On 2/14/24 12:06 AM, Oğuz wrote:
On Tuesday, February 13, 2024, Chet Ramey via austin-group-l at The Open
Group <[email protected] <mailto:[email protected]>>
wrote:
`continue' is a builtin; continue has a return status; `!' says to
negate it. It seems easy to come to the conclusion that the script
should return 1.
The same can be said about `return'. But bash disagrees:
$ bash -c 'f(){ ! return 1;}; f; echo $?'
1
$
Does POSIX allow this or is it another case where bash diverges from POSIX?
POSIX requires this, since it says that return sets $? to 1 here. If
you find cases where you believe bash differs from POSIX, and it's not
documented in POSIX mode, please report them.
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU [email protected] http://tiswww.cwru.edu/~chet/
