Hi..
I was reading the LDD3's DMA chapter (chapter 15). The author talks
about the splitting of address space between 1GB and 3GB. I couldn't
make myself crystal clear as to why it is done.
because things like code and data must stay and operate in different
modes. Kernel code stays in kernel space, that way user can not directly
execute arbitrary kernel codes without first elevating its privilege.
Also, some codes must be put on kernel space so they can manage the
hardwares. I/O, setting up page tables, managing CPU clock.
Of course, there's nothing stopping you from creating OS with
all-in-kernel-mode programming style, but IMHO that will make that OS
very very vulnerable to local attack and any code can control anything
it wants without coordination.
I got even more
confused when he said kernel can't access physicalmemory address which
is not mapped into kernel space. Why doesn't kernel has full access to
all the address space?
In 32 bit, you have 4 GB address space. This 4GB space is splitted
between kernel and user space, usually 1 GB for kernel space and 3GB for
user space. Within this 1GB space, the kernel must map the RAM so they
become accessible. Unfortunately, not all that 1 GB space can freely
used. Normally, the upper 128 MB space is reserved for high memory
mapping. So, it leaves you the rest (896 MB) for normal mapping.
In 64 bit, you have all the room to map the entire RAM, because
theoritically 64 bit gives you "unlimited space".
Why you need to map first before you can access them? Because if you
don't map them (by setting the approriate page directory, page middle
directory, page table entry), the MMU (memory management unit) can not
translate your requested address (logical address) into physical
address. . This is especially needed since Linux operates in protected
mode. In real mode addressing, you just need to deal with segmentation.
read Intel documentation in www.intel.com for deeper explanation about
all these hardware internals. maybe, LDD forgot to mention that the
explanations are based on Intel x86 32 bit... that's why you got all
those confusions.
regards,
Mulyadi
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