使用 Hendrik Vogt 的“\widebar”时下标输出错误

Question

当我尝试使用亨德里克·沃格特的“\widebar”时我可以不使用 mathabx 包来获得 \widebar 吗？，输出恰好

\widebar{\mathcal{I}}_{x}=\widebar{\mathcal{I}_{x}}

变为如下：

。

内部索引“x”会发生变化。但如果我添加一个额外的“{ }”，例如\widebar{{\mathcal{I}}_{x}}，我将获得正确的形式

我不知道如何修复代码，使其在没有额外“{ }”的情况下工作。有人能帮忙吗？

Answer 1

您将获得与以下相同的输出

\mathcal{I_x}

只需在上方添加上划线即可。您可以使用

\widerbar{{\mathcal{I}}_{x}}

要解决这个问题。

我在下面的示例代码中重现了该行为，以表明除了使用了错误的符号之外，输出与 相同\overline。

\documentclass{article}
\usepackage{amsmath}

\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\[email protected]\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

\linespread{1.44}

\begin{document}

$\mathcal{I_x}$

$\widebar{\mathcal{I}}_{x}=\widebar{\mathcal{I}_{x}}$
\quad
$\widebar{\mathcal{I}}_{x}=\overline{\mathcal{I_x}}$

$\widebar{\mathcal{I}}_{x}=\widebar{{\mathcal{I}}_{x}}$
\quad
$\widebar{\mathcal{I}}_{x}=\overline{\mathcal{I}_{x}}$

\end{document}

\widebar{{\mathcal{I}}_{x}}在第三行中，您可以看到和 的输出\overline{\mathcal{I}_{x}}是相同的。

与以下内容进行比较，基于https://tex.stackexchange.com/a/364929/4427

\documentclass{article}
\usepackage{amsmath}

\DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
\DeclareFontShape{U}{mathx}{m}{n}{ <-> mathx10 }{}
\DeclareSymbolFont{mathx}{U}{mathx}{m}{n}
\DeclareFontSubstitution{U}{mathx}{m}{n}
\DeclareMathAccent{\widebar}{\mathalpha}{mathx}{"73}

\linespread{1.44}

\begin{document}

$\widebar{\mathcal{I}}_{x}=\widebar{\mathcal{I}_{x}}$

$\widebar{\mathcal{I}}_{x}=\overline{\mathcal{I}_{x}}$

\end{document}