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diff --git a/tex/.texmf/tex/generic/pgfplots/util/pgfplotsbinary.code.tex b/tex/.texmf/tex/generic/pgfplots/util/pgfplotsbinary.code.tex new file mode 100644 index 0000000..ca2c74c --- /dev/null +++ b/tex/.texmf/tex/generic/pgfplots/util/pgfplotsbinary.code.tex @@ -0,0 +1,603 @@ +%-------------------------------------------- +% $Header: /cvsroot/pgfplots/pgfplots/generic/pgfplots/util/pgfplotsbinary.code.tex,v 1.13 2009/07/21 18:18:48 ludewich Exp $ +% +% Package pgfplots +% +% Provides a user-friendly interface to create function plots (normal +% plots, semi-logplots and double-logplots). +% +% It is based on Till Tantau's PGF package. +% +% Copyright 2007/2008 by Christian Feuersänger. +% +% This program is free software: you can redistribute it and/or modify +% it under the terms of the GNU General Public License as published by +% the Free Software Foundation, either version 3 of the License, or +% (at your option) any later version. +% +% This program is distributed in the hope that it will be useful, +% but WITHOUT ANY WARRANTY; without even the implied warranty of +% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the +% GNU General Public License for more details. +% +% You should have received a copy of the GNU General Public License +% along with this program. If not, see <http://www.gnu.org/licenses/>. +% +%-------------------------------------------- + +% This file provides a self-contained package which does only need +% pgfkeys. +% +% It provides a method to convert TeX numbers (integers and +% dimensions) into binary format (macros with catcode 11 or 12). +% + +\edef\pgfplotsbinaryatcode{\the\catcode`\@ } +\catcode`\@=11 + + +% Returns a single character, which has the +% binary ASCII code '#1', with catcode 11. +% +% #1 (expands to) a number between 0 and 255 (inclusive). +% +% @see \pgfplotsgetchar Note that \pgfplotsgetchar is more powerful, +% but can't be used inside of \edef (it is not expandable) whereas +% \pgfplotscharno is. +\def\pgfplotscharno#1{\csname pgfp@bin@#1\endcsname}% +\let\pgfplotscharno@bincatcode=\pgfplotscharno + +\def\pgfplotscharno@lualatex#1{#1,} + +\input pgfplotsbinary.data.code.tex + +% Defines the LUA (!) value pgfplotsretval to be a binary string +% containing the pgfplots binary value #1. +% +% #1 a pgfplots binary value collected with \pgfplotscharno. +% More precisely, it should be a comma-separated sequence of numbers +% of the form '0,255,2,128,' (can be terminated by comma). It will be +% converted to the respective binary numbers 0x0, 0xff, 0x02,.. +% +% example: +% \pgfplotsbinarytoluabinary{0, 255,2,128} +% \directlua{ +% pdf.immediateobj{"stream", pgfplotsretval,"/DataWithBinaryStream"} +% } +\def\pgfplotsbinarytoluabinary#1{% + % lualatex does not support binary chars as pdftex does - so we have to resort to LUA + % methods. The idea is to use + % string.char(1,2,3) which results in a binary string with chars 0x01, 0x02, 0x03 etc. + % I only need to get the integer numbers. To this end, I patch \pgfplotscharno + % and create the binary string here: + \directlua{% + pgfplotsretval = pgfplotsGetLuaBinaryStringFromCharIndices({#1}); + }% +}% + +% Defines \pgfplotsretval to be the ASCII character for #1, with +% catcode 11. +% +% #1: either a number between 0 and 255 (inclusive) or a description +% of the character. +% +% Examples: +% \pgfplotsgetchar{35} +% \pgfplotsgetchar{`\#} % code for '#' +% \pgfplotsgetchar{`\^^M} % Newline +% \pgfplotsgetchar{`\^^ff}% 255 +% +% @see \pgfplotscharno +\def\pgfplotsgetchar#1{% + \begingroup + \count0=#1\relax + \edef\pgfplotsretval{\pgfplotscharno{\the\count0 }}% + \pgfmath@smuggleone\pgfplotsretval + \endgroup +}% + +\def\pgfplotsbinary@apphighorderbytes@BIGENDIAN#1{\xdef\pgfplotsbinaryresult{#1\pgfplotsbinaryresult}}% +\def\pgfplotsbinary@apphighorderbytes@LITTLEENDIAN#1{\xdef\pgfplotsbinaryresult{\pgfplotsbinaryresult#1}}% + +\def\pgfplotsbinaryencode@badic@unsigned@PAD@LITTLEENDIAN{% + % pad with zeros: + \ifcase\c@pgfplotsbin@byteno + % ok. + \or + % one byte missing. + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST}% + \or + % two bytes missing. + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO@HIGHEST}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO@HIGHEST}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO@HIGHEST}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO@HIGHEST}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO@HIGHEST}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO@HIGHEST}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO@HIGHEST}% + \else + \pgfplots@error{Sorry, I can't process byte no \the\c@pgfplotsbin@byteno... you may need to change bytes=\pgfplotsbinary@bytes.}% + \fi +}% +\def\pgfplotsbinaryencode@badic@unsigned@PAD@BIGENDIAN{% + % pad with zeros: + \ifcase\c@pgfplotsbin@byteno + % ok. + \or + % one byte missing. + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST}% + \or + % two bytes missing. + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST\pgfplotsbinary@ZERO}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO}% + \or + \pgfplotsbinary@apphighorderbytes{\pgfplotsbinary@ZERO@HIGHEST\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO}% + \else + \pgfplots@error{Sorry, I can't process byte no \the\c@pgfplotsbin@byteno... you may need to change bytes=\pgfplotsbinary@bytes.}% + \fi +}% + +\expandafter\def\csname pgfplotsbinarysetbytes@1\endcsname{% + \def\pgfplotsbinary@add@signed@largest@absolute{\advance\c@pgfplotsbin@input by 127 }% + \def\pgfplotsbinaryencodesignedmaplinearly@prepare{% + % warning: \pgfplotsbinary@bytes is NOT necessarily 1 (ASCII + % encoding features) + \c@pgfplotsbin@byteno=\pgfplotsbinary@bytes\relax + \c@pgfplotsbin@basis=\pgfplotsbinary@basis\relax + \divide\c@pgfplotsbin@input by16909320 % ~= (2^31-1) / (2^(8*1-1) -1) + }% +}% +\expandafter\def\csname pgfplotsbinarysetbytes@2\endcsname{% + \def\pgfplotsbinary@add@signed@largest@absolute{\advance\c@pgfplotsbin@input by 32767 }% + \def\pgfplotsbinaryencodesignedmaplinearly@prepare{% + \c@pgfplotsbin@byteno=\pgfplotsbinary@bytes\relax + \c@pgfplotsbin@basis=\pgfplotsbinary@basis\relax + \divide\c@pgfplotsbin@input by65538 % ~= (2^31-1) / (2^(8*2-1) -1) + }% +}% +\expandafter\def\csname