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Re: aliases not getting expanded inside functions?
- X-seq: zsh-users 5672
- From: "Bart Schaefer" <schaefer@xxxxxxxxxxxxxxxx>
- To: zsh-users@xxxxxxxxxx
- Subject: Re: aliases not getting expanded inside functions?
- Date: Sat, 11 Jan 2003 18:40:19 +0000
- In-reply-to: <15903.36155.716460.639226@xxxxxxxxxxxxxx>
- Mailing-list: contact zsh-users-help@xxxxxxxxxx; run by ezmlm
- References: <15893.44217.393956.262362@xxxxxxxxxxxxxx> <20030103164552.A28966@xxxxxxxxxxx> <15893.50996.646711.184945@xxxxxxxxxxxxxx> <20030103184455.A5692@xxxxxxxxxxx> <15893.53780.524763.695176@xxxxxxxxxxxxxx> <20030103185407.GA11836@xxxxxxxx> <15897.15986.562636.628562@xxxxxxxxxxxxxx> <1030106125404.ZM4660@xxxxxxxxxxxxxxxxxxxxxxx> <15903.36155.716460.639226@xxxxxxxxxxxxxx>
On Jan 11, 1:19am, Carlos Carvalho wrote:
>
> Sorry I couldn't follow up earlier :-( Anyway the below told me how to
> pass parameters by name to a function so that it changes the values.
> Good!
That isn't the only way to do that, and not even the most readable in my
opinion. For scalars, each of the following has the same effects:
: ${(P)1::=$2}
eval $1='$2'
typeset -g $1=$2
For arrays:
eval $1='( $3 $2 )'
set -A $1 $3 $2
The array form ${(P)=1::=$3 $2} is almost but not quite the same, because
field splitting is applied *after* expanding $3 and $2 in that case, even
if they're quoted, so you can't preserve embedded whitespace.
> Bart Schaefer (schaefer@xxxxxxxxxxxxxxxx) wrote on 6 January 2003 12:54:
> >Nearly as often, the right thing is instead to ask the list how to solve
> >problem X, because there's a better solution than Y.
>
> Agreed, so here's the story, with two questions. I read a csv file
> that comes from a spreadsheet and need to split the fields to
> different variables. Instead of doing the full parsing of the data
> line by hand, it's easier to have zsh do the split:
>
> fields=( ${(s:;:)dataline} )
How did you get the data into $dataline in the first place?
Rather than:
while read dataline
do
fields=( "${(@s:;:)dataline}" ) # Answers your other question
rate=$fields[1] capital=$fields[2] etc.
# manipulate $rate $capital and so on ...
done
You can simply do:
while IFS=';' read rate capital etc.
do
# manipulate $rate $capital and so on ...
done
> rate_prev=$rate capital_prev=$capital etc.
>
> Instead of copying manually I'd like to do
>
> rate_prev=$fields_prev[1] capital_prev=$fields_prev[2] etc.
>
> only once, and then just do fields_prev=( $fields ) whenever I have to
> copy the values.
A much better way to do this is to use two associative arrays:
typeset -A fields fields_prev
while IFS=';' noglob read fields[rate] fields[capital] etc.
do
fields_prev=( ${(kv)fields} ) # Save all keys and values
# manipulate $fields[rate] $fields[capital] and so on ...
done
The "noglob" above is to avoid having to quote all the square brackets
in the "read" command (otherwise they'd be treated as file patterns).
> I mentioned some variant of a loop like
>
> for ((i=1; i<= num_fields; i++)) {
> : ${(P)${fields_prev[i]}::=${(P)${fields[i]}}}
> }
>
> However this doesn't work because I cannot assign to the individual
> variables (ex. capital=$((capital+interest)) ) without losing the
> connection with the fields array.
If you're unwilling to use $fields[capital] everywhere -- that is, if
you insist on being able to write $capital in some cases -- then there
is no solution I can suggest. However, if it's OK to write e.g.
fields[capital]=$((fields[capital]+fields[interest]))
or, more succinctly,
(( fields[capital] += fields[interest] ))
then there is no connection to worry about being lost.
--
Bart Schaefer Brass Lantern Enterprises
http://www.well.com/user/barts http://www.brasslantern.com
Zsh: http://www.zsh.org | PHPerl Project: http://phperl.sourceforge.net
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