"Martin Dalecki wrote:"
> "Peter T. Breuer" wrote:
> >
> > Let me put it more plainly. Martin Daleki + rumour assures me that the
> > blk_size array nowadays measure in blocks not KB, yet to me it seems that
>
> sectors = 512 per default
> blocks = 1024 per default.
I know that! But it's irrelevant. What I need to know is if blk_size is
still counting in KB, or if it has switched to blocks.
> Never said anything else.
Err .. you said that blk_size is now measured in blocks, not KB. You
said thet the rumour is true.
"A month of sundays ago Martin Dalecki wrote:"
> "Peter T. Breuer" wrote:
> > Is blk_size[][] supposed to contain the size in KB or blocks?
> There is no rumor it's in blocks.
Maybe I misinterpret what you write. I interpret it as meaning "the
rumour is not a rumour but a fact. It is in blocks".
> Look at the initialization point for the arrays. They all use constants
> which you can look up in the kernel headers.
I _know_ that. It's irrelevant.
The point is that if blk_size counts in KB, then the size of a device
cannot reach more that 2^32 * 2^10 = 2^42 = 4TB. I'd personally say
2TB, becuase the int blk_size number is signed.
That's rumoured not to be the case, and the max size of a device is
supposed to be about 8 to 16TB. Let's suppose the rumour is true ..
So we deduce that one has to assign a different meaning for the blk_size
array. "count in blocks" is how the rumour goes. That way you can get
4 times higher sizes .. all the way to 8 or 16TB per device. And this
is what is rumoured to be the case.
Is it or is it not so? A straight answer from the list would be nice!
> ./linux/fs.h:#define BLOCK_SIZE_BITS 10
> ./linux/fs.h:#define BLOCK_SIZE (1<<BLOCK_SIZE_BITS)
These are _defaults_ for _blksize_. Sure you can change it as you like,
but according to the "blk_size is in KB" hypothesis, this matters not one
iota to the size limit on devices. Change blksize and size does not
change. But according to the "blk_size is in blocks" hypothesis, yes
changing blksize will change the size of the device. Testing shows
that scenario "blk_size is in KB" is true.
Am I making plain the difference between blk_size and blksize?
blk_size is the number of blocks or KB (which?) in a device. blksize is
the size of the blocks. Is blk_size in KB or blocks?
It should be in blocks if the size of a device is to reach 8 or 16TB.
If it is in KB, we are limited to 2 or 4TB.
> Which means 1024 bytes for blk_size as default value.
But so what? That doesn't answer the question of whether blk_size
is in blocks or not.
> > it doesn't. Look at this code from ll_rw_blk.c in 2.4.[14]:
Look at it:
if (blk_size[major])
minorsize = blk_size[major][MINOR(bh->b_rdev)];
if (minorsize) {
unsigned long maxsector = (minorsize << 1) + 1;
This clearly hardcodes blk_size as measuring in units of 2 sectors, no
matter what we set for blksize. It should be, in my view
unsigned long maxsector =
minorsize * blksize_size[major][MINOR(bh->b_rdev] + 1;
or no device can be larger than 4TB. And neither can a filesystem, and
neither can a file ...
Now, I know I can write my own generic_make_request() code, but I have
no intention of maintaining it through different kernel versions just
to get the right size measurement. Besides, it's everyone's problem.
Persuade me that this is not a bug, and an important one at that :-)
Hellloooooo everybody! Linux cannot manage partitions greater than
4TB, ha ha ha hhhhhaaaa! ;-)
I at least am getting up to devicesizes at the 8TB range.
Best wishes!
Peter
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This archive was generated by hypermail 2b29 : Thu Nov 15 2001 - 21:00:40 EST