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I'm currently working on a software for an graphic display, so I recreated it in TikZ for documentation purposes:

image of the display

I currently use this to set individual pixels on the display:

\setpixel{x}{y}

Where x and y are coordinates between 0 and 132/64.

The real display is separated into 8 pages, each 8 pixels high, and a byte sent to the display is displayed as a column. See this image for details.

Since I don't want to always calculate the individual pixels, I'd like the LaTeX variant to behave like the real display, that is, I want some command like

\dispbyte{0x01} \dispbyte{0x03} \dispbyte{0x07} \dispbyte{0x0F} \dispbyte{0x1F} \dispbyte{0x3F} \dispbyte{0x7F} \dispbyte{0xFF}

(this would create one of the triangles above) I have an idea how I'd implement counting of the current column and page switching and all that -- shouldn't be hard with some counters after all -- but I just can't find out how to parse the hexadecimal values and then iterate over every bit inside them.

I've found fmtcount and binhex.tex, but they only display a LaTeX counter in another format, and I can't understand their code at all.

The finished source code (display.sty) and some examples are now available at http://cmpl.cc/downloads/disp/

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  • That sounds fascinating. Do you intent to publish the source to the pixel -> TikZ conversion? Commented Apr 25, 2013 at 21:13
  • @Alexander I will release the complete source code in a few weeks, when the project I'm writing the documentation for is finished. Commented Apr 26, 2013 at 7:37
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    It is not absolutely clear to me what you want, hence I write this as a comment: Take a look at the bitset package by Heiko Oberdiek. It provides the abstraction of bitsets of various lenght, which can be initialized from a hex value by \bitsetSetHex. Individual bits can be tested (\bitsetGet) or all set lists transformed into a comma-separated list (\bitsetGetSetBitList) to be then iterated with, e.g., \foreach. Commented Apr 26, 2013 at 10:07
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    Note also this related question. Commented Apr 26, 2013 at 10:09
  • @TheCompiler: For \foreach you just have to expand it first, e.g., with \edef. I have now posted this as an answer. Commented Apr 26, 2013 at 14:53

5 Answers 5

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The following is less impressive than the other answers from the visual point of view (Mark, I like yours!), but addresses the actual question of the OP: Bit-wise iteration over hexadecimal values, which becomes fairly easy when using the bitset package by Heiko Oberdiek:

\documentclass{article} \usepackage{bitset} \usepackage{pgf,pgffor} \begin{document} \bitsetSetHex{mybitset}{AA} % use \bitsetGetSetBitList % expand first \edef\mybits{\bitsetGetSetBitList{mybitset}} \noindent \foreach \bit in \mybits {% Bit \bit{} is set! \\ } % just itereate all bits \noindent \foreach \i in {0,...,7} {% Bit \i: \bitsetGet{mybitset}{\i} \\ } \end{document}

enter image description here

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  • That was by far the most easy to use way. Took me like 5 minutes to implement what I wanted based on that. Thanks! Commented Apr 27, 2013 at 1:24
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Not sure this is robust, it only works within the limited numerical range of PGFMath, and clearly I've gone for something a bit more over-the-top than the requirements.

EDIT: following Daniels example of the bitset package the code has been updated to be a bit more like that.