pgfplotsbinarysetbytes@3\endcsname{% + \def\pgfplotsbinary@add@signed@largest@absolute{\advance\c@pgfplotsbin@input by 8388607 }% + \def\pgfplotsbinaryencodesignedmaplinearly@prepare{% + \c@pgfplotsbin@byteno=\pgfplotsbinary@bytes\relax + \c@pgfplotsbin@basis=\pgfplotsbinary@basis\relax + \divide\c@pgfplotsbin@input by256 % ~= (2^31-1) / (2^(8*3-1) -1) + }% +}% +\expandafter\def\csname pgfplotsbinarysetbytes@4\endcsname{% + \def\pgfplotsbinary@add@signed@largest@absolute{% + \advance\c@pgfplotsbin@input by 2147483647 % this is the *absolute* largest int that TeX can handle. + }% + \def\pgfplotsbinaryencodesignedmaplinearly@prepare{% + \c@pgfplotsbin@byteno=\pgfplotsbinary@bytes\relax + \c@pgfplotsbin@basis=\pgfplotsbinary@basis\relax + }% +}% +\def\pgfplotsbinarysetbytes@@{% + \def\pgfplotsbinaryencodesignedmaplinearly@prepare{% + \c@pgfplotsbin@byteno=\pgfplotsbinary@bytes\relax + \c@pgfplotsbin@basis=\pgfplotsbinary@basis\relax + \pgfplots@error{Sorry, but I can't perform \string\pgfplotsbinaryencodesignedmaplinearly\space for bytes=\pgfplotsbinary@bytes\space yet... bytes=4 is the maximum.}% + }% + \def\pgfplotsbinary@add@signed@largest@absolute{ + \advance\c@pgfplotsbin@input by 2147483647 + }% +}% +\expandafter\let\csname pgfplotsbinary@bytes@5\endcsname=\pgfplotsbinarysetbytes@@ +\expandafter\let\csname pgfplotsbinary@bytes@6\endcsname=\pgfplotsbinarysetbytes@@ +\expandafter\let\csname pgfplotsbinary@bytes@7\endcsname=\pgfplotsbinarysetbytes@@ +\expandafter\let\csname pgfplotsbinary@bytes@8\endcsname=\pgfplotsbinarysetbytes@@ + +\pgfqkeys{/pgfplots/bin}{% + % ordering not yet implemented; uses always BIG ENDIAN. + ordering/.is choice,% + ordering/big endian/.code={% + \def\pgfplotsbinary@byteorder{0}% + \let\pgfplotsbinary@apphighorderbytes=\pgfplotsbinary@apphighorderbytes@BIGENDIAN + \let\pgfplotsbinaryencode@badic@unsigned@PAD=\pgfplotsbinaryencode@badic@unsigned@PAD@BIGENDIAN + },% + ordering/net/.style={/pgfplots/bin/ordering/big endian},% + ordering/little endian/.code={% + \def\pgfplotsbinary@byteorder{1}% + \let\pgfplotsbinary@apphighorderbytes=\pgfplotsbinary@apphighorderbytes@LITTLEENDIAN + \let\pgfplotsbinaryencode@badic@unsigned@PAD=\pgfplotsbinaryencode@badic@unsigned@PAD@LITTLEENDIAN + }, + ordering/big endian,% + % + % The standard method - it results in binary encoded numbers. + binary encoding/.code={% + \pgfutil@IfUndefined{directlua}{% + \let\pgfplotscharno=\pgfplotscharno@bincatcode + }{% + % Ah - we use LuaTeX! + % At the time of this writing, LUA does not allow binary output which has been + % created by means of catcode modifications & TeX string concatenation. + % binary output in LUA needs to be (re)implemented in LUA (see inline code + % comments below). + % + % There are two possible work-arounds: + % (a) Base64 encoding + % (b) binary encoding using special LUA handling. + % This is what I do. Set the 'encode filter' such that it reinitializes the encoder: + % we patch \pgfplotscharno with a special routine which collects + % only the integer indices: + \let\pgfplotscharno=\pgfplotscharno@lualatex + % later, the user has to convert this list into a binary lua + % string before