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Re: [Help-bash] Is there a particular reason that $@ is expanded to a st
Re: [Help-bash] Is there a particular reason that $@ is expanded to a string *using a space as separator instead of $IFS* in contexts where word splitting is not performed?
Tue, 8 Sep 2015 09:51:27 -0400
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On 9/6/15 12:11 AM, ziyunfei wrote:
> $ set -- 1 2 3
> $ IFS=:
> $ # in a [[...]] command
> $ [[ $@ == "1 2 3" ]]; echo $?
> 0 # print 1 in zsh
> $ [[ $@ == "1:2:3" ]]; echo $?
> 1 # print 0 in zsh
> $ # in the rhs of an assignment statement
> $ foo=$@
> $ IFS=
> $ echo $foo
> 1 2 3 # print 1:2:3 in zsh
> $* use the first character of $IFS as separator in these cases but $@
> doesn’t. Does POSIX standard mention this, or this is just a historical
> behavior of Bourne shell?
The key is that $@ expands to multiple fields regardless of whether or not
field splitting is performed, so you have to decide how to join those
fields together into a string.
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRU address@hidden http://cnswww.cns.cwru.edu/~chet/