How do I do this:
set foo bar baz
unset # Something else here
echo $# # Should ouput 0
By doing set "", $# will still be 1 and not 0.
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2 Answers
Ok this was just stupid, and I found out using set -- will work.
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@Gilles: According to the bash man page,
set -with no arguments leaves the positional parameters unchanged.camh– camh2011-08-19 03:07:34 +00:00Commented Aug 19, 2011 at 3:07 -
@camh I must disagree. I see this in the manual: set -- If no arguments follow this option, then the positional parameters are unset. Otherwise, the positional parameters are set to the args, even if some of them begin with a -.kdubs– kdubs2019-05-08 12:33:46 +00:00Commented May 8, 2019 at 12:33
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@kdubs you are quoting the man page for
set --. I have usedset -(single hyphen), which says: "If there are no args, the positional parameters remain unchanged."camh– camh2019-05-08 13:00:14 +00:00Commented May 8, 2019 at 13:00 -
you are right. I got them confused. my screen made your - look like --kdubs– kdubs2019-05-14 01:31:51 +00:00Commented May 14, 2019 at 1:31
Easily unset all positional parameters with:
shift $#
To remove the first 3 positional parameters:
shift 3
Shift "pops" off the first N positional parameters. By using $# we get the total number of positional parameters. The command thus pops off all the positional parameters.