he can use it. See \pgfplotsbinarytoluabinary + }% + \edef\pgfplotsbinary@ZERO{\pgfplotscharno0}% + \edef\pgfplotsbinary@ZERO@LINEARMAP{\pgfplotscharno{128}}% + \let\pgfplotsbinary@ZERO@HIGHEST=\pgfplotsbinary@ZERO + \def\pgfplotsbinary@basis{256}% + \let\pgfplotsbinary@hook=\relax + \def\pgfplotsbinary@hook@signed@linearmap{% + \ifnum\c@pgfplotsbin@byteno=0 + \advance\c@pgfplotsbin@input by128 + \ifnum\c@pgfplotsbin@input>255 + \pgfplotsbinary@hook@signed@linearmap@error + \fi + \fi + }% + \def\pgfplotsbinarysetbytes##1{% + \pgfutil@ifundefined{pgfplotsbinarysetbytes@##1}{% + \pgfplots@error{Sorry, I can't write binary output with '##1' bytes yet...}% + }{% + \edef\pgfplotsbinary@bytes{##1}% + \csname pgfplotsbinarysetbytes@##1\endcsname + }% + }% + },% + % + % This applies 'binary encoding' and encodes the resulting bytes + % in Hex. It corresponds to the ASCIIHexEncode in postscript or + % pdf. + % Please note that 'bytes' sets the number of binary bytes - the + % actual encoding length is exactly twice as large. + ASCIIHexEncode/.code={% + \let\pgfplotscharno=\pgfplotscharno@bincatcode + \edef\pgfplotsbinary@ZERO{\pgfplotscharno{48}}% + \edef\pgfplotsbinary@ZERO@LINEARMAP{\pgfplotscharno{56}}% + \let\pgfplotsbinary@ZERO@HIGHEST=\pgfplotsbinary@ZERO + \pgfkeysalso{/pgfplots/bin/ordering/big endian}% + \def\pgfplotsbinary@basis{16}% + \def\pgfplotsbinary@hook@hex{% + \ifnum\c@pgfplotsbin@input<10 + \advance\c@pgfplotsbin@input by48 + \else + \advance\c@pgfplotsbin@input by55 + \fi + }% + \let\pgfplotsbinary@hook=\pgfplotsbinary@hook@hex + \def\pgfplotsbinary@hook@signed@linearmap{% + \ifnum\c@pgfplotsbin@byteno=0 + \advance\c@pgfplotsbin@input by8 + \ifnum\c@pgfplotsbin@input>16 + \pgfplotsbinary@hook@signed@linearmap@error + \fi + \fi + \pgfplotsbinary@hook@hex + }% + \def\pgfplotsbinarysetbytes##1{% + \pgfutil@ifundefined{pgfplotsbinarysetbytes@##1}{% + \pgfplots@error{Sorry, I can't write binary output with '##1' bytes yet...}% + }{% + \csname pgfplotsbinarysetbytes@##1\endcsname + \begingroup + \count0=##1\relax + \multiply\count0 by2 + \xdef\pgfplotsbinary@glob@TMP{\the\count0 }% + \endgroup + \let\pgfplotsbinary@bytes=\pgfplotsbinary@glob@TMP + }% + }% + },% + % + % + % This applies 'binary encoding' and encodes the resulting bytes + % using a base 85 encoding. It corresponds to the ASCII85Encode in postscript or + % pdf. + % Handle this method with care - it works just for ONE number, not for a stream of + % numbers as in pdf. Therefore, it might not be useful at all. + % Please note that 'bytes' will be ignored; ASCII85Encode assumes + % 4 binary bytes and uses 5 bytes to encode them. + % + % @ATTENTION bytes is ALWAYS 4, regardless of the setting of + % 'bytes'! + ASCII85Encode/.code={% + \let\pgfplotscharno=\pgfplotscharno@bincatcode + \edef\pgfplotsbinary@ZERO{\pgfplotscharno{33}}% + \edef\pgfplotsbinary@ZERO@LINEARMAP{\pgfplotscharno{42}}% + \let\pgfplotsbinary@ZERO@HIGHEST=\pgfplotsbinary@ZERO + \pgfkeysalso{/pgfplots/bin/ordering/big endian}% + \edef\pgfplotsbinary@ASCII@specialzero{\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO\pgfplotsbinary@ZERO}% + \expandafter\def\expandafter\pgfplotsbinaryencode@badic@unsigned@PAD\expandafter{% + \pgfplotsbinaryencode@badic@unsigned@PAD + \ifx\pgfplotsbinaryresult\pgfplotsbinary@ASCII@specialzero + % PDF standard: 0 is encoded as 'z': + \gdef\pgfplotsbinaryresult{z}% + \fi + }% + \def\pgfplotsbinary@basis{85}% + \def\pgfplotsbinary@hook{% + \advance\c@pgfplotsbin@input by33 + }% + \def\pgfplotsbinary@hook@signed@linearmap{% + \advance\c@pgfplotsbin@input by33 + \ifnum\c@pgfplotsbin@byteno=0 + \advance\c@pgfplotsbin@input by42 + \ifnum\c@pgfplotsbin@input>85 + \pgfplotsbinary@hook@signed@linearmap@error + \fi + \fi + }% + % I know, that does only work efficiently if bytes=4 for every + % encoded number. + \def\pgfplotsbinarysetbytes##1{% + \def\pgfplotsbinary@bytes{5}% + \csname pgfplotsbinarysetbytes@4\endcsname + }% + \pgfplotsbinarysetbytes4% + },% + binary encoding,% + % + bytes/.code={\pgfplotsbinarysetbytes{#1}},% + bytes=4, + % + % Irreversibly change to VERBATIM output for debugging: + debug mode/.code={% + \let\pgfplotsbinary@apphighorderbytes@ORIG=\pgfplotsbinary@apphighorderbytes + \def\pgfplotsbinary@ZERO{[Pad-0]}% + \let\pgfplotsbinary@ZERO@HIGHEST=\pgfplotsbinary@ZERO + \def\pgfplotsbinary@ZERO@LINEARMAP{[Pad-128]}% + \def\pgfplotsbinary@apphighorderbytes##1{% + \pgfutil@ifnextchar\pgfplotscharno{% + \pgfplotsbinary@apphighorderbytes@DEBUG@csname + }{% + \pgfplotsbinary@apphighorderbytes@DEBUG@normal + }% + ##1\relax + }% + },% + % Write pdf objects in binary form. This does only work with + % pdftex, and its output is only useful in conjunction with + % \pdfcompresslevel=0 + % and a text editor. + % Usage: + % \pgfkeys{/pgfplots/bin/debug to pdf={\pgfplotsbinaryencodeunsigned}{1,2,3,...,16}} + % + % works only with pdftex + debug to pdf/.code 2 args={% + \foreach \num in {#2} {% + #1{\num}% + \immediate \pdfobj stream attr { + /Decimal \num\space + /Routine (\string#1) + } {% + \pgfplotsbinaryresult + }% + }% + },% +} + + +\def\pgfplotsbinary@hook@signed@linearmap@error{% + \pgfplots@error{Sorry, there are not enough bytes to store the current number. I tried to write \the\c@pgfplotsbin@input...}% +}% + +\def\pgfplotsbinary@apphighorderbytes@DEBUG@csname\pgfplotscharno#1\relax{% + \pgfplotsbinary@apphighorderbytes@ORIG{[#1]}% +}% +\def\pgfplotsbinary@apphighorderbytes@DEBUG@normal#1\relax{% + \pgfplotsbinary@apphighorderbytes@ORIG{#1}% +}% + +%\pgfkeys{/pgfplots/bin/debug mode} + + +\countdef\c@pgfplotsbin@input=0 +\countdef\c@pgfplotsbin@tmpa=1 +\countdef\c@pgfplotsbin@tmpb=2 +\countdef\c@pgfplotsbin@byteno=3 +\countdef\c@pgfplotsbin@basis=4 +\def\pgfplotsbinaryempty{} + +%-------------------------------------------------- +% input: unsigned int x, b, n; +% output: unsigned int y[n]; +% for (i=0; i<n; i++) y[i] = 0; +% i=0; +% while (x > 0) +% { y[i] = x % b; /* entspricht x mod b */ +% x = x / b; /* ganzzahlige Division */ +% i++; +% } +%-------------------------------------------------- +% with x = #1 +% b = basis +% will store stuff into \pgfplotsbinaryresult in binary format +% +% PRECONDITION: +% - \pgfplotsbinaryresult= empty! +% - \c@pgfplotsbin@byteno=\pgfplotsbinary@bytes +\def\pgfplotsbinaryencode@badic@unsigned@{% + \ifnum\c@pgfplotsbin@input>0 + \c@pgfplotsbin@tmpa=\c@pgfplotsbin@input + \divide\c@pgfplotsbin@tmpa by\c@pgfplotsbin@basis\relax + \c@pgfplotsbin@tmpb=\c@pgfplotsbin@tmpa\relax + \multiply\c@pgfplotsbin@tmpa by\c@pgfplotsbin@basis\relax + \advance\c@pgfplotsbin@input by -\c@pgfplotsbin@tmpa\relax + \advance\c@pgfplotsbin@byteno by-1 + \pgfplotsbinary@hook% hooks for modifications. + \pgfplotsbinary@apphighorderbytes{\pgfplotscharno{\the\c@pgfplotsbin@input}}% + \c@pgfplotsbin@input=\c@pgfplotsbin@tmpb +%\message{RESULT SO FAR byte no \the\c@pgfplotsbin@byteno: \pgfplotsbinaryresult}% + \expandafter\pgfplotsbinaryencode@badic@unsigned@ + \else + \pgfplotsbinaryencode@badic@unsigned@PAD +%\message{RESULT SO FAR byte no \the\c@pgfplotsbin@byteno: \pgfplotsbinaryresult}% + \fi +}% + + +% Defines \pgfplotsbinaryresult to be the binary representation of an +% unsigned integer. +% +% The representation will use unsigned dual number representation. +% +% The assignment to \pgfplotsbinaryresult will be globally. +% #1: an unsigned integer. It won't be transformed in any way, so make +% sure it fits into the configured number of bytes. It is an error if +% the number is too large or too small. Please note that only unsigned +% numbers are supported with this method. +% +% FIXME : fix > 2^30 +\def\pgfplotsbinaryencodeunsigned#1{% + \begingroup + \global\let\pgfplotsbinaryresult=\pgfplotsbinaryempty + \c@pgfplotsbin@input=#1 % + \c@pgfplotsbin@byteno=\pgfplotsbinary@bytes\relax + \c@pgfplotsbin@basis=\pgfplotsbinary@basis\relax + \pgfplotsbinaryencode@badic@unsigned@% + \endgroup +}% + +% An implementation for signed integers which maps the signed integer linearly into +% the unsigned data range before it proceeds. +% +% The idea is thus, to first introduce a linear mapping +% +% phi : [- smallest_possible, +largest_possible ] -> [0, 256^bytes-1 ] +% +% A signed integer in TeX is in [ - (2^31-1), 2^31-1 ]. +% Thus, we should map +% +% phi : [ -(2^31-1), 2^31-1 ] -> [ 0, 2^32-1 ]. +% +% A simpler case is to employ the symmetry in TeX's registers and +% leave one out, i.e. to map to 2^32-2: +% +% phi : [ -(2^31-1), 2^31-1 ] -> [ 0, 2^32-2 ]. +% +% Then, +% +% phi(x) = ( x + 2^31 -1 ) / (2^31-1 + 2^31-1) * (2^32-2) = x+ 2^31-1. +% +% The same map phi(x) = x + 2^31 -1 with target space [0, 2^32-1 ] +% could be realized with the input space [- (2^31-1), 2^31 ]. +% +% I am using this encoding procedure, phi(x) = x + 2^31 -1. +% +% As a consequence, the binary pattern FF FF FF FF does never occur as +% result of the mapping. +% +% To invert the mapping (i.e. to decode the result), set up the unique +% linear map +% +% psi : [ 0, 2^32-1 ] -> [ -(2^31-1), 2^31 ]. +% +% Then, psi( phi(x) ) = x and the decoding procedure is correct. +% +% This doesn't need TeX register arithmetics on the whole range. +% +% REMARK: the whole operation does also work if bytes<4 (i.e. we have +% less than 32 bits in the target range). In this case, the mapping is +% phi : [ -(2^31-1), 2^31-1 ] -> [ 0, 2^{8*bytes}-1 ] +% and a further, *lossy* quantization still will be applied. The +% quantization step is an integer division performed in