Raghu Vadapalli wrote:
On 9/2/05, Rahul Iyer <idlisambar@xxxxxxxxx> wrote:
Hi All,
I followed the kernel address space thread, and unfortunately, by the
flow of it, found it rather confusing. So at the risk of repitition,
I'll try to explain this from scratch. Some of it might be very
elementary, so bear with me. Also, since I know precious little about
any other architecture, what I write pertains to x86 ONLY.
ok, so an x86 is a 32 bit processor. Hence, the maximum memory that can
be addressed is 2^32 = 4GB. This 4GB set of addressable addresses
(redundant, i know) is called the address space and the addresses are
called virtual addresses. Now, to access physical memory, every process
must go through the paging system (provided paging is turned on, which
it is in Linux). Also, in order to be able to access *any* physical
page, that page *must* be in your page tables. Every process has it's
own page tables.
Now, the kernel has it's own code and data in some *physical* pages, and
each user process has it's code, data etc, in some *physical* pages. In
order to be able to access these physical pages, these physical pages
must be mapped in the page tables. How is this done?
The first 3G worth of virtual addresses are mapped to the physical pages
of the process, i.e, the userland pages. The remaining 1G is used to map
the physical pages containing the kernel's stuff. So, every process has
in it's page tables, the first 3G of virtual addresses mapped to it's
own pages and the last 1G mapped to the kernel's pages. This is commonly
called the 3G/1G split.
Now, since the kernel needs to be able to access all of physical memory
and the kernel has only 1G of address space, the kernel can access only
1G of physical memory. This places an upper limit on the amount of RAM a
machine could have. Some people use a 2G/2G split.. that is the first 2G
is userspace and the next 2G is kernel space.
Does it mean that if I have RAM > 1GB on my m/c it's not of any use ?
Can somebody please clarify. or Only kernel space code + data can't
be more than 1GB ?. User space can always get the rest of the RAM ?
The answer is below! It is > 2G for the 2G/2G split and > 1G for the
3G/1G split.
A way to access > 2GB is to reserve the last 128MB of the 1G of kernel
space addresses for temporary mappings. These are done on the fly, as
and when needed. 1024MB - 128MB = 896MB. That's where the magical 896 MB
comes from.
Hope that helped
-rahul
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