Re: Help in understanding rules of parameter expansion

[Mikael Magnusson <mikachu@xxxxxxxxx>]

On Fri, Aug 23, 2024 at 12:33 AM Bart Schaefer
<schaefer@xxxxxxxxxxxxxxxx> wrote:
>
> On Thu, Aug 22, 2024 at 3:37 AM Roman Perepelitsa
> <roman.perepelitsa@xxxxxxxxx> wrote:
> >
> >     % foo=AxB
> >     % bar=${(s.x.)foo}
> >     % baz=${${(s.x.)foo}}
> >     % typeset -p bar baz
> >     typeset bar=AxB
> >     typeset baz='A B'
> >
> > I might be missing something but I expect $bar to be the same as $baz.
>
> Hrm.  ${(s.x.)foo} produces an array result.  bar=${(s.x.)foo} is a
> scalar assignment.  There's no implicit join, because (per the
> "Rules") all joining takes place before splitting.  So the result of
> this assignment is unspecified; I suppose it could have generated an
> "inconsistent type for assignment" error, but the internals just throw
> away the array value and use the scalar value.
>
> Conversely ${${(s.x.)foo}} has an implicit join step after the split
> because of the inside-out ordering of nested expansions.

I was thinking something along these lines too, but shouldn't it at
least act the same as double quoted "${(s.x)foo}" (not in an
assignment)?

-- 
Mikael Magnusson