signed number +% arithmetics (i.e. it is symmetric around 0). +\def\pgfplotsbinaryencodesignedmaplinearly#1{% + \begingroup + \global\let\pgfplotsbinaryresult=\pgfplotsbinaryempty + \c@pgfplotsbin@input=#1\relax% + \pgfplotsbinaryencodesignedmaplinearly@prepare + \ifnum\c@pgfplotsbin@input<0 % + % compute + 2^31 - 1 + \pgfplotsbinary@add@signed@largest@absolute + \else + % change zero padding such that positive numbers + % get the EFFECT of + 2^31. + \let\pgfplotsbinary@ZERO@HIGHEST=\pgfplotsbinary@ZERO@LINEARMAP + \let\pgfplotsbinary@hook=\pgfplotsbinary@hook@signed@linearmap + % and compute the -1 explicitly here: + \advance\c@pgfplotsbin@input by-1 % + \fi + \pgfplotsbinaryencode@badic@unsigned@% + \endgroup +}% + +% Encodes a dimen (like 1pt or \dimen0) in binary form. +% +% The encoding works by mapping #1 linearly into the allowed integer +% range using a quantization technique to respect the (possibly) +% restricted number of bytes. +% +% The implementation is fast and uses only integer arithmetics. +% It relies on \pgfplotsbinaryencodesignedmaplinearly and a scale. +% +% So, what we do is to setup a linear map into binary range with k +% bytes. The range of a TeX dimen is precisely (in units of pt) +% [ -(2^30-1) / 2^16, (2^30 -1) / 2^16 ] = [-16383.99998, 16383.99998] +% +% Thus, for an input dimen x, we set up the mapping +% phi(x) = 2^16 * x * 2 +% which maps +% phi: [ -(2^30-1) / 2^16, (2^30-1) / 2^16 ] -> [-(2^31 -2), 2^31-2]. +% +% I simply use the \pgfplotsbinaryencodesignedmaplinearly to process +% this further. To simplify the computation, I simply compute +% phi_signed( phi(x) ), +% where phi_signed denotes an application of +% \pgfplotsbinaryencodesignedmaplinearly: +% phi_signed( y ) = y +2^31 -1, +% phi_signed( phi(x) ) = 2^16 * 2 * x + 2^31 - 1. +% This is NOT a linear map to [0,2^32-1] as promised. +% But, we can setup an inverse transformation PHI (which is linear) anyway +% such that +% PHI( phi_signed(phi(x)) ) = x +% and that's all I want. Do do that, we use the unique linear decoder map +% PHI : [ 0,2^32-1 ] -> [ -16383.999992, 16384 ]. +% +% This is not exacty the input range of before, but using it results +% in a proper decoder. The difference is due to the non-unique zero +% representation in TeX's arithmetics. +% +% +% REMARK: the whole operation does also work if bytes<4 (i.e. we have +% less than 32 bits in the target range). In this case, a further +% *lossy* quantization step is applied in phi_signed. The inverse +% transformations are the same, however. See +% \pgfplotsbinaryencodesignedmaplinearly for details about the +% quantization step (or try it out). +% + % + %% DEBUG NOTE: This mapping appears to work correctly according to + %% my tests. + %% For bc -l test codes: + %% ibase=16; + %% -4000 + 809658FA. / (2^20) * 8000 +\def\pgfplotsbinaryencodedimenmaplinearly#1{% + \begingroup + \dimen0=#1\relax + \c@pgfplotsbin@input=\dimen0 + \multiply\c@pgfplotsbin@input by2 +%\message{LOWLEVEL ENCODING '\the\c@pgfplotsbin@input' with linear map}% + \pgfplotsbinaryencodesignedmaplinearly\c@pgfplotsbin@input + \endgroup +}% + +\catcode`\@=\pgfplotsbinaryatcode +\endinput |