Generating a table of glyphs with XeTeX

Question

I am trying to get XeTeX to generate a table of all the possible glyphs of a given font, but can not figure out how to make XeTeX know how many glyphs the font has, so I have to specify an exact number.

This is what I have got so far:

\documentclass[border=0.5cm,10pt,landscape]{standalone}
\usepackage{geometry}
\usepackage{xunicode, xltxtra}
\usepackage{fontspec}
\usepackage{multicol}
\setmainfont[Mapping=tex-text, Ligatures={Historical}, Alternate=0]{Linux Libertine O}
\setlength{\columnsep}{0.3cm}
\setlength{\columnseprule}{1pt}
\begin{document}

\begin{multicols}{10}
\newcount\charcount
\charcount=0
\loop
\number\charcount \hspace{1ex} \XeTeXglyph\charcount
\par
\ifnum\charcount<2500
\advance\charcount1
\repeat
\end{multicols}

\end{document}

Fig. 1: Example of the output.

How can it automatically know the exact number of accessible glyphs?

Bonus: Is anyone willing to waste his time by making it put everything inside a table with 10 columns? :-) It also reports an error probably caused by multicol which a I could not fix yet.

Answer 1

To answer your question:

1. Use \iffontchar\font number ... \fi to test if a glyph is in the font.

2. multicol is OK for me. Set \parindent to 0pt to get proper result.

Here is a neat example:

...

\documentclass[landscape]{article}
\usepackage{geometry}
\usepackage{fontspec}
\setmainfont{Linux Libertine O}
\usepackage{multicol}
\setlength{\columnseprule}{0.4pt}
\usepackage{multido}
\setlength{\parindent}{0pt}
\begin{document}

\begin{multicols}{10}
\multido{\i=0+1}{"10000}{% from U+0000 to U+FFFF
  \iffontchar\font\i
    \makebox[3em][l]{\i}%
    \symbol{\i}\endgraf
  \fi
}
\end{multicols}

\end{document}

Note: I tried my best to keep the code simple. But there're still two unusual points: I use \iffontchar to test the font characters, and use \endgraf instead of \par to cheat \multido.

Answer 2

I know that you are asking for a solution with XeTeX (and egreg already gave one), but here is a LuaTeX solution:

\documentclass[10pt,landscape]{article}
\usepackage{luacode}
\usepackage[margin=0.5cm]{geometry}
\usepackage{fontspec}
\usepackage{multicol}
\setlength{\columnsep}{0.3cm}
\setlength{\columnseprule}{1pt}

\newfontface\thefont{Linux Libertine O}

\begin{document}
\section*{Linux Libertine O}
\begin{multicols}{7}\noindent
\begin{luacode*}
local f = fontloader.open('/usr/local/texlive/2011/texmf-dist/fonts/opentype/public/libertine/fxlr.otf')

local glyphs = {}
for i = 0, f.glyphmax - 1 do
   local g = f.glyphs[i]
   if g then
       table.insert(glyphs, {name = g.name, unicode = g.unicode})
   end
end

table.sort(glyphs, function (a,b) return (a.unicode < b.unicode) end)

for i = 1, #glyphs do
   tex.sprint(glyphs[i].unicode .. ': ')
   if (glyphs[i].unicode > 0) then
       tex.sprint('{\\thefont\\char' .. glyphs[i].unicode .. '}');
   end
   tex.sprint(' {\\tiny (')
   tex.sprint(-2, glyphs[i].name)
   tex.sprint(')}\\\\')
end

fontloader.close(f)
\end{luacode*}
\end{multicols}
\end{document}

Answer 3

\documentclass{article}
\usepackage[a4paper,landscape,textwidth=239mm,textheight=480pt]{geometry}
\usepackage{fontspec}
\setmainfont{Linux Libertine O}
\newcount\charcount
\parindent=0pt
\begin{document}

\chardef\highest=\XeTeXcountglyphs\font
\offinterlineskip
\loop
\makebox[15mm][l]{\strut\vrule\,\number\charcount\hfill \XeTeXglyph\charcount}\hskip1mm plus 1mm
\ifnum\charcount<\numexpr\highest-1\relax
\advance\charcount1
\repeat

\end{document}

With Linux Libertine the last glyph produces an error, that's the reason of the strange limitation. This does not happen with other fonts.

If one wants a table with the actual codes for the characters, a slower routine must be used:

\documentclass{article}
\usepackage[a4paper,landscape,textwidth=239mm,textheight=480pt]{geometry}
\usepackage{fontspec}
\setmainfont{Old Standard}
\newcount\charcount
\parindent=0pt
\begin{document}

\offinterlineskip
\def\dochar{\iffontchar\font\charcount
  \makebox[15mm][l]{\strut\vrule\,{\tiny\number\charcount}\hfill\char\charcount}\hskip1mm plus 1mm
  \fi}
\loop\dochar
\ifnum\charcount<"10FFFF\relax
\advance\charcount1
\repeat

\end{document}

Answer 4

Supplemental Answer

I was looking at some of the answers, and I decided to write a similar one using PGF/TikZ. So, for the sake of the community, here it is!

The fast-to-typeset version:

\documentclass{article}
\usepackage{fontspec}
\setmainfont{Linux Libertine O}
\usepackage{multicol}
\usepackage{pgffor}
\usepackage{tikz}
\begin{document}
\pgfmathtruncatemacro\maxstep{\the\XeTeXcountglyphs\font-1} % https://tex.stackexchange.com/a/235688/13552
\typeout{maxstep is \maxstep} % Prints maxstep to log
Number of glyphs: \maxstep\\
\begin{multicols}{5}
\foreach \charstep in {1,...,\maxstep}{%
    %\iffontchar\font\charstep
        \makebox[3em][l]{\small\charstep}%
        \XeTeXglyph\charstep
        \endgraf
    %\else
        %\charstep{} not exist
    %\fi
}%
\end{multicols}
\end{document}

The slow-to-typeset, \tikzpicture version:

\documentclass{article}
\usepackage{fontspec}
\setmainfont{Linux Libertine O}
\usepackage{multicol}
\usepackage{pgffor}
\usepackage{tikz}
\begin{document}
\begin{multicols}{5}
\pgfmathsetmacro\maxstep{\the\XeTeXcountglyphs\font-1} % https://tex.stackexchange.com/a/235688/13552
\typeout{maxstep is \maxstep} % Prints maxstep to log
\foreach \charstep in {1,...,\maxstep}{%
    %\iffontchar\font\charstep
        \makebox[3em][l]{\small\charstep}%
        \begin{tikzpicture}[anchor=base, baseline]
            \node [red,font=\fontsize{12}{17}\selectfont] {\XeTeXglyph\charstep};
        \end{tikzpicture}\endgraf
    %\else
        %\charstep{} not exist
    %\fi
}%
\end{multicols}
\end{document}

The great thing about using \XeTeXglyph as Paul suggested is that you do not need to check whether or not the characters exist. This is why I commented out the if statement (\iffontchar\font\charstep).

Note that instead of \pgfmathsetmacro, you could use \edef:

\edef\maxstep{\the\numexpr\XeTeXcountglyphs\font-1\relax} % https://tex.stackexchange.com/a/235686/13552

It also works with Leo Liu's answer as mentioned by Paul. Here is an example:

\begin{multicols}{5}
\multido{\i=0+1}{\XeTeXcountglyphs\font}{
    \makebox[3em][l]{\i}%
    \XeTeXglyph\i\endgraf
}%
\end{multicols}

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