Why does “page allocation failure” occur whereas there are still “58*4096kB (C)” could be used?

[Valdis Kl=?utf-8?Q?=c4=93?=tnieks]

On Fri, 19 Jun 2020 14:56:20 +0800, 孙世龙 sunshilong said:

> Why doesn't the kernel use two memory blocks whose size is 2048KB(i.e.*oder 9 *)
> instead of one block *order 10 *(you see, there are still three free blocks and
>  2048KB*2=4096KB equivalent to the memory size of order 10)?

Most parts of the kernel, when asking for very high-order allocations, *will*
have a fallback strategy to use smaller chunks. So, for instance,  if a device
need a 1M buffer and supports scatter-gather operations, if 1M of contiguous
memory isn't available, the kernel can ask for 4 256K chunks and have the I/O
directed into the 4 areas.  However, if the memory *has* to be contiguous (for
example, no scatter/gather available, or it's for an array data structure),
then it can't do that.

And in fact, that fallback could very well have happened in this case - I
didn't bother chasing back to see if the gadget driver does recovery by
allocating multiple smaller chunks.

(That's a good "exercise for the student"... :)
-------------- next part --------------
A non-text attachment was scrubbed...
Name: not available
Type: application/pgp-signature
Size: 832 bytes
Desc: not available
URL: <http://lists.kernelnewbies.org/pipermail/kernelnewbies/attachments/20200619/7242a10d/attachment.sig>