On Wed, 06 Jun 2007 19:42:26 -0400
Aditya Mahajan <[EMAIL PROTECTED]> wrote:
> Hi,
>
> How can I make \framed ignore the width of the rule? I am trying to
> typeset internet protocol headers, but I cannot fully get what I want.
> Here is my attempt:
>
> \unprotect
>
> \makecounter{v!protocolheaderlength}
>
> \defineframed[protocolfieldheader]
> [location=lohi,height=3\lineheight,offset=none,strut=yes]
>
>
> \def\startprotocolheader
> {\dodoubleempty\dostartprotocolheader}
>
> \def\dostartprotocolheader[#1][#2]%
> {\getparameters[c!protocol][#1]%
> \let\field\protocolheaderfield
> \resetcounter{v!protocolheaderlength}%
> \vbox\bgroup
> % \ruledvbox\bgroup
> \removeunwantedspaces
> \dontleavehmode
> \protocolfieldheader[frame=on,width=\c!protocolbits\c!protocolunitsize]%
\protocolfieldheader[frame=on,width=\dimexpr\c!protocolunitsize*\c!protocolbits\relax]%
> {\let\\\crlf\bold #2}\crlf}
>
> \def\stopprotocoloheader
> {\egroup}
>
> \def\protocolheaderfield#1#2%
> {\doif{\countervalue{v!protocolheaderlength}}{\c!protocolbits}
> {\resetcounter{v!protocolheaderlength}\crlf}%
> \removeunwantedspaces
> \dontleavehmode
> \protocolfieldheader[width=#1\c!protocolunitsize]{\let\\\crlf #2\\(#1)}%
\protocolfieldheader[width=\dimexpr\c!protocolunitsize*#1\relax]{\let\\\crlf
#2\\(#1)}%
> \incrementcounter{v!protocolheaderlength}{#1}}
>
>
> \protect
>
> % \showstruts
> % \showboxes
>
> \starttext
>
>
> % Why should unitsize=1mm. Are things being scaled by another factor?
> \startprotocolheader[bits=16,unitsize=1mm][General MAC Headers]
\startprotocolheader[bits=16,unitsize=1cm][General MAC Headers]
> \field {1} {HT}
> \field {1} {EC}
> \field {6} {Type}
> \field {1} {ESF}
> \field {1} {CI}
> \field {2} {EKS}
> \field {1} {Rsv}
> \field {3} {LEN MSB}
> \field {8} {LEN LSB}
> \field {8} {CID MSB}
> \field {8} {CID LSB}
> \field {8} {HCS}
> \stopprotocoloheader
>
>
> \stoptext
>
>
> Notice that all the lines are not of the same size. It seems that the
> rules of framed are causing the frame to be of a larger width.
>
> Also am I missing a scaling factor in width= expressions above.
> Everything looks fine with unitsize=1mm, but the final figure is
> clearly larger than 16mm. I am confused as to what is happening.
>
> Is there any better way to do this?
>
> Thanks,
> Aditya
Hi Aditya,
you wanted something like this
\newdimen\scale \scale\1mm
\def\rule#1{\hrule width #1\scale}
\rule{10}
but you made this
\def\scale{1mm}
\def\rule#1{\hrule width #1\scale}
\rule{10}
The length of the \hrule in the first example is 10pt (10 \times 1pt = 10pt),
while in the second example it is 101pt width (10 & 1pt = 101pt).
Wolfgang
___________________________________________________________________________________
If your question is of interest to others as well, please add an entry to the
Wiki!
maillist : ntg-context@ntg.nl / http://www.ntg.nl/mailman/listinfo/ntg-context
webpage : http://www.pragma-ade.nl / http://tex.aanhet.net
archive : https://foundry.supelec.fr/projects/contextrev/
wiki : http://contextgarden.net
___________________________________________________________________________________
