Re: ${!variable@operator} does not work for variables without values; inconsistencies between present and absent [@] for @A and @a

[Eli Schwartz]

On 2/19/20 6:18 PM, Arfrever Frehtes Taifersar Arahesis wrote:
> While described problems exist, there were typos in original reproduction 
> steps.
> 
> Corrected output also reveals another bug:
> ${!variable[@]@A} for (indexed or associative) arrays does not include 
> indexes.
Your examples are all (still) broken. Do not try to reproduce bugs using:

VAR1=(aaa)

Instead, use:

VAR1=(aaa bbb)

Reformulate all test cases where you expect the bbb to be printed, but
it isn't.

$ VAR1=(aaa bbb)
$ ref=VAR1
$ echo "${!ref[@]}"
0
$ echo "${!ref[@]@A}"
declare -a VAR2='aaa'

$ ref2=VAR1[@]
$ args() { printf '<%s> ' "$@"; printf \\n; }
$ args "${!ref2}"
<aaa> <bbbb>
$ args "${!ref2@A}"
<>

You cannot use ${!ref[@]}, you need the array subscript as part of the
set value of "ref" and then indirectly refer to ref.

$ declare -A VAR2=([foo]=bar [baz]="how interesting")
$ args "${!VAR2[@]}"
<foo> <baz>


From man bash:

```
${!name[@]}
${!name[*]}
       List  of array keys.  If name is an array variable, expands to
the list of array indices (keys) assigned in name.  If name is not an
array, expands to
       0 if name is set and null otherwise.  When @ is used and the
expansion appears within double quotes, each key expands to a separate word.
```

And, as predicted, if instead of using ${!ref[@]} you use
${!varname[@]}, you get meaningful information. If you try to look at
the array keys of "ref" itself, all it prints is the number 0, because:

ref=VAR1

and since it is set and non-null...

-- 
Eli Schwartz
Arch Linux Bug Wrangler and Trusted User

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