>> 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.
Thank you for the clarification.
I understand it on a deeper level with your help.
How can I know whether scatter/gather is available or not?
>> 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.
Thank you for the clarification.
I understand it on a deeper level with your help.
How can I know whether scatter/gather is available or not?
In another word, when it's available and when it's not?
I do not intend to ask the behavior of gadget driver.
I just wonder how I can confirm it in general.
I do not intend to ask the behavior of gadget driver.
I just wonder how I can confirm it in general.
Thank you for your attention to this matter.
Look forward to hearing from you.
Best regards.
Valdis Klētnieks <valdis.kletnieks@xxxxxx> 于2020年6月19日周五 下午3:14写道：
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"... :)
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