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Re: v. 6, third argument to rectangle

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75937] Re: v. 6, third argument to rectangle
  • From: rip pelletier <bitbucket at comcast.net>
  • Date: Sun, 13 May 2007 05:59:44 -0400 (EDT)
  • References: <f21e5e$6e7$1@smc.vnet.net> <200705111030.GAA08108@smc.vnet.net> <f23oii$mvm$1@smc.vnet.net>

In article <f23oii$mvm$1 at smc.vnet.net>,
 Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:

> It's true that it works, but the fact that it works seems to be now  
> undocumented. Mathematica 5.2 documentation states:
> 
> "In Rectangle[{a, a}, {a, a}, graphics], graphics can be any graphics  
> object."
> 
> but there does not seem to be any such statement in 6.0 (?)
> 
> Andrzej Kozlowski
> 
> > rip pelletier wrote:
> >> hi,
> >>
> >> thru 5.2 we could display, for example, 2 graphs side-by-side by  
> >> using a
> >> third argument to Rectangle, the name of a graphics object.
> >>
> >> 6.0 tells me that i cannot call Rectangle with 3 arguments. (true, it
> >> was a bit of a kludge, and in principle i approve of it going away,
> >> but....)
> >>
> >> what is the replacement?
> >>
> >> thanks,
> >>    rip
> >>
> >
> 
> 

well, thanks. your answer makes it clear that i have some debugging to 
do. i didn't post because the documentation changed; i posted because i 
got the error message

"Rectangle called with 3 arguments; 1 or 2 arguments are expected."

somehow it must not mean what it says.

i also see that i forgot to say i am using a macintosh, OS X 10.4.9.

since i was using dave park's drawgraphics package, i need to construct 
an example without the package. i'll get to it. given a choice between 
being productive or crawling thru shattered notebooks, i think i'll keep 
working for a while.

vale,
   rip

-- 
NB eddress is r i p 1 AT c o m c a s t DOT n e t


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