I was trying to use Hendrik Vogt’s \widebar in my re-definitions of some macros in the actuarialsymbol package. The package offers commands that come with * variants which produce \bar{<principal symbol>}, and some that come with ** variants which produce \ddot{<principal symbol>}.
When I replaced the * variant with \overline{<principal symbol>}, everything worked just fine. But when I tried \widebar{<principal symbol>}, the starred command seemed to gobble the first optional argument and left some weird spaces in front of the principal symbol.
Here is a MWE:
\documentclass{article}
%\usepackage{amsmath}
\usepackage[nobracket]{actuarialangle}
\usepackage{actuarialsymbol} % actuarialsymbol loads amsmath internally
% Hendrik Vogt's \widebar: https://tex.stackexchange.com/a/60253/164314
% Depends on the amsmath package!
\makeatletter
\let\save@mathaccent\mathaccent
\newcommand*\if@single[3]{%
\setbox0\hbox{${\mathaccent"0362{#1}}^H$}%
\setbox2\hbox{${\mathaccent"0362{\kern0pt#1}}^H$}%
\ifdim\ht0=\ht2 #3\else #2\fi
}
%The bar will be moved to the right by a half of \macc@kerna, which is computed by amsmath:
\newcommand*\rel@kern[1]{\kern#1\dimexpr\macc@kerna}
%If there's a superscript following the bar, then no negative kern may follow the bar;
%an additional {} makes sure that the superscript is high enough in this case:
\newcommand*\widebar[1]{\@ifnextchar^{{\wide@bar{#1}{0}}}{\wide@bar{#1}{1}}}
%Use a separate algorithm for single symbols:
\newcommand*\wide@bar[2]{\if@single{#1}{\wide@bar@{#1}{#2}{1}}{\wide@bar@{#1}{#2}{2}}}
\newcommand*\wide@bar@[3]{%
\begingroup
\def\mathaccent##1##2{%
%Enable nesting of accents:
\let\mathaccent\save@mathaccent
%If there's more than a single symbol, use the first character instead (see below):
\if#32 \let\macc@nucleus\first@char \fi
%Determine the italic correction:
\setbox\z@\hbox{$\macc@style{\macc@nucleus}_{}$}%
\setbox\tw@\hbox{$\macc@style{\macc@nucleus}{}_{}$}%
\dimen@\wd\tw@
\advance\dimen@-\wd\z@
%Now \dimen@ is the italic correction of the symbol.
\divide\dimen@ 3
\@tempdima\wd\tw@
\advance\@tempdima-\scriptspace
%Now \@tempdima is the width of the symbol.
\divide\@tempdima 10
\advance\dimen@-\@tempdima
%Now \dimen@ = (italic correction / 3) - (Breite / 10)
\ifdim\dimen@>\z@ \dimen@0pt\fi
%The bar will be shortened in the case \dimen@<0 !
\rel@kern{0.6}\kern-\dimen@
\if#31
\overline{\rel@kern{-0.6}\kern\dimen@\macc@nucleus\rel@kern{0.4}\kern\dimen@}%
\advance\dimen@0.4\dimexpr\macc@kerna
%Place the combined final kern (-\dimen@) if it is >0 or if a superscript follows:
\let\final@kern#2%
\ifdim\dimen@<\z@ \let\final@kern1\fi
\if\final@kern1 \kern-\dimen@\fi
\else
\overline{\rel@kern{-0.6}\kern\dimen@#1}%
\fi
}%
\macc@depth\@ne
\let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar
\mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax}%
\macc@set@skewchar\relax
\let\mathaccentV\macc@nested@a
%The following initialises \macc@kerna and calls \mathaccent:
\if#31
\macc@nested@a\relax111{#1}%
\else
%If the argument consists of more than one symbol, and if the first token is
%a letter, use that letter for the computations:
\def\gobble@till@marker##1\endmarker{}%
\futurelet\first@char\gobble@till@marker#1\endmarker
\ifcat\noexpand\first@char A\else
\def\first@char{}%
\fi
\macc@nested@a\relax111{\first@char}%
\fi
\endgroup
}
\makeatother
% End of \widebar
\begin{document}
\linespread{1.5}\selectfont % More visible
% Default:
$\Ax[u|][2]{\termxn}$, $\Ax*[u|]{\termxn}$, $\Ax*[u|][2]{\termxn}$, works fine.
% Change the one starred version to \overline:
\makeatletter
\renewcommand\@actinssc[1]{%
\@ifstar{\acts@actsc{\overline{#1}}}%
{\acts@actsc{#1}}}
\makeatother
$\Ax[u|][2]{\termxn}$, $\Ax*[u|]{\termxn}$, $\Ax*[u|][2]{\termxn}$, works fine as well.
% Change the one starred version to \widebar:
\makeatletter
\renewcommand\@actinssc[1]{%
\@ifstar{\acts@actsc{\widebar{#1}}}%
{\acts@actsc{#1}}}
\makeatother
$\Ax[u|][2]{\termxn}$, $\Ax*[u|]{\termxn}$, $\Ax*[u|][2]{\termxn}$, what happened?
\end{document}

@actsymboldoes not help. The bad interaction between\widebarand actuarialsymbol must be at this line inactuarialsymbol.stythat positions the left subscript and superscript next to an invisible symbol:\mathopen{\vphantom{\textstyle #4}^{\box\tw@}\sb{\box\@ne}}. I don't know how\widehatgobbles the subscript, though. @hendrik-vogt? – Vincent Goulet Jun 05 '18 at 02:30