Some misunderstanding about macro expansion

Question

I have the following code which doesn't work:

\documentclass[12pt]{article}

\tracingcommands=2
\tracingmacros=2
\tracingall

\makeatletter

\let\sep\relax

\def\put@stack@#1#2{\edef#2{#1\sep#2}}

\def\put@queue@#1#2{\edef#2{#2\sep#1}}

\def\get@#1#2{\expandafter\get@@#2\endget@@#1#2}

\def\get@@#1\sep#2\endget@@#3#4{\edef#3{#1}\edef#4{#2}}

\def\empty@stack{\sep}

\newtoks\piecetoks

\def\leftrightbr{
    \let\o@left\left
    \let\o@right\right
    \let\piece@nd\relax
    \let\pieces@stack\empty@stack
    \let\delims@stack\empty@stack
    \let\on@end@line\empty@stack
    \let\on@begin@line\empty@stack
    \def\left##1{
        \piece@nd
        \let\piece@nd\endgroup
        \def\on@first@right{
            \put@stack@{\the\toks0}{\pieces@stack}
        }
        \put@stack@{##1}{\delims@stack}
        \let\repl\relax
        \let\vphantomer####1{}
        \put@stack@{\the\piecetoks}{\pieces@stack}
        \put@stack@{\right.}{\on@end@line}
        \put@stack@{\left.}{\on@begin@line}
        \piecetoks=\begingroup
    }
    \def\right##1{
        \endgroup
        \on@first@right
        \let\on@first@right\relax
        \get@\@dummy\on@end@line
        \get@\@dummy\on@begin@line
        \get@\@lpart\pieces@stack
        \get@\@ldelim\delims@stack
        \ifx\@lpart\empty
            \repl
        \else
            \edef\repl{\o@left\@ldelim\@lpart\vphantomer{\the\piecetoks}\repl\the\piecetoks\vphantomer{\@lpart}\o@right##1}
        \fi
    }
    \def\mathbr{
        \piece@end
        \put@stack@{\piece}{\pieces@stack}
        \let\vphantomer\vphantom
        \edef\repl{\on@end@line \\ \on@begin@line}
        \def\piece\begingroup
    }
}

\begin{document}

\begin{equation}
    \leftrightbr
    \left(a\right)
\end{equation}

\end{document}

So what's the catch?

EDIT: The error is

! Undefined control sequence.
<argument> \@dummy 

l.85     \left(
               a\right)

Looking into log gets us to conclusion that the problem is caused by call to \get@\@dummy\on@end@line. This seems to be strange since direct call to \get@ on \empty@stack results in no error.

EDIT: The code can be narrowed down:

\documentclass[12pt]{article}

\tracingcommands=2
\tracingmacros=2
\tracingall

\newtoks\piecetoks

\def\leftrightbr{
    \def\left##1{
        \piecetoks=\begingroup
    }
    \def\right##1{
        \endgroup
    }
}

\begin{document}

\begin{equation}
    \leftrightbr
    \left(a\right)
\end{equation}

\end{document}

This is shorter, isn't it? :-) It gives the other error, and the other question arises: how can we achieve behavior in which TeX captures tokens between two macros' calls?

Answer 1

You can't assign tokens to a token register with

\piecetoks=\begingroup...\endgroup`

Nor you can start the assignment with a macro and end it with another one, because

<toks register>={...}

needs an explicit brace at the end and no expansion takes place during the evaluation of the <balanced text> to be stored.

In the TeXbook, page 276, you find what a <variable assignment> is; one of the alternatives is

<token variable> <equals> <general text>

and some lines above you see

<general text> → <filler> { <balanced text> <right brace>

In the syntax rules, { means an implicit or explicit character token of category code 1, while <right brace> means an explicit character token of category code 2.

Answer 2

Delimited arguments need to look for a fixed token, but you don't need to look ahead for \\-or-\right you can look ahead for \right and then look in the tokens you collected to see if there was a \\.

so

\documentclass{article}


\makeatletter
\newtoks\tA
\newtoks\tB

\def\left#1\right{%
  \zz#1\\\right}

\def\zz#1\\#2\right#3{%
\ifx\relax#2\relax
  \tA{\left#1\right#3}%
  \typeout{^^J^^Jno \space\string\\:^^J\the\tA^^J}%
  \expandafter\remove@to@nnil
\else
  \tA{\left#1\right.\\}%
  \expandafter\zzb
 \fi
 #2\right#3\@nnil}

\def\zzb#1\\\right#2\@nnil{%
  \tB{\left.#1\right#2}%
  \typeout{^^J^^Jhas \string\\:^^J\the\tA\space \the\tB^^J}}
\makeatother

\begin{document}

\left( a b\right)

\left( a \\ b\right)

\end{document}

produces

no  \\:
\left ( a b\right )



has \\:
\left ( a \right .\\ \left . b\right )

However I would never use this in production. redefining \left and \right to take macro arguments changes their scanning rules and will introduce all kinds of incompatibilities. Even if you avoid that by using new names rather than redefining the primitives, using \right.\\ \left. at the line break does not ensure that the paired delimiters are the same size so can not produce an acceptable result.

It is better really just to use \bigl( in one cell and \bigr) in a later cell.