[SOLVED] timeout core utility

poojithas · 10-12-2010, 09:10 AM

Hi,

I'm using timeout core utility in my bash shell script. The problem I'm facing is I want the PID of the command which is supplied to timeout. Let me explain with the code I'm using:

Code:

 #!/bin/bash

 timeout 120 ./prog1 & 
 echo "@$!"

the above script gives me PID of timeout process but not of the prog1, so how do I get that.

What I observed is: prog1 PID = timeout PID + 1
Is it always the case that if I add 1 to timeout PID, I get prog1 PID ?

Tinkster · 10-12-2010, 04:22 PM

I don't think that's a feasible assumption - it can (and frequently will)
be correct, but it doesn't have to.

What you CAN do is to ps -ef, and find the
process which has the timeout PID as its PPID.

Code:

ps -ef|awk '$3 == PID {print $2}'

Cheers,
Tink

poojithas · 10-12-2010, 06:53 PM

Great, it works.

Between, Is there any C language function to get parent PID given a child PID ? . I know getppid() but this only gives current process parent pid. I'm looking for a function which can take child pid and gives me parent pid.

Tinkster · 10-12-2010, 07:25 PM

Quote:

Originally Posted by poojithas

Great, it works.

Between, Is there any C language function to get parent PID given a child PID ? . I know getppid() but this only gives current process parent pid. I'm looking for a function which can take child pid and gives me parent pid.

I don't think that's exposed via a function anywhere - but
you can easily implement it yourself by traversing /proc/<PID>/

Either of "stat" and "status" will give you the ppid.

Cheers,
Tink

poojithas · 10-12-2010, 07:54 PM

Modified script:

#!/bin/bash
timeout 120 ./prog1 &
PID=`ps -eo pid,ppid | awk -v pid="$!" '$2 == pid {print $1}'`
echo "@$PID"

here I'm only supplying pid and ppid from ps command to awk to filter. This is somewhat neat I think.

Tinkster · 10-12-2010, 08:03 PM

Yup - looks good :}