On 8/7/07, Neil Brown <[email protected]> wrote:
> On Monday August 6, [email protected] wrote:
> > On Tue, 2007-08-07 at 16:07 +1000, Neil Brown wrote:
> > > Suppose that in a program I have an open file descriptor for a device,
> > > and I want to find the /sys/block information for this device.
> > > There is currently no direct way to do this. I need to read
> > > /sys/block/*/dev, /sys/block/*/*/dev
> > > and match major/minor numbers with the result from fstat.
> > >
> > > I would like a more direct mechanism.
$ udevinfo --query=path --name=sda
/block/sda
> > > The following patch is a proposal for such a mechanism.
> > >
> > > It provides an 'ioctl' which returns then 'name' of the device, as
> > > generated by bdevname. This is the same name that is used to create
> > > entries in sysfs.
> > > For a partition of a device, it returns 'device/partition'.
> >
> >
> > how about returning the entire path relative to the start of sysfs? That
> > way, if things move or something you're tolerant against that....
>
> That makes a lot of sense.
> So it would return "block/sda/sda1" now,
Every devpath always starts with a '/' in the kernel, its kind of
weird, but we should not introduce a new variation of it. :)
> but one day that might change
> to "class/block/sda/sda1" or some-such.
It will start with /devices/..., like:
/sys/devices/pci0000:00/0000:00:1f.2/host0/target0:0:0/0:0:0:0/sda/sda1
> > (I'd not be against making this a generic IOCTL for every device, a
> > SYSFSLOCATION kind of ioctl... it's by no means block specific...)
> >
>
> That too seems very sensible. Only it's harder to choose a 'generic'
> ioctl request number than to choose a block-specific one :-)
>
> We would also need a somewhat longer buffer. The longest
> /sys/**/dev
> path on my test machine is
> /sys/devices/pci0000:00/0000:00:1d.0/usb2/2-1/2-1:1.1/usb_endpoint/usbdev2.2_ep82/dev
>
> So at least 80 chars. Probably 256 would do....
They can be definitely longer than than 256, probably not longer than
512 today, but you can't be sure about that.
Thanks,
Kay
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