Re: Questions about the zoned page frame allocator and fix mapped addresses
On Mon, Mar 30, 2015 at 10:46 PM, wrote: > On Mon, 30 Mar 2015 20:25:38 +0530, Sunny Shah said: > > >- The book says, about releasing page frames to the per CPU cache - > "no > >page frame is ever released to the cold cache: the kernel always > assumes > >the freed page frame is hot with respect to the hardware cache". What > is > >the reason for this decision ? > > What can go wrong if the page is cold but the kernel assumes it's hot? > > What can go wrong if the kernel assumes it's cold but it's actually hot? > > (Hint - in which cases will it do a cache flush? In which cases is a > cache flush needed? What happens in each case if the kernel guesses wrong?) > Here's my understanding: - If a page is cold but kernel assumes it to be hot, a 'hot' allocation of that page would need it to be loaded into the CPU cache which means a little overhead. - If a page is hot but kernel assumes it to be cold, a future 'hot' allocation would eventually cause the invalidation of the CPU cache location corresponding to that page. But, even if the page were assumed to be hot, this might have anyway happened due to another page occupying the same cache location. This has confused me even more! I'm not sure if this is correct, but if it is, wouldn't this mean treating hot pages as cold is better ? I apologise for anything that I might be looking over or not thinking in the right direction. Please feel free to rebuke. Also, any help on my other questions would be greatly appreciated: - It is possible for a page to be in ZONE_NORMAL and yet have it's PG_reserved flag cleared. Is this correct ? - The function "fix_to_virt" for fix-mapped linear addresses does the following: return (0xf000UL - (idx << PAGE_SHIFT)); Why are the upper 4096 bytes not used, and the addressing starts from the top of the virtual address space - 4096 ? - The book says "each fix-mapped linear address maps one page frame of the physical memory". Shouldn't it be "maps one*physical location* of memory" rather than one page frame ? - My understanding is that the kernel page table entries for addresses > 896 MB would be empty and those addresses would be mapped using separate data structures used for temporary and permanent kernel mappings and non-contiguous page frame allocation. Is this wrong ? Thanks, Sunny ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Questions about the zoned page frame allocator and fix mapped addresses
Hello, I have few more questions from my reading of "Understanding the Linux Kernel", chapter "Memory Management". - The book says, about releasing page frames to the per CPU cache - "no page frame is ever released to the cold cache: the kernel always assumes the freed page frame is hot with respect to the hardware cache". What is the reason for this decision ? - It is possible for a page to be in ZONE_NORMAL and yet have it's PG_reserved flag cleared. Is this correct ? - The function "fix_to_virt" for fix-mapped linear addresses does the following: return (0xf000UL - (idx << PAGE_SHIFT)); Why are the upper 4096 bytes not used, and the addressing starts from the top of the virtual address space - 4096 ? - The book says "each fix-mapped linear address maps one page frame of the physical memory". Shouldn't it be "maps one *physical location* of memory" rather than one page frame ? - My understanding is that the kernel page table entries for addresses > 896 MB would be empty and those addresses would be mapped using separate data structures used for temporary and permanent kernel mappings and non-contiguous page frame allocation. Is this wrong ? Thanks, Sunny ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Re: Understanding the mapping of physical memory to kernel address space
Thank you so much !! On Sun, Mar 15, 2015 at 8:48 PM, Arun KS wrote: > > On Mar 15, 2015 8:27 PM, "Sunny Shah" wrote: > > > > Hi Arun, > > > > Thanks for that excellent explanation. It's more or less clear to me now. > > > > However, quoting what you said: > > > > Because we have plenty of kernel virtual > > address(3GB) and can easily map 2GB of RAM in to it. > > > > Why should we have to map the whole RAM into the KVA? Shouldn't it be > only LOW_MEM? > > We dont need to. I was just telling we can do that aswell. When you go > with 2:2 split, you are changing user space virtual memory layout. Which > will bring in lot if other problems. So very common approach is to use high > mem. > > > > > I also read on a stack overflow thread that