Re: ${(P)${foo}} (Re: Associative array ordering)

["Bart Schaefer" <schaefer@xxxxxxxxxxxxxxxx>]

On Feb 3,  9:02am, Sven Wischnowsky wrote:
} Subject: Re:  ${(P)${foo}} (Re: Associative array ordering)
}
} Bart Schaefer wrote:
} 
} > There are actually several ways we can go from here.
} > 
} > We can keep Sven's syntax as is.
} > 
} > We can modify Sven's syntax so that ${(P)foo} is the same as ${(P)${foo}},
} > and make ${!foo} a synonym for it in ksh compatibility mode.  This is
} > almost like ksh namerefs except that they don't get their own namespace.
} 
} I first thought about implementing ${(P)foo}, but implementing only the one 
} I did, seemed easier.

I suspect that's at least in part how we ended up with ${${param}} instead
of simply ${{param}} in the first place, but ....

} Now that I had a deeper look into the substitution 
} code again, making ${(P)foo} work doesn't look that complicated, too.
} Although that would be less powerful, so keeping the thing I
} implemented may still be useful.

Yes, particularly ${(P)$(...)} is probably useful, even if ${(P)${...}}
was not.

} Making ${!foo} a synonym may be useful, and as far as I can this we
} wouldn't have to restrict this to ksh compatibility mode (am I missing 
} something?).

You're missing `setopt banghist`, which is unsetopt in ksh mode.  ${!foo}
will have substituted the history item beginning with "foo" long before
it makes it to the parameter code.  That's why the ${!assoc[@]} syntax
is accepted only in ksh mode, too.

Let's go this way:  Implement ${(P)foo}, the same as ${(P)${foo}} where
the inner ${...} has no flags or modifiers, and leave ${!foo} undefined
until we do real namerefs (if we ever do).

-- 
Bart Schaefer                                 Brass Lantern Enterprises
http://www.well.com/user/barts              http://www.brasslantern.com