Hi
hi all,
For a 32-bit machine, there is total 4GB of memory(Physiacl
addresses). So
in that 1 GB for Kernel and 3GB for User Space.
Correct, assuming you're using 3:1 VM (virtual memory) split, that is
3GB for user space and 1GB for kernel space.
So, kernel won't access that 3GB of memory, right?
Kernel space has no problem accessing user space, but not the other way
around. This is because of page protection.
All the kernel code is exist in 1 GB of memory.... CMIIW
Correct.
So, my doubt is the DMA(0-16MB),NORMAL(16-896MB),HIGH(896MB-) memory
divided within this 4GB of memory ,right?
Not really, what you see above (the VM split) is in virtual address
space, while memory zone splits the RAM physical address into different
parts. For example, 0-16 MB is named as DMA zone because they are
adddressable for DMA operation.
So, both has different context (virtual VS physical) although in
practice they are connected since you will get free pages from those
zones. But once you are speaking virtual address space, there is no
zone. Zones are in physical addressing context.
if we allocate the pages using alloc_pages, from which region the
pages are
allocated for the kernel modules. is it from Physical RAM memory or
from 1GB
of kernel space ??
It can be from any zone as requested, but normally IIRC it is from
either normal or DMA zone. In some cases, to relieve pressure toward
those two zones, requests can be satisfied from highmem zone (if there
is such available, i.e on machine with RAM more than 896 MB in x86 32 bit).
Remember that request is always served by looking into free pages from
RAM (physical context). And yes, if you bring the context into virtual
addressing, it would be seems as being "satisfied" by looking into
kernel space, but this is no more than just identity mapping comes to play.
PS: Sorry if I directly reply to your original post, since I think here
I can understand what is your real confusion.
regards,
Mulyadi
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