase{\egroup\def~}}

\mathdef +{\mathbin{\Red\mathchar`+}}
\mathdef ={\mathrel{\Green\mathchar`=}}

aha + $a+b+c=d$

\bye

The \input opmac is here only for setting colors \Red, \Green etc. If you are using another macro for colors then use only the second line \def\mathdef... in your macros.

Answer 3

There are a few problems:

\MakeActiveChar does \def#1{#2} which is not permitted since #1 is a character that is not active. Here's an alternative with The \lowercase trick

\NewDocumentCommand\MakeActiveChar{mm}
  {\begingroup\lccode`\~=`#1\lowercase{\endgroup\def~}{#2}%
   \catcode`#1=12 \mathcode`#1="8000 }

Apart from that, you have a missing = in a \let. The first equal is optional, so \let\a=\b is correct, the equal, however, is necessary when you want to let to = because if not, it would be treated as the optional one, so

\let\originalequal=

should be

\let\originalequal==

And the last part is that by making a character active in math mode, and then doing \newcommand\plus{\mathbin{\originalplus}} you get an endless loop like \def\a{\a}, so you need to use for instance \mathchar`\+ or \mathchar`+ instead.

Complete solution

\documentclass{article}
\usepackage{mathtools}
\usepackage{xparse}
\usepackage{xcolor}

\NewDocumentCommand\MakeActiveChar{mm}
  {\begingroup\lccode`\~=`#1\lowercase{\endgroup\def~}{#2}%
   \catcode`#1=12 \mathcode`#1="8000 }

\newcommand*\mathcolor[2]{\begingroup\color{#1}#2\endgroup}

\newcommand*\equal[1][green]{\mathrel{\mathcolor{#1}{\mathchar`\=}}}
\newcommand*\plus[1][blue]{\mathbin{\mathcolor{#1}{\mathchar`\+}}}
\newcommand*\factorial{\mathord{\mathcolor{brown}{\mathchar`\!}}}

\everymath{%
  \MakeActiveChar{=}{\equal}%
  \MakeActiveChar{+}{\plus}%
  \MakeActiveChar{!}{\factorial}%
}

\begin{document}
The ``math'' outside here is just to test that characters are active \emph{only} in math mode:

a=+b $c=+4+d+3!$ e=+f 
\end{document}

By the way, I defined \mathcolor because \textcolor works with \bgroup/\egroup which create an ord atom.