Philip Taylor wrote:
> I can see you have not lost your Wizard's touch, David, but will > that work (as-is) in an IniXeTeX regime ? It looks to me as if > it makes use of some plain-defined control words such as \z@ and > \@ne. We need it to work in an Ini(Xe)TeX regime so that we can > introduce different patterns to compare the behaviour. Hmmm. I did my best to take out all Plain dependencies, but I failed and it will no longer compile. David, can you identify where I screwed up ? ** Phil. -------- % !TeX Program=IniXeTeX \catcode `\{=1 \catcode `\}=2 \catcode `\# = 6 \catcode `\% = 14 \catcode `\ = \catcode `\ % tab !!! \let \bgroup = { \let \egroup = } \input unicode-letters.tex \input hyph-de-1996.tex \font \tenrm = "Times New Roman" at 10pt \hyphenchar \tenrm = `\- \tenrm \parfillskip = 0 pt \catcode `\@ = 11 \def \showhyphenspace #1 {% \ifx \valign #1\valign \else #1\vadjust {\penalty 1 } \expandafter \showhyphenspace \fi } \def \showhyphens #1% {% \bgroup \showboxbreadth = 16383.99999 pt \showboxdepth = 16383.99999 pt \pretolerance = -1 \tolerance = -1 \setbox 2 = \hbox {}% \setbox 0 = \vbox \bgroup \pretolerance = -1 \tolerance = -1 \hbadness = 0 \parfillskip = 0 pt \relax \hsize = 1 sp \noindent \hskip 0 pt \hfuzz = 16383.99999 pt \hbadness = \dimexpr {16383.99999 pt }% \showhyphenspace #1 {} \par \loop \count 0 = 1 \ifnum 1 = \lastnodetype \setbox 4 = \lastbox \setbox 2 = \hbox {\unhbox 4 \unhbox 2}% \count 0 = 0 \fi \ifnum 11 = \lastnodetype \unskip \count 0 = 0 \fi \ifnum 13 = \lastnodetype \count 2 = \lastpenalty \unpenalty \count 0 = 0 \ifnum \count 2 = 1 \setbox 2 = \hbox {\unhbox 2 }% \count 0 = 0 \fi \fi \ifnum 0 = \count 0 \repeat \hsize = 16383.99999 pt \hfuzz = 0 pt \hbadness = 0 \par \unhbox 2 \par \egroup \egroup } \showhyphens {wußte geißeln} \end -------------------------------------------------- Subscriptions, Archive, and List information, etc.: http://tug.org/mailman/listinfo/xetex