Am 21.03.2012 22:46, schrieb Moritz Bunkus:
> Hey,
>
> what about
>
> $ declare -A foo bar
> $ foo=(a 42 b 54)
> $ bar=(${(kv)foo})
> $ print ${(kv)bar}
> a 42 b 54
>
>> using this, bar only contains the values of foo (and using ${(kv)foo}
>> instead still only makes it a string).
>
> Yes, because you didn't put it into parenthesis, bar=${(kv)foo} vs
> bar=(${(kv)foo})
This makes sense. Thanks for the pointer :)
- René
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