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Re: Word splitting for $@ in variable assignment

From: Ilkka Virta
Subject: Re: Word splitting for $@ in variable assignment
Date: Thu, 24 Jun 2021 11:09:49 +0300

On Thu, Jun 24, 2021 at 10:53 AM Alvin Seville <alvinseville7cf@gmail.com>

>  Hello! I want to understand why the following code doesn't produce any
> error:
> set -- "a b" "c"
> a="$@"
> ? I expected smth like: main.sh: line 2: b: command not found due to word
> splitting according to documentation
> <
> https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Shell-Parameters
> >.
> What am I missing?

It's a bit oddly put in the manual. It doesn't actually get word-split as
such, it just gives the individual positional parameters joined with spaces
to a single string.

3.4.2 Special Parameters explains that: "$@: In contexts where word
splitting is not performed, this expands to a single word with each
positional parameter separated by a space."

But 3.4 Shell Parameters is a bit confusing: "Word splitting is not
performed, with the exception of "$@" as explained below." I'm not sure if
this is meant to refer to that sentence 3.4.2, but it's a bit confusing
since that one refers to cases without word splitting. It would probably be
clearer to not call it an exception and just refer to Special Parameters
for how $@ behaves when not word-split.

Anyway, it may be better to use a="$*" as it's treated more consistently
between different shells. a="$@" uses the first character of IFS as the
separator in some shells (Zsh, Busybox, Debian/Ubuntu's Dash), and a space
always in others (Bash, Ksh), while a="$*" consistently uses the first
character of IFS. With the default IFS, the result is of course the same.

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