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From: Zoltan Hidvegi <hzoli@frontiernet.net>
Message-Id: <199707290736.DAA03285@hzoli.home>
Subject: Re: RC_EXPAND_PARAM bug
In-Reply-To: <970729000902.ZM15018@candle.brasslantern.com> from Bart Schaefer at "Jul 29, 97 00:09:02 am"
To: schaefer@nbn.com
Date: Tue, 29 Jul 1997 03:36:48 -0400 (EDT)
Cc: zsh-workers@math.gatech.edu (Zsh hacking and development)
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> On Jul 29,  2:04am, Zoltan Hidvegi wrote:
> > 
> > % echo 1${^a}1${^^x}
> > 1a1x 1ay 1b1x 1by
> > 
> > The logic is that the string after the rc-param, 1${^^x}, is expanded,
> > producing two strings, 1x y, which is combined with 1a 1b.  It is true
> > that this is incompatible with 2.6-beta16 and older, which first expanded
> > it to 1a1${^^x} 1b1${^^x} and later this was expanded to 1a1x y 1b1x y.
> > Similarily, let i=0; echo ${^a}$[i++] expanded to a$[i++] b$[i++] and
> > later to a0 b1, while in zsh-3.0.4 it expands to a0 b0.
> 
> Can you generalize this rule for us?  E.g.
> 
> % echo ${^a}$[i++]$[++j]${^x}....
> 
> where .... is some arbitary number of other substitutions?  Is it just that
> it now does everything from right to left instead of left to right?  Why?

No, it is left to right.  ${^a} is expanded first, then the remaining
part, $[i++]$[++j]${^x} is expanded separately, and the result is
combined with the expansion of ${^a}.  You can see it if you try

let i=0; echo $[i++]${^a}$[i++]

which gives

0a1 0b1

If it were right to left, it would be 1a0 1b0.

Zoltan