This is comparable to the problem of bundling several key-value arguments (for instance for \includegraphics
) in a macro: At the time of executing (the amsmath version of) \sqrt
, which is now \OldSqrt
, the content of []
is parsed for \leftroot
&c. At that time only \DHLindex
is found, and later \leftroot
is useless.
Changing
\setbox0=\hbox{$#1\OldSqrt[\DHLindex]{#2\,}$}\dimen0=\ht0\relax%
to
\setbox0=\hbox{$#1\expandafter\OldSqrt\expandafter[\DHLindex]{#2\,}$}\dimen0=\ht0\relax%
makes the \leftroot
visible and the first attempt goes through.
Dunno whether it does what you want though.
This should also work for attempt 2. Didn't look at attempt 3.
EDIT: What exactly do you want to know about the \DHLhksqrt
macro? AFAICS it takes two arguments, builds a root expression from them, which it measures and tries (with debatable success) to decorate with an additional vertical rule.
EDIT2: To make this look a bit better, at least \DHLhksqrt
needs to take into account when the root index is moved above the root symbol, otherwise the additional rule is placed too high.
EDIT3: Ok, here's my take at a correction for the displacement problem. The root is formatted twice, so there might be a performance problem.
\def\DHLhksqrt#1#2{%
\setbox0=\hbox{$#1\OldSqrt{#2\,}$}\dimen0=\ht0\relax%
\advance\dimen0-0.2\ht0\relax%
\setbox2=\hbox{\vrule height\ht0 depth -\dimen0}%
{\hbox{$#1\expandafter\OldSqrt\expandafter[\DHLindex]{#2\,}$}\lower0.4pt\box2}%
}
EDIT4: Unfortunately the root symbol has more whitespace above in bold math mode. Look at
\fboxsep0pt
\fbox{$f(x) = \OldSqrt{e^{2x}}$}
\fbox{\boldmath$ f(x) = \OldSqrt{e^{2x}}$}

Hence, this has to be considered when placing the "closing rule":
\makeatletter
\newcommand*\bold@name{bold}
\def\DHLhksqrt#1#2{%
\setbox0=\hbox{$#1\OldSqrt{#2\,}$}\dimen0=\ht0\relax%
\advance\dimen0-0.2\ht0\relax%
\setbox2=\hbox{\vrule height\ht0 depth -\dimen0}%
{%
\hbox{$#1\expandafter\OldSqrt\expandafter[\DHLindex]{#2\,}$}%
\lower\ifx\math@version\bold@name0.6pt\else0.4pt\fi\box2%
}%
}
\makeatother

Best Answer
If you want commands with at most one optional argument, LaTeX's built in
\newcommand
and\renewcommand
are an easy way to define them. The syntax isIf you specify the second optional argument, then the first argument to
\name
will be optional. So in your case one can simply useNote that (as in your question), we do not need to specify the second argument, because
\oldsqrt
will look for it anyway, so that this definition is functionally equivalent toIf you need several optional arguments or an argument other than the first one should be optional, you can either chain
\@ifnextchar
s (as in Yiannis' answer), or usexparse
(as in Peter's answer).