by When installing OpenBSD, I assumed that my swap drive had been set up -
I used fdisk to create a partition of x bytes in swap formatted it
during the install, and it's now shown as /dev/wd0b under disklabel.
When I type swapctl -l it spits this out:
#swapctl -l
Device 512-blocks Used Avail Capacity Priority
swap_device 316065 8 316057 0% 0
#
This is the correct size for my wd0b partition. However, reading the
FAQs at openbsd.org I found a section on swap files
(http://www.openbsd.com/faq/faq14.html#14.4) that talked about setting
up a vnode in order to swap. Is this required? When I try to do it it
just creates two devices to swap to, one the original 'swap_device' and
the other my newly created '/dev/vnd0c', both with the same properties.
So it would seem to me that swap_device is already set up to swap
properly, but in this case why the whole thing in the FAQ about setting
up /dev/vnd0? The document says:
This shows the devices currently being used for swapping and their
current statistics. In the above example there is only one device named
"swap_device". This is the predefined area on disk that is used for
swapping. (Shows up as partition b when viewing disklabels) As you can
also see in the above example, that device isn't getting much use at the
moment. But for the purposes of this document, we will act as if an
extra 32M is needed.
They then go on, however, to set up /dev/wd0b as the device they
associate with a vnode and thereby swap to. What's the point of this?
I'm just confused as to how the whole swapping scheme works, any help's
appreciated.
--
Matthew L Creech