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On Tue, 19 Jan 2010, Eric Sandeen wrote:
This has been in the back of my mind for a while, but maybe
it's worth addressing now that disks getting awfully large.
class XFS(FS):
""" XFS filesystem """
_type = "xfs"
_mkfs = "mkfs.xfs"
_modules = ["xfs"]
_labelfs = "xfs_admin"
_defaultFormatOptions = ["-f"]
_defaultLabelOptions = ["-L"]
_maxLabelChars = 16
_maxSize = 16 * 1024 * 1024
...
XFS can actually go much bigger than this, as can some other
filesystems. However, there is a VM limit on some architectures;
basically the page cache can't address > (2^(long bits) * (page size))
So for x86, it's 2^32 * 4096 = 16T hence the limit above.
But for 64-bit arches, the max fs size is well beyond 16T.
A lot of the _maxSize values were filled in by me, and I was looking around
wherever I could to find what the maximum sizes were of these filesystems. If
there is a way to programatically set it for a filesystem, I'm all for that.
We just needed fs experts to weigh in on that, so thanks.
This stuff is available in C-land:
#include <unistd.h>
long sz = sysconf(_SC_PAGESIZE);
(most systems allow the synonym _SC_PAGE_SIZE for _SC_PAGESIZE), or
or:
#include <unistd.h>
int sz = getpagesize();
and sizeof(long) I guess; not sure how to get there from Python.
import resource
resource.getpagesize()
Anyway if Anaconda wants to be more correct, it'd be something like:
_maxSize = min((2^bits_per_long * page_size), fs_max_size)
Does that sound doable?
Sounds reasonable. There's probably a better way to do this, but here's
something:
import resource
import struct
bitsPerLong = struct.Struct("l").size * 8
fsMaxSize = ???
_maxSize = min((2^bitsPerLong * resource.getpagesize()), fsMaxSize)
A struct.Struct() object has the size property which gives the size in bytes.
"l" is the format char for a long. Tried on x86_64 and bitsPerLong becomes 64
and on x86_32 it becomes 32.
- --
David Cantrell <dcantrell@xxxxxxxxxx>
Red Hat / Honolulu, HI
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