On Wed, 26 Aug 2015 17:22:25 +0530
Pratyush Anand <pan...@redhat.com> wrote:

> Hi Jeff,
> 
> Thanks for your review comments.
> 
> On 26/08/2015:06:31:03 AM, Jeff Layton wrote:
> > On Tue, 25 Aug 2015 11:34:19 +0530
> > Pratyush Anand <pan...@redhat.com> wrote:
> > >   TP_fast_assign(
> > >           __entry->xprt = xprt;
> > > +         __entry->addr =
> > > +                 xprt ? (struct sockaddr *)&xprt->xpt_remote : NULL;
> > 
> > I don't get it. It's not safe to save a pointer to xprt and dereference
> > that in a tracepoint (and I understand why that is), but it is safe to
> > save a pointer to a structure embedded inside of xprt? Shouldn't you be
> > saving a copy of the entire sockaddr struct instead?
> 
> As far as "saving of a pointer to a structure embedded inside of xprt" is
> concerned, I do not see any issue.
> 
> Its your take that what you want to print as your tracepoint print output. I
> might be missing something..  However, I do not see any value addition in
> printing address of a structure(located in ring buffer) where xprt->xpt_remote
> has been copied.
> 
> > 
> > Ahh, ok -- I think I see the confusion. %pIScp does not print the
> > address of the sockaddr, but instead dereferences the pointer and
> > prints it as a formatted address string. See pointer() in
> > lib/vsprintf.c. You do want to save off a copy of the structure instead.
> 
> In my opinion, saving of structure would only be necessary if you want to 
> access
> element of the structure xprt->xpt_remote.
> 
> ~Pratyush

That's exactly what %pI does. Look at the pointer() function.

For this tracepoint, we want to print out the address of the remote end
of the socket. That's saved in xprt->xpt_remote.

%pIScp doesn't print the pointer's address. It dereferences that
pointer and formats and prints the socket address that it points to. If
you pass a NULL pointer into anything that uses %pI, it's going to oops.

-- 
Jeff Layton <jlay...@poochiereds.net>
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