Date: Tue, 27 Apr 2021 05:57:40 +0000
From: "Beer, Mathis" <mathis.b...@funkwerk.com>
Message-ID: <3024777.xmhlYjkYgV@ka-nl-mbeer>
| Given a background process that has exited before the script got to wait -n:
|
| function foo() { return 1; }
| foo & FOO_PID=$!
|
| function bar() { sleep 1; }
| bar & BAR_PID=$!
|
| sleep 0.1
| # should exit with 127 since FOO_PID is non-existent now
| wait -n $FOO_PID $BAR_PID
That's not the way it works, pid's "exist" for these purposes until
waited upon (just as they do in the process table for processes and
the wait() sys call).
The exit status should be 1, as that's the exit status of foo.
| Then wait -n will wait on BAR_PID to exit, despite the fact that FOO_PID is
| invalid. (Tested with 5.0.17.)
A second wait -n (in bash) generates a diagnostic about the pid that
no longer exists (the NetBSD sh which has the same options doesn't do that)
but waits for anything valid.
| The way it currently works makes it hard to make a script that waits for one
| of a set of background processes to exit
Not if you code it properly. Use -p to help. It all just works.
| (ie. that fails if one of a set of
| background processes fails), since any error that happens before the script
| reaches the wait will simply be ignored.
No, it won't.
kre
