Re: `printf -v foo ""` does not set foo=

[Linda Walsh]

Mike Frysinger wrote:
> simple test code:
>         unset foo
>         printf -v foo ""
>         echo ${foo+set}
>
> that does not display "set".  seems to have been this way since the feature 
> was added in bash-3.1.
> -mike
----
Indeed:
> set -u
> unset foo
> printf -v foo ""
> echo $foo       
bash: foo: unbound variable
> foo=""
> echo $foo
----
I have a feeling this would be hard to fix, since how can printf
tell the difference between
    printf -v foo ""
and
    printf -v foo
??

(with nothing after it?)  it seems the semantic parser would have already
removed the quotes by the time the args are passed to printf, even this:
> set -u
> printf -v foo "$(echo "$'\000'")"
> echo $foo

still leaves foo gutless: without content (even if were null)