10

Is it possible to unset the $1 variable? If not, I can't find out where it is explained in man.

[root@centos2 ~]# set bon jour
[root@centos2 ~]# echo $1$2
bonjour
[root@centos2 ~]# unset $1
[root@centos2 ~]# echo $1$2
bonjour
[root@centos2 ~]#

EDIT:

Finaly, here is what I found out in man (man set option double-dash) to empty all the positional parameters (and the man used the word "unset"!):

If no arguments follow this option, then the positional parameters are unset.

[root@centos2 ~]# echo $1

[root@centos2 ~]# set bon jour
[root@centos2 ~]# echo $1$2
bonjour
[root@centos2 ~]# set --
[root@centos2 ~]# echo $1$2

[root@centos2 ~]#

It's @Jeff Schaller's answer that helped me understand that.

Improve this question
2
  • 4
    I'm not sure what you're asking but you could use 'shift' to make $2 take the place of $1, therefore removing $1. EDIT: Kusalananda has a more expansive answer below. Commented May 17, 2017 at 20:45
  • You are right my question is not well formed. I was looking for how to empty a positional variable if I had set it. Commented May 17, 2017 at 21:25

4 Answers 4

Reset to default
21

You can't unset a positional parameter, but you may shift $2 into $1, effectively removing the old value of $2.

$ set -- bon jour
$ echo "$1$2"
bonjour

$ shift
$ echo "$1$2"  # $2 is now empty (i.e. "unset" or "not set")
jour

The shift command will shift all positional parameters one step lower. Command-line parsing loops (that does not use getopt/getopts) commonly shift the positional parameters over each iteration while repeatedly examining the value of $1. It is uncommon to want to unset a positional parameter.

By the way, unset takes a variable name, not its value, so unset $1 will, in effect, unset the variable bon (had it been previously set).

Improve this answer
6
  • What about if I want to empty $1? Also, good point on the fact that it would be unset bon. I didn't think about that. Commented May 17, 2017 at 21:30
  • @Pozinux That would be quite an unusual thing to want to do. What is the issue you're trying to solve? Commented May 17, 2017 at 21:33
  • In fact, I'm learning. I'was trying to understand how setworks. I tried this example with "bonjour" but couldn't find a way how to empty this variable. I mixed a bit with unset. Probably because I did't get yet all the concept about set, unset, env, export etc. Commented May 17, 2017 at 21:36
  • 1
    @Pozinux Well, setting them again would overwrite them, like Jeff shows in his answer. Commented May 17, 2017 at 21:39
  • When you use getopt or getopts you do use shift. Commented May 18, 2017 at 20:55
16

You could empty it, by re-setting the parameters to: "the empty string" followed by "the parameters, starting from the 2nd one":

$ set -- 1 2 3 4
$ printf -- "->%s<-\n" "$@"
->1<-
->2<-
->3<-
->4<-
$ set -- "" "${@:2}"
$ printf -- "->%s<-\n" "$@"
-><-
->2<-
->3<-
->4<-
Improve this answer
1
  • Your anwser is nice and right aswell but it took me more time to understand it because of my little shell level... In fact, it helped me to find out about the "--" option that I didn't know (see my edit on my question). Thanks. Commented May 18, 2017 at 17:05
6

Besides all the correct technical answers you should notice that there is a logical flaw in your question. $1, $2,... are not the values of named variables like $A, $B, ..., but the 1st, 2nd,... value of a list of values. Maybe the first value of a list is an empty string but it does not make sense to say my list has no first value. If a list has no first value then it has no value at all and so it is an empty list.

So you cannot "unset" the first value of a list. You can only remove the first value from a list. Then the former 2nd value of the list is now the 1st value of the list, the former 3rd is now the 2nd and so on. This is exactly what the shift operator does.

Improve this answer
3
  • Though note that actual arrays don't need to be indexed consecutively from zero, you could go unset A; A[4]=foo; A[7]=bar; and leave with A[0] among others unset (not just empty). And then unset A[4]. But I'm digressing, this doesn't apply to the positional parameters. Commented May 18, 2017 at 11:26
  • @miracle173, you are right it doesn't make sense. I thought that unset was also used to empty a variable. Commented May 18, 2017 at 11:49
  • @ilkkachu A is an array and not a list of values. A[0], A[1], ... are similar to variables and therefore can be empty Commented May 18, 2017 at 12:48
4

No, at least not directly. This syntax should work but yields an error:

$ set bon jour
$ echo $1 $2
$ unset 1
1: not a valid identifier

unset $variable is usually a mistake and unset variable was what was intended. Unfortunately it doesn't work here.

Improve this answer
0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.