On Sep 11, 2008, at 6:20 PM, Thomas A. Schmitz wrote:
>> \starttext
>>
>> \start
>>
>> \setbox\scratchbox\vbox{\externalfigure[mill]}
>>
>> \dimen0=\wd\scratchbox
>> \dimen2=\ht\scratchbox
>>
>> \framed[frame=on,strut=no,width=8cm,height=2cm]
>> {\dimen1=\hsize \divide\dimen1 by \dimen0
>> \dimen3=\vsize \divide\dimen3 by \dimen2
>> \ifdim\dimen1>\dimen3
>> \externalfigure[mill][height=\vsize]
>> \else
>> \externalfigure[mill][width=\hsize]
>> \fi}
>>
>> \framed[frame=on,strut=no,width=2cm,height=8cm]
>> {\dimen1=\hsize \divide\dimen1 by \dimen0
>> \dimen3=\vsize \divide\dimen3 by \dimen2
>> \ifdim\dimen1>\dimen3
>> \externalfigure[mill][height=\vsize]
>> \else
>> \externalfigure[mill][width=\hsize]
>> \fi}
>>
>> \stop
>>
>> \stoptext
>>
>> Wolfgang
>
> Wolfgang,
>
> as always, you're a source of wisdom and knowledge... Just I
> understand this correctly and can adapt it to my macro: \dimen1=
> \hsize: here \hsize refers to the size of the \framed inside which
> we're operating, right?
And another question: I get an error "! Illegal unit of measure (pt
inserted)." Is it really possible to divide a dimension by another
dimension? Not according to what I read here:
http://www.tug.org/utilities/plain/cseq.html#divide-rp
"must be a nonzero integer." See, I'm far from being a native
speaker of TeX...
Best
Thomas
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