Custom-length arrows, text over and under

Question

Expanding on this question I'd like to know if it's possible to modify this macro to allow for text under the arrow as well, as the \xrightarrow[below]{under}. I don't have any experience with macros but using the answer to the aforementioned question I tried this to no effect

\newlength{\arrow}
\settowidth{\arrow}{\scriptsize$1000$}
\newcommand*{\myrightarrow}[2]{\xrightarrow[#1]{\mathmakebox[\arrow]{#2}}}

What would be the right way to do this?

Answer 1

Here is a \xxrightarrow that has a first mandatory argument, the desired subscript to which the arrow should adapt (see the examples in the test document below).

\documentclass[a4paper]{article}
\usepackage{amsmath,mathtools}
\makeatletter
\newlength\min@xx
\newcommand*\xxrightarrow[1]{\begingroup
  \settowidth\min@xx{$\m@th\scriptstyle#1$}
  \@xxrightarrow}
\newcommand*\@xxrightarrow[2][]{
  \sbox8{$\m@th\scriptstyle#1$}  % subscript
  \ifdim\wd8>\min@xx \min@xx=\wd8 \fi
  \sbox8{$\m@th\scriptstyle#2$} % superscript
  \ifdim\wd8>\min@xx \min@xx=\wd8 \fi
  \xrightarrow[{\mathmakebox[\min@xx]{\scriptstyle#1}}]
    {\mathmakebox[\min@xx]{\scriptstyle#2}}
  \endgroup}
\makeatother

\begin{document}
$A\xxrightarrow{1000}{1}A$

$A\xxrightarrow{1000}{1000}A\xrightarrow{1000}A$

$A\xxrightarrow{1000}[1]{1}A\xrightarrow[1]{1}A$

\end{document}

So you call it as

\xxrightarrow{<sample>}[<below>]{<above>}

where <below> is optional.

Answer 2

The syntax that you have written means that the arrow has to be called with two arguments. So even if you don't have anything below the arrow, you still need to say \myrightarrow{}{above}. To get an optional argument, define the command as:

\newcommand*{\myrightarrow}[2][]{\xrightarrow[#1]{\mathmakebox[\arrow]{#2}}}

However, this treats the two arguments differently. To see this, try putting a really long text in the upper and lower arguments. The arrow will stretch for the lower one but not for the upper one. To ensure equal treatment, we need to put both arguments in a \mathmakebox. However, for spacing reasons we should only put the lower one in a box if it is defined. So we test for that. Here's one way to do it, probably not the most elegant (but then I'm not know for the elegance of my code!).

\def\empty{}
\newcommand*{\myrightarrow}[2][]{%
  \def\temp{#1}%
  \ifx\temp\empty
   \def\mycmd{\xrightarrow}%
  \else
   \def\mycmd{\xrightarrow[{\mathmakebox[\arrow]{#1}}]}%
  \fi
  \mycmd{\mathmakebox[\arrow]{#2}}%
 }

What this does is the following: the #1 is a placeholder for our first argument. We want to test if this is empty, which we do by defining a temporary macro (\temp) to hold it and testing that against a pre-defined \empty macro (note to TeXperts: I'm deliberately avoiding @s). If it is empty, we define a temporary command to be just \xrightarrow. If it is not, we add in the lower part with the surrounding \mathmakebox. Then whatever that was set to, we execute it together with the command for the upper code.

Answer 3

This answer takes the ideas in the two other answers and combines them. The code is mostly copied from @egreg. When there is no subscript given, there will not be reserved space below the arrow. This causes the lines to be more compact, if there is no subscript given.

\documentclass[a4paper]{article}
\usepackage{amsmath,mathtools,xparse}
\makeatletter
\newlength\min@xx
\DeclareDocumentCommand\xxrightarrow{ m o m }{%
    \IfNoValueTF%
    {#2}
    {%
        \begingroup
        \settowidth\min@xx{$\m@th\scriptstyle#1$}
        \sbox8{$\m@th\scriptstyle#3$} % superscript
        \ifdim\wd8>\min@xx \min@xx=\wd8 \fi
        \xrightarrow{\mathmakebox[\min@xx]{\scriptstyle#3}}
        \endgroup
    }%
    {%
        \begingroup
        \settowidth\min@xx{$\m@th\scriptstyle#1$}
        \sbox8{$\m@th\scriptstyle#2$}  % subscript
        \ifdim\wd8>\min@xx \min@xx=\wd8 \fi
        \sbox8{$\m@th\scriptstyle#3$} % superscript
        \ifdim\wd8>\min@xx \min@xx=\wd8 \fi
        \xrightarrow[{\mathmakebox[\min@xx]{\scriptstyle#2}}]
        {\mathmakebox[\min@xx]{\scriptstyle#3}}
        \endgroup
    }%
}
\makeatother

\begin{document}    
    $A\xxrightarrow{1000}{1}A$

    $A\xxrightarrow{1000}{1000}A\xrightarrow{1000}A$

    $A\xxrightarrow{1000}[1]{1}A\xrightarrow[1]{1}A$

    $A\xxrightarrow{1000}[1]{1}A\xrightarrow[1]{1}A$
\end{document}