On Saturday, January 12, 2013 02:35:34 AM John Kearney wrote:
> so there is always at least one word or one arg, just because its "${@}"
> should not affect this behavior.
...
> printf "%q" "${@}"
> becomes
> printf "%q" ""
>
> which is correct as ''
No, "${@}" doesn't always become at least one word. "$@" can expand to zero or
more words. It can become nothing, one word, or more, possibly concatenated
with adjacent words. That's completely beside the point.
BTW, your wrappers won't work. A wrapper would need to implement format string
parsing in order to determine which argument(s) correspond with %q so that
they could be removed from the output if not given an argument. It also would
have to work around handling -v, which can take up either 1 or 2 args, plus
possibly --. It isn't feasible to wrap printf itself this way.
I just used "$@" as an example of something that can expand to zero words and
preserves exactly the positional parameters present (possibly none). This is
important because printf %q is often used to safely create a level of escaping
that guarantees getting out exactly what you put in.
$ ksh -c 'f() { cmd=$(printf "%q " "$@"); }; f; eval x=moo "$cmd"; echo "$x"'
moo
Bash yields a "command not found" error for obvious reasons, but it goes
against the spirit of what %q is supposed to do IMHO. It's a minor detail.
This isn't even a particularly good example.
See Chet's previous message for the actual explanation. The reason is to be
consistent with the defaults for other format specifiers which act like they
were given a null or zero argument if more formats than arguments are given.
We already had pretty much this same discussion here:
http://lists.gnu.org/archive/html/bug-bash/2012-12/msg00083.html
It somehow slipped my mind.
--
Dan Douglas
