Path: utzoo!utgpu!watmath!iuvax!rutgers!bellcore!spectral!sjs
From: sjs@spectral.ctt.bellcore.com (Stan Switzer)
Newsgroups: comp.windows.news
Subject: Re: How do you get rid of the current point (SPOILER)
Message-ID: <17351@bellcore.bellcore.com>
Date: 7 Aug 89 16:07:19 GMT
References: <17344@bellcore.bellcore.com> <39596@sgi.SGI.COM>
Reply-To: sjs@ctt.bellcore.com (Stan Switzer)
Organization: Bellcore
Lines: 31

It hardly seems worth the net bandwidth to continue this thread, but I
seem to have generated a bit of confusion.  Of course, as several
people have pointed out, "moveto" is the obvious solution.  The
trouble is that I'd have to figure out WHERE moveto should move to.
"Arc" figures this out for itself (but doesn't tell you).  I just
wanted to avoid duplicating that work.

My solution:

/Xarc { % xc yc r a0 a1 -> xc yc r a0 a1
    4 index 4 index moveto 1 index cos 3 index mul
    2 index sin 4 index mul rmoveto
} def

Then I just plop an Xarc in front of the "arc" or "arcn" in question.
This moves the point to the start of the arc and leaves everything set
up for the arc itself.

So the answer to the original question is that (in standard
PostScript) you can't get rid of the current point without otherwise
affecting the current path.  Consequently, it's a bit more difficult
than it should be to construct a path consisting of unconnected "arc"
segments.

In NeWS (1.1), closepath DOES get rid of the current point.

It would have been easier if there were a "nukepoint" operator.  I was
surprised that PostScript didn't anticipate this particular need.  I
just assumed I was missing something obvious.

Stan Switzer  sjs@ctt.bellcore.com