2012/10/08 の失敗例2

numtohex.tex

%#! luatex
\catcode`\@=11
\def\len@begin{\directlua{
    local len_count = 0
    local len_flag = true
    while len_flag do
      local t = token.get_next()
      if token.csname_name(t) == 'len@end' then
        len_flag = false
        print(); print()
        print('len(yourcode)=' .. tostring(len_count))
      end
      len_count = len_count + 1
    end
}}
\def\length#1{\len@begin#1\len@end#1}
\length{% ここから
\def\numtohex#1{%
    \count@#1\divide\count@\sixt@@n
    \uccode`<\numtohex@aux
    \count@\numexpr#1-\count@*\sixt@@n
    \uccode`>\numtohex@aux
    \uppercase{\def\result{<>}}}%
\def\numtohex@aux{\numexpr\count@+`0\ifnum\count@>9+7\fi}% ここまで
}
\catcode`\@=12
\numtohex{042}\show\result %==>2A
\numtohex{`\\}\show\result %==>5C
\count0=3 \numtohex{\count0}\show\result %==>03
\bye

http://d.hatena.ne.jp/zrbabbler/20121008/1349677570