Word splitting in zsh

[Deborah Ariel Pickett <debbiep@mail.csse.monash.edu.au>]

Hiya,

I've come across this little problem in zsh when I run it under
setopt SHWORDSPLIT (not that this is something I normally do).
What it boils down to is that in constructs like ${variable-word}
(where "-" is any of the characters that can go in that place - e.g.,
"-", "+", "=") the shell doesn't seem to be honouring double quotes the
way the manpage says.

Here's the relevant sections in the manpage:
[...] (Parameters)
      @      In double quotes, array elements are put into sepa-
              rate  words.   E.g.,  "${(@)foo}"  is equivalent to
              "${foo[@]}" and "${(@)foo[1,2]}"  is  the  same  as
              "$foo[1]" "$foo[2]".
[...] (Parameter expansion)
      ${name:+word}
              If name is set  and  is  non-null  then  substitute
              word; otherwise substitute nothing.
[...]
       Note that double quotes may appear around nested substitu-
       tions, in which case only the part inside  is  treated  as
       quoted;  for  example, ${(f)"$(foo)"} quotes the result of
       $(foo), but the flag `(f)' (see below)  is  applied  using
       the  rules  for unquoted substitutions.  Note further that
       quotes are themselves nested in this context; for example,
       in  "${(@f)"$(foo)"}",  there  are two sets of quotes, one
       surrounding the whole expression,  the  other  (redundant)
       surrounding the $(foo) as before.

With these two bits together, the POSIX and Bourne shells can make a
generic "any number of arguments, including zero" by using the form
${1+"$@"} (i.e., if $1 is set, substitute "$@", otherwise substitute
nothing).

Trying that in zsh, I get this:

bruce ~ % echo $ZSH_VERSION
3.1.6
# (But this also applies in 3.1.9)
bruce ~ % setopt
alwaystoend
noappendhistory
autocd
noautomenu
autonamedirs
autopushd
autoresume
braceccl
noclobber
completeinword
correct
extendedglob
noflowcontrol
globdots
histexpiredupsfirst
histignoredups
histnostore
ignoreeof
interactive
nolistbeep
longlistjobs
monitor
numericglobsort
promptsubst
pushdignoredups
pushdminus
pushdsilent
pushdtohome
shinstdin
zle
bruce ~ % setopt shwordsplit
# Now splitting is done like in other shells.
bruce ~ args()
function >{
function >for x
function for >do
function for >echo "'$x'"
function for >done
function >}
bruce ~ % args "a1 a2 a3" b c
'a1 a2 a3'
'b'
'c'
# Ok, that's what we expected.
bruce ~ % args "$@"
# Acceptable, but Bourne sh would print a single blank entry here, since
# there's a pair of quotes.
bruce ~ % args ${1+"$@"}
# This is the Proper Bourne Shell way to do it.
bruce ~ % set "a1 a2 a3" b c
bruce ~ % args "$@"
'a1 a2 a3'
'b'
'c'
# Fine . . .
bruce ~ % args ${1+"$@"}
'a1'
'a2'
'a3'
'b'
'c'
# What??  The $@ was in quotes, why was "a1 a2 a3" split?
bruce ~ % bash --posix
# Let's try the same under a POSIX shell.
bash-2.03$ args()
> {
> for x
> do
> echo "'$x'"
> done
> }
bash-2.03$ set "a1 a2 a3" b c
bash-2.03$ args ${1+"$@"}
'a1 a2 a3'
'b'
'c'
# Here it does it the correct way.

So . . am I misunderstanding how double quotes are propagated through
${} constructs in zsh?  Or is this a bona fide bug?  Whatever the
answer, this is something that doesn't work the same as in the Bourne
shell.  Should it?

The only hint of an answer comes in this section of the manpage:
       1. Nested Substitution
              If multiple nested ${...} forms are  present,  sub-
              stitution  is  performed  from the inside outwards.
              At each level, the substitution  takes  account  of
              whether  the current value is a scalar or an array,
              whether the whole substitution is in double quotes,
              and what flags are supplied to the current level of
              substitution, just as if  the  nested  substitution
              were  the  outermost.  The flags are not propagated
              up to enclosing substitutions; the nested substitu-
              tion  will  return  either  a scalar or an array as
              determined by  the  flags,  possibly  adjusted  for
              quoting.   All the following steps take place where
              applicable at all  levels  of  substitution.   Note
              that,  unless  the `(P)' flag is present, the flags
              and any subscripts apply directly to the  value  of
              the nested substitution; for example, the expansion
              ${${foo}} behaves exactly the same as ${foo}.

This makes some sense, since these produce different results:
bruce ~ % args "${@}"
'a1 a2 a3'
'b'
'c'
bruce ~ % args "${${@}}"
'a1 a2 a3 b c'

though I wonder if they ought to, since this disagrees with the last
sentence about ${${foo}} and ${foo}.

-- 
Debbie Pickett http://www.csse.monash.edu.au/~debbiep debbiep@csse.monash.edu.au
"Look at me; I will never pass for a perfect bride, or a perfect daughter.  Can
  it be I'm not meant to play this part? Now I see that if I were truly to be
        myself, I would break my family's heart." - Reflection, _Mulan_