Re: GNU Parallel Bug Reports Bug with version 20120322: --timeout gives ps error

[parallel]

I'm a sleepy idiot.

The code should be:

 my $script = q{
        family_pids() {
            for CHILDPID in `ps -o ppid=,pid= -A | gawk '$1=='$@'{print
 $2}' `; do family_pids $CHILDPID &
            done
            echo "$@"
        }
    } .


On Tue, 27 Mar 2012 23:27:00 -0700
address@hidden wrote:

> Failing command:
> # parallel -S : --nonall --timeout 1 sleep 2
> ps: illegal option -- -
> usage: ps [-AaCcEefhjlMmrSTvwXx] [-O fmt | -o fmt] [-G gid[,gid...]]
>           [-g grp[,grp...]] [-u [uid,uid...]]
>           [-p pid[,pid...]] [-t tty[,tty...]] [-U user[,user...]]
>        ps [-L]
> 
> As the comment in the source indicates, this needs to be tested with
> other operating systems.
> 
> For MacOSX 10.6.8, the correct ps command is: 
>   ps -o ppid= -p $$
> Note, there is nothing after the equal sign; this deletes the header.