Re: 'declare +a -g' destroys local arrays

[Chet Ramey]

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On 8/22/13 4:52 AM, Arfrever Frehtes Taifersar Arahesis wrote:
 I use bash 4.2.45.
 
 `man bash` says:
 Using `+' instead of `-' turns off the attribute instead, with the 
 exceptions that +a may not be
 used to destroy an array variable and +r will not remove the readonly 
 attribute.
 
 However 'declare +a -g' destroys local arrays, which do not shadow global 
 arrays:

Thanks for the report.  This has been fixed for bash-4.3, which is now in
beta.

Chet
- -- 
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Chet Ramey, ITS, CWRUc...@case.eduhttp://cnswww.cns.cwru.edu/~chet/
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'declare +a -g' destroys local arrays

[Arfrever Frehtes Taifersar Arahesis]

I use bash 4.2.45.

`man bash` says:
Using `+' instead of `-' turns off the attribute instead, with the exceptions 
that +a may not be
used to destroy an array variable and +r will not remove the readonly 
attribute.

However 'declare +a -g' destroys local arrays, which do not shadow global 
arrays:

$ f() { local ARRAY=(a b c); declare -p ARRAY; declare +a -g ARRAY; declare -p 
ARRAY; }
$ f
declare -a ARRAY='([0]=a [1]=b [2]=c)'
bash: declare: ARRAY: cannot destroy array variables in this way
bash: declare: ARRAY: not found
$ ARRAY=(x y z)
$ f
declare -a ARRAY='([0]=a [1]=b [2]=c)'
bash: declare: ARRAY: cannot destroy array variables in this way
declare -a ARRAY='([0]=a [1]=b [2]=c)'

--
Arfrever Frehtes Taifersar Arahesis


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