Am i correct that If we have 1 G of physical RAM on our system, 128 MB will be reserved for temporary mappings? But this won't be used for any purpose since we don't need any temporary mapping(which is required for >1 GB). Then is it that 128 MB physical memory doesn't serve any purpose? Please clarify.
Thayumanavar S.
On 9/3/05, Rahul Iyer <idlisambar@xxxxxxxxx> wrote:
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
