Re: Question on $@ vs $@$@

[Marc Chantreux <mc@unistra.fr>]

On Wed, Aug 14, 2024 at 09:23:49AM -0400, Greg Wooledge wrote:
> The most obvious would be to treat "$@$@" as if it were "$@" "$@",
> generating exactly two words for each positional parameter.
> 
>     <A> <B> <C> <A> <B> <C>

Thanks for this. My two cents:

if I want "$@" "$@", i write it litterally. "$@$@" kinda tricks me
in that case because it's not consistent with the other cases I already
used ("$foo$@", for exemple).

> If there are *two* instances of $@ within the same word, then the final
> parameter of the *first* $@ is supposed to be "joined with the last
> part of the original word".  But the "last part of the original word"
> is another list expansion, not a string!  What does it even mean for
> the final parameter to be "joined" with a list expansion?

good point: that's ambigious. zsh has the explicit twigil $^ (inspired
by the rc operator, I think) to make the distribution explicit (like a
brace expansion) but as dash lack both those features, I still think
my proposal both more consistent and doing something different.

> So, neither of these results would shock me:
>     <A> <B> <CA B C>    (treat it like "$@$*")
>     <A B CA> <B> <C>    (treat it like "$*$@")

well. "$@" "$@", $@$* and $*$@ are clear syntaxes that proves an intention
so "$@$@" should be another one related to the fact that something is
joined at the boundary. and to me "something" means "anything left between"
(possibly empty).

so: set A B
        "$@ x $@"       =>   "A" "B x A" "B"
        "$@$@"          =>   "A" "BA"    "B"

> I'd still love to know what the script's intent is.

I my guess would be based on "nothing but what the other syntaxes
provides more expicitly".

regards

-- 
Marc Chantreux
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