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Groups > gnu.bash.bug > #12133
| From | Andreas Schwab <schwab@linux-m68k.org> |
|---|---|
| Newsgroups | gnu.bash.bug |
| Subject | Re: ${1+"$@"} does not generate multiple words if IFS is empty |
| Date | 2015-12-30 10:02 +0100 |
| Message-ID | <mailman.1334.1451466172.843.bug-bash@gnu.org> (permalink) |
| References | <56835243.0Dp3As7Vk0rP8qDy%martijn@inlv.org> |
martijn@inlv.org writes:
> The substitution ${1+"$@"} should resolve to "$@" if there is at
> least one parameter -- i.e. one word per parameter. This works fine
> if IFS contains any character or is unset. If IFS is empty, it
> instead resolves to the equivalent of "$*", i.e. a single word
> concatenating all the parameters without a separator. IFS should
> not influence the behaviour of "$@" under any circumstances.
http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06_05
"If the value of IFS is null, no field splitting shall be performed."
Andreas.
--
Andreas Schwab, schwab@linux-m68k.org
GPG Key fingerprint = 58CA 54C7 6D53 942B 1756 01D3 44D5 214B 8276 4ED5
"And now for something completely different."
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Re: ${1+"$@"} does not generate multiple words if IFS is empty Andreas Schwab <schwab@linux-m68k.org> - 2015-12-30 10:02 +0100
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