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? 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. Thank you for your attention to this matter. Look forward to hearing from you. Best regards. Valdis Klētnieks <valdis.kletnieks@vt.edu> 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"... :)