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 ? > > 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. > > Some kernels, like Redhat's from what I read on this list, have a > separate page table for the kernel. That is, the kernel has a *separate* > page table with 4G addresses and the user processes too have page tables > with all 4GB belonging to the user process. The problem, IMHO, with this > is that every switch from userspace to kernel space involves a TLB > flush. This is bad for performance. Any comments? Rik, Arjan? > > Hope that helped > Thanks > Rahul > > > -- > Kernelnewbies: Help each other learn about the Linux kernel. > Archive: http://mail.nl.linux.org/kernelnewbies/ > FAQ: http://kernelnewbies.org/faq/ > > -- Kernelnewbies: Help each other learn about the Linux kernel. Archive: http://mail.nl.linux.org/kernelnewbies/ FAQ: http://kernelnewbies.org/faq/
