BUG: zsh-3.1.5-pws-14: parameter expansion not working properly

BUG: zsh-3.1.5-pws-14: parameter expansion not working properly

[Geoff Wing]

Here's a fragment of the expansion similar to that used in
cvsentries() in Misc/compctl-examples 

% zsh-3.1.5
% a=("${(f@)$(echo foo; echo bar)}"); print -l $a
foo
bar
% a=("${${(f@)$(echo foo; echo bar)}:#f*}"); print -l $a
bar
% zsh-3.1.5-pws-14
% a=("${(f@)$(echo foo; echo bar)}"); print -l $a
foo
bar
% a=("${${(f@)$(echo foo; echo bar)}:#f*}"); print -l $a

%
-- 
Geoff Wing   <gcw@pobox.com>            Mobile : (Australia) 0412 162 441
Work URL: http://www.primenet.com.au/   Ego URL: http://pobox.com/~gcw/

RE: zsh-3.1.5-pws-14: parameter expansion not working properly

[Andrej Borsenkow]

Yes, as I understand, it was changed in pws-14. It includes full description
of subst rules in Parameters expansion section of manual.

Basically, in quotes, the value of inner substitution is always forced to be
scalar for outer substitution unless some flags or modifiers (notably, =,
(@), (s)) prevent it.
> % zsh-3.1.5-pws-14
> % a=("${(f@)$(echo foo; echo bar)}"); print -l $a
> foo
> bar

(@) is not needed. (f) is already enough - first, the value of $(...) is
forced to scalar and then it is splitted using (f) flag. The result is
*always* array - in this case it is top-level one so nothing more happens.

> % a=("${${(f@)$(echo foo; echo bar)}:#f*}"); print -l $a
>

Yep. The first ${(f)$(...)} yields array value. The outer ${...} first
converts array to scalar (because the whole is quoted) and then applies # to
scalar. To preserve "arrayness" you need (@) for outer subst:

bor@itsrm2:~%> a=("${(@)${(f@)$(echo foo; echo bar)}:#f*}"); print -l $a
bar

So, it is not a bug, it is by desing. There was a long thread here, where I
tried to understand what ZSH substitution rules actually are. I admit (now
:-) that current state is more or less consistent. The problem is, it is
near to impossible to explain difference between pws-14 and earlier versions
because it is the first time these rules are explicitly stated.

/andrej

Re: zsh-3.1.5-pws-14: parameter expansion not working properly

[Geoff Wing]

Andrej Borsenkow <borsenkow.msk@sni.de> typed:
:Yes, as I understand, it was changed in pws-14. It includes full description
:of subst rules in Parameters expansion section of manual.
:Basically, in quotes, the value of inner substitution is always forced to be
:scalar for outer substitution unless some flags or modifiers (notably, =,
:(@), (s)) prevent it.
:> % zsh-3.1.5-pws-14
:> % a=("${(f@)$(echo foo; echo bar)}"); print -l $a
:> foo
:> bar
:So, it is not a bug, it is by desing. There was a long thread here, where I
:tried to understand what ZSH substitution rules actually are. I admit (now
::-) that current state is more or less consistent. The problem is, it is
:near to impossible to explain difference between pws-14 and earlier versions
:because it is the first time these rules are explicitly stated.

OK, thanks, so in Misc/compctl-examples, we now have something like the
following and I can happily (yeah!) complete in cvs again:

cvsentries () {
        setopt localoptions nullglob unset
        if [[ -f ${pref}CVS/Entries ]]
        then
		reply=("${pref}${(@)^${(@)${(@)${(f)$(<${pref}CVS/Entries)}:#D*}#/}%%/*}")
        fi
}

-- 
Geoff Wing   <gcw@pobox.com>            Mobile : (Australia) 0412 162 441
Work URL: http://www.primenet.com.au/   Ego URL: http://pobox.com/~gcw/