On Fri, 2013-02-15 at 00:58 +0100, Samuel Thibault wrote:
> Svante Signell, le Fri 15 Feb 2013 00:37:39 +0100, a écrit :
> > > No, it doesn't. There are quite a few ways in which it will break.
>
> if (d[i].io_port == MACH_PORT_NULL)
> {
> _hurd_port_free (&d[i].cell->port,
>
> Doesn't make sense either. You haven't understood what the original code
> was doing (cleaning the previously allocated ports, not this newly one).
I did understand the code, but maybe the change went wrong. I would
really appreciate if you could help me to understand what's the problem
with the new code.
Old code: (cleaning previously allocated ports when one FD is bogus and
exit out of the loop)
if (fd < _hurd_dtablesize)
{
...
if (d[i].io_port != MACH_PORT_NULL)
continue;
}
/* If one descriptor is bogus, we fail completely. */
while (i-- > 0)
if (d[i].type != 0)
_hurd_port_free (&d[i].cell->port,
&d[i].ulink, d[i].io_port);
break;
New code: (cleaning the ports allocated for bogus FDs, and continue to
next i)
/* If one descriptor is bogus, mark and remove it. */
if (d[i].io_port == MACH_PORT_NULL)
{
_hurd_port_free (&d[i].cell->port,
&d[i].ulink, d[i].io_port);
continue; /* Next i */
}
Is the test for MACH_PORT_NULL wrong? Maybe it is not needed at all, the
ports for bogus FDs can be kept, doing no harm?
