`printf -v foo ""` does not set foo=
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
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Re: `printf -v foo ""` does not set foo=
On 17 June 2013 13:27, 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.
Interesting. It also won't change it if it already exists, it seems:
$ foo=bar
$ printf -v foo ""
$ echo "$foo"
bar
$ echo "$BASH_VERSION"
4.2.45(2)-release
Re: corrupted input after size function (input that's not recorded by bash)
Dave Gibson wrote:
Trial and error suggests it's something to do with new-style command
substitution. Try backticks:
local s=`stty size`
Yes... you are right. This works... while
local s=$(stty size) does not.
That's icky! I thought they were identical, they appear not
to be...umm
Chet??? Hello? Um... why are these not the same?
local s=`stty size`
local s=$(stty size)
the latter seems to eat a param or similar when used in a function
at an input prompt...now I wonder what other side effects "$()" has over "``"
showsize()
{
local o="(${LINES}x${COLUMNS})" ; local s="${o//?/\\b}" ; printf "$o$s"
}
---
Well ain't that sweet. I thought I tried that at one point, though,
and they weren't updating hmmm...proof's in the pudding, I suppose..
Re: `printf -v foo ""` does not set foo=
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)
