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Groups > gnu.bash.bug > #14427
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|---|---|
| From | Ilkka Virta <itvirta@iki.fi> |
| Newsgroups | gnu.bash.bug |
| Subject | Unquoted array slice ${a[@]:0} expands to just one word if IFS doesn't have a space |
| Date | Wed, 1 Aug 2018 14:43:27 +0300 |
| Lines | 39 |
| Approved | bug-bash@gnu.org |
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On both Bash 4.4.12(1)-release and 5.0.0(1)-alpha, a subarray slice like
${a[@]:0} expands to just one word if unquoted (and if IFS doesn't
contain a space):
$ a=(aa bb); IFS=x; printf ":%s:\n" ${a[@]:0}
:aa bb:
I expected it would expand to separate words, as it does without the
slice, and just like $@ does, sliced or not:
$ a=(aa bb); IFS=x; printf ":%s:\n" ${a[@]}
:aa:
:bb:
$ set -- aa bb; IFS=x; printf ":%s:\n" $@
:aa:
:bb:
$ set -- aa bb; IFS=x; printf ":%s:\n" ${@:1}
:aa:
:bb:
It's as if it first joins the picked elements with spaces, and then
splits using IFS, instead of producing multiple words and word-splitting
them individually.
The same thing happens with ${a[*]:0} (but not with ${*:1}):
the array elements get joined with spaces to a single word. If IFS is
empty, unset, or contains a space the result is multiple words as
expected with both [@] and [*].
An expansion like that should in most cases be quoted,
but the current behaviour still seems a bit inconsistent.
--
Ilkka Virta / itvirta@iki.fi
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Unquoted array slice ${a[@]:0} expands to just one word if IFS doesn't have a space Ilkka Virta <itvirta@iki.fi> - 2018-08-01 14:43 +0300
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