\documentclass[border=5pt]{standalone} \usepackage{tikz} \newcount\bitcount \tikzset{ zeros/.style={ draw=black, insert path={ (-\nbit-1/2, -1/2) rectangle ++(1,1) } }, ones/.style={ draw=black, fill=gray, insert path={ (-\nbit-1/2, -1/2) rectangle ++(1,1) } }, max bits/.store in=\maxbits, max bits=0 } \newcommand\dispbyte[2][]{% \begingroup% \tikzset{#1}% \pgfmathsetcount\bitcount{#2}% \pgfmathparse{int(\maxbits)}\let\maxbits=\pgfmathresult% \pgfmathloop% \ifnum\bitcount>0\relax% \ifodd\bitcount% \expandafter\def\csname bit\pgfmathcounter\endcsname{1}% \else% \expandafter\let\csname bit\pgfmathcounter\endcsname=\relax% \fi% \divide\bitcount by2\relax% \repeatpgfmathloop% \pgfmathparse{int(\maxbits>\pgfmathcounter?\maxbits+1:\pgfmathcounter+1)}% \let\nbits=\pgfmathresult% \pgfmathloop% \ifnum\pgfmathcounter=\nbits\relax% \else% \let\nbit=\pgfmathcounter% \expandafter\ifx\csname bit\pgfmathcounter\endcsname\relax% \path [zeros]; \else% \path [ones]; \fi% \repeatpgfmathloop% \endgroup% } \begin{document} \begin{tikzpicture}[x=10pt, y=10pt] \foreach \d [count=\c from 0, evaluate={\x=floor(\c/8)*8; \y=-mod(\c,8);}] in {% 0x7f,0x49,0x08,0x08,0x08,0x08,0x08,0x1c, 0x00,0x00,0x08,0x00,0x18,0x08,0x08,0x1c, 0x08,0x08,0x10,0x12,0x14,0x28,0x24,0x22, 0x7f,0x02,0x04,0x08,0x08,0x10,0x20,0x7f}{ \dispbyte[max bits=8,shift={(\x,\y)}]{\d} } \foreach \d [count=\c from 0, evaluate={\x=floor(\c/8)*8; \y=-mod(\c,8);}] in {% 0x7f,0x49,0x08,0x08,0x08,0x08,0x08,0x1c, 0x00,0x00,0x08,0x00,0x18,0x08,0x08,0x1c, 0x08,0x08,0x10,0x12,0x14,0x28,0x24,0x22, 0x7f,0x02,0x04,0x08,0x08,0x10,0x20,0x7f}{ \dispbyte[zeros/.style={}, ones/.style={ fill=gray, insert path={ (-\nbit-3/8, -3/8) rectangle ++(0.75,0.75) } },shift={(\x,\y-9)}]{\d} } \foreach \d [count=\c from 0, evaluate={\x=floor(\c/8)*8; \y=-mod(\c,8);}] in {% 0x7f,0x49,0x08,0x08,0x08,0x08,0x08,0x1c, 0x00,0x00,0x08,0x00,0x18,0x08,0x08,0x1c, 0x08,0x08,0x10,0x12,0x14,0x28,0x24,0x22, 0x7f,0x02,0x04,0x08,0x08,0x10,0x20,0x7f}{ \dispbyte[max bits=8, zeros/.style={ fill=black, insert path={ (-\nbit, 0) circle [radius=0.5] } }, ones/.style={ fill=orange, insert path={ (-\nbit, 0) circle [radius=0.5] } },shift={(\x,\y-18)}]{\d} } \end{tikzpicture} \end{document}

enter image description here

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You can use LaTeX3 that has a function for converting integers to binary. Instead of \fbox use your preferred macro.

\documentclass{article} \usepackage{xparse} \ExplSyntaxOn \NewDocumentCommand{\dispbyte}{m} { \compiler_dispbyte:n { #1 } } \tl_new:N \l_compiler_bits_tl \tl_new:N \l_compiler_byte_tl \cs_new_protected:Npn \compiler_dispbyte:n #1 { % we need to remove the 0x \tl_set:Nn \l_compiler_byte_tl { #1 } \tl_remove_once:Nn \l_compiler_byte_tl { 0x } % convert the number to a string of bits \tl_set:Nx \l_compiler_bits_tl { \int_to_binary:n { "\l_compiler_byte_tl } } % loop through the list of bits \tl_map_inline:Nn \l_compiler_bits_tl { \fbox{ ##1 } } } \ExplSyntaxOff \begin{document} \dispbyte{0x01} \dispbyte{0x03} \dispbyte{0x07} \dispbyte{0x0F} \dispbyte{0x11} \dispbyte{0x13} \dispbyte{0x17} \dispbyte{0x1F} \end{document}

If you need the bits in reverse order, just add

 \tl_set:Nx \l_compiler_bits_tl { \tl_reverse:V \l_compiler_bits_tl }

before the \tl_map_inline:Nn line.

enter image description here

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5

I didn't get exactly what you want to do but pgfmath already understands Hex syntax. 

\documentclass{article} \usepackage{tikz} \begin{document} \pgfmathparse{bin(0x1F)} \pgfmathresult \pgfmathparse{bin(0X1F)} \pgfmathresult \end{document}

Both leads to 11111

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enter image description here

This loops over the binary digits from the hex, for testing it just prints them out (in reverse order to keep the code simple)

\documentclass{article} \makeatletter \def\dispbyte#1{\@displaybyte#1\relax} \def\@displaybyte#1x#2\relax{\count@\string"#2\relax \loop \fbox{\ifodd\count@ 1\else0\fi}% \divide\count@\tw@ \ifnum\count@>\z@ \repeat} \makeatother \begin{document} \dispbyte{0x17} \end{document}
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  • It can't:-) Do you have a non standard format making " active even if babel is notloaded? Commented Apr 25, 2013 at 21:01
  • add \tracingall before \dispbyte and post the log (or email it me: google my name for my gmail account) Commented Apr 25, 2013 at 21:03
  • I didn't try it yet, but two things come to mind by looking at it: 1) how would I use that with babel? Wouldn't " be active as well then? 2) shouldn't the second \makeatletter be a \makeatother? Commented Apr 26, 2013 at 5:42
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    @TheCompiler you can use \string" to be safe if " is active and yes the second make... is a typo (although it doesn't really matter, I'll fix both:-) Commented Apr 26, 2013 at 6:45

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