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Re: aliases not getting expanded inside functions?

On 2003-01-03 at 15:24 -0200, Carlos Carvalho wrote:
> Phil Pennock (Phil.Pennock@xxxxxxxxxxx) wrote on 3 January 2003 16:45:
>  >% bar='Fred Bloggs'
>  >% foo='bar'
>  >% function escr { print ${(P)foo} }
>  >% escr
>  >Fred Bloggs
>  >
>  >Is the above sufficient?
> Unfortunately no, because I don't need it only for printing, but for
> referencing as well. I have an array and I'd like to access the
> elements by more meaningful names, so I tried to do
> % alias -g guess_what='array[1]', etc.
> Your suggestion doesn't work however when an assignment has to be made
> to guess_what.
>  >You might wish to read zshexpn(1), since there are many other weird and
>  >wonderful ways in which zsh can pervert your data on demand.
> I did but didn't see this possibility. In fact I still don't see...
> What's the relation between recognizing the same escape sequences as
> print and making a substitution? I'd be grateful for an explanation.

print is merely what was used for showing the results.  The variable
substitution does not rely upon print.

% set -A array alpha beta gamma delta epsilon zeta
% guess_what='array[3]'
% echo ${(P)guess_what}
% guess_what='array[4]'
% echo ${(P)guess_what}
% guess_what='array[1]'
% for i in $(perl -le "print foreach split //, '${(P)guess_what}'") ; echo $i

My suggestion appears to work, independent of echo/print.  :^)

I think that you're confusing "print -P", which recognises the same
escape sequences as are used in prompt substitution, with the parameter
expansion flag 'P', which is described in zshexpn(1).
"We've got a patent on the conquering of a country through the use of force.
 We believe in world peace through extortionate license fees." -- Andy Forster

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