Re: Re: How to get the pid of all children, grand children..../... of a process

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On 31 May 2006 04:54:23 -0000, rohit  hooda <rohit13hooda@xxxxxxxxxxxxxx> wrote:

 On Wed, 31 May 2006 Taha Hafeez Siddiqi wrote :

 >On Tue, 2006-05-30 at 17:20 +0530, Gopala Krishna wrote:
 > >
 > >
 > >         Fortunately, here comes the proc filesystem !
 > >         So, you have the pid of the father process : look at the file
 > >         /proc/'ppid'/status there should be a fild ppid : you have the
 > >         pid of
 > >         your grand parent.
 > >         You can do a recursive function to find all the ancestors of
 > >         your
 > >         process.
 > >         This solution is quite annoying since you have to handle
 > >         files, but it
 > >         works.
 > >
 > >         If this solution is not acceptable for you, you can look at
 > >         the pstree
 > >         source code.
 > >
 > >         I don't know if you wanted to do that in your own programm or
 > >         just
 > >         wanted some soft to give you the entire tree (like pstree).
 > >         Anyway... :
 > >
 > >
 > > In My case, I knew only pid of the first process. I want the top down
 > > approach rather than the bottom up (i.e if I know child, I know
 > > parent. but, my requirement is to find out child and it's grand
 > > children, If I know the parent pid). Currently I am going through
 > > pstree code.
 > >
 > > Thanks and regards,
 > > Gopal.
 > >
 > >
 >
 >Hi all,
 >
 >I think there is another way.*
 >
 >You can get hold of your current, use its list to get hold of the
 >process structure and the up you go using the process->parent....
 >
 >It may not be the correct way, but I suppose surely it is one way of
 >doing it*.


 Well in kernel mode, if you have the PID of a process, from that you can have
its task_struct and then you can iterate down to get the lists of its children and
grand children and so on ... but over here we have to do things in user space n
not in kernel space

Then please take it off this list (KERNELnewbies).

Thanks,
Nish

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