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Created Oct 8, 2012
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2012/10/08 の失敗例
%#! luatex
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
len_count = len_count+2 % \newcount\@tempcnta の分を足す
print(); print()
print('len(yourcode)=' .. tostring(len_count))
len_count = len_count + 1
\length{% ここから数える
\def\@numtohex#1{\ifcase\numexpr#1-10 A\or B\or C\or D\or E\or F\else\number#1\fi}% ここまで
\numtohex{042}\show\result %==>2A
\numtohex{`\\}\show\result %==>5C
\count0=3 \numtohex{\count0}\show\result %==>03

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@kmaed kmaed commented Oct 8, 2012

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