Here's a code example
$ sleep 1 & killall -d sleep
sig:15, cmd:sleep, pid:13393, dev:0x2e6 uid:1001
 Done sleep 1
As we can see, sleep is not getting killed if -d is specified.
It works fine with -v, however:
$ sleep 1 & killall -v sleep
kill -TERM 19067
 Terminated sleep 1
Is this a desired behavior?
According to killall source, it seems to be expected:
May be man page needs to specify that -d implies -s? And it looks like -d and -v need separate descriptions while here.
If there is no objection, I can give the man page a try.
(In reply to Fernando Apesteguía from comment #2)
Changes in review D25413
A commit references this bug:
Date: Sat Jun 27 11:28:12 UTC 2020
New revision: 362678
killall(1): Clarify -d, -s and -v options
-d and -v are not equivalent options. The former is more verbose than the
latter and the former does not actually send the signals while the latter does.
Let them have their own paragraphs.
From the point of view of the output, -v is equivalent to -s, so describe them
close to each other. The difference is that former actually sends the signals
and the latter doesn't.
Approved by: manpages(0mp)
Differential Revision: https://reviews.freebsd.org/D25413