In Bash, using "$@" in an assignment, (as in a="$@" ) concatenates the
positional parameters to single string, joined with spaces. Somewhat
similarly to what "$*" does, except that $* uses the first letter of
IFS, but $@ always uses a space.
However, I can't see this documented in the manual, is it somewhere?
"3.4 Shell Parameters" [1] discusses assignments to variables, and
there's the phrase: "Word splitting is not performed, with the exception
of "$@" as explained below." But the actual explanation seems to be
missing, as there's no other mention of $@ on the page.
There's also no mention of assignments under "3.4.2 Special Parameters"
[2]. It simply states that "$@" expands to separate words.
I'd suggest adding something like this to the description of $@ in 3.4.2:
""
If $@ or "$@" is used on the right hand side of an assignment to a
variable, it instead expands to a single word with the value of each
positional parameter separated by a space. That is, a="$@" is equivalent
to a="$1 $2 ...".
""
(It seems the online manual is an older version than that in git, but I
didn't see this mentioned in the devel version of the manual either.)
[1] https://www.gnu.org/software/bash/manual/html_node/Shell-Parameters.html
[2]
https://www.gnu.org/software/bash/manual/html_node/Special-Parameters.html
--
Ilkka Virta / itvi...@iki.fi