LOW_MEM is memory that is > permanently mapped into KVA, while HIGH_MEM is mapped as required. Is this > true? > Absolutely > > Thanks, > Arun > > > > > Thanks, > > Sunny > > > > On Sun, Mar 15, 2015 at 1:17 PM, Arun KS wrote: > >> > >> Hello Sunny, > >> > >> On Sat, Mar 14, 2015 at 8:25 PM, Sunny Shah > wrote: > >> > Thank you guys! > >> > > >> > I have two more questions from your replies: > >> > > >> > I thought I had understood HIGH_MEM and LOW_MEM, but it appears I was > wrong. > >> > Does the concept of high memory/low memory correspond to physical > address > >> > space or virtual address space? Also, does LOW_MEM always have to be > 1 GiB > >> > maximum? > >> > >> Physical memory is divided into HIGH_MEM and LOW_MEM. > >> Why do we need two memory? To understand this, we need to know who > >> all are consumers of kernel virtual address(KVA) which is limited to > >> 1GB(in case of 3:1 split). > >> 1. low mem( physical ram which has linear mapping to KVA). > >> 2. IO memory address(for eg a DMA controller registers, Memory > >> controller register, etc). When MMU is enabled all the address > >> generated by the cpu are virtual address. Hence io memory should have > >> a valid virtual to physical memory mapping. > >> 3. Vmalloc address space. A dynamic kernel memory allocation > >> mechanism, which only guarantees continuity in virtual address space. > >> 4. Persistent kernel map. (if kernel want to use HIGH memory, it maps > >> high memory to this portion of virtual address). > >> 5. vector table. > >> > >> Let me give a rough calculation for a better understanding. Lets say a > >> system with configuration as follows, > >> 2GB of physical RAM, 40 MB of physical io address space, 240 MB of > >> vmalloc address space, 32 MB for persistent kernel map > >> The maximum RAM which can be mapped as low mem = 1GB - (40 MB + 240 MB > >> + 32MB) = 712MB. > >> > >> Rest of RAM 1336MB( 1GB - 712MB) will fall as HIGH_MEM. > >> > >> Now how system uses HIGH memory. Major user of HIGH mem is user space > >> code. Kernel directly maps high mem to user space virtual address. > >> Hight mem is also used by kernel though PK mappings. Even vmalloc > >> allocation can also fall from HIGH mem region. > >> > >> Now if we decides to use 1:3 user space to kernel space split, high > >> memory is not required. Because we have plenty of kernel virtual > >> address(3GB) and can easily map 2GB of RAM in to it. > >> > >> HTH. > >> > >> Thanks, > >> Arun > >> > >> > For a RAM of 896 MiB - 4096 MiB, the book says: > >> > "In this case, the RAM cannot be mapped entirely into the kernel > linear > >> > address space. The best Linux can do during the initialization phase > is to > >> > map a RAM window of size 896 MB into the kernel linear address space." > >> > > >> > Why is there a need to map the whole RAM into the kernel space (the > usage of > >> > the word "entirely") ? Shouldn't it be only LOW_MEM ? Or am I > confusing the > >> > two things here ? > >> > > >> > > >> > I believe all doubts are pointing to the concepts of LOW_MEM and > HIGH_MEM, > >> > but I'm still not being able to wrap my head around them. > >> > > >> > Thanks, > >> > Sunny > >> > > >> > On Thu, Mar 12, 2015 at 11:49 PM, Jeff Haran > wrote: > >> >> > >> >> -Original Message-
Re: Understanding the mapping of physical memory to kernel address space
Hi Arun, Thanks for that excellent explanation. It's more or less clear to me now. However, quoting what you said: Because we have plenty of kernel virtual address(3GB) and can easily map 2GB of RAM in to it. Why should we have to map the whole RAM into the KVA? Shouldn't it be only LOW_MEM? I also read on a stack overflow thread that LOW_MEM is memory that is permanently mapped into KVA, while HIGH_MEM is mapped as required. Is this true? Thanks, Sunny On Sun, Mar 15, 2015 at 1:17 PM, Arun KS wrote: > Hello Sunny, > > On Sat, Mar 14, 2015 at 8:25 PM, Sunny Shah > wrote: > > Thank you guys! > > > > I have two more questions from your replies: > > > > I thought I had understood HIGH_MEM and LOW_MEM, but it appears I was > wrong. > > Does the concept of high memory/low memory correspond to physical address > > space or virtual address space? Also, does LOW_MEM always have to be 1 > GiB > > maximum? > > Physical memory is divided into HIGH_MEM and LOW_MEM. > Why do we need two memory? To understand this, we need to know who > all are consumers of kernel virtual address(KVA) which is limited to > 1GB(in case of 3:1 split). > 1. low mem( physical ram which has linear mapping to KVA). > 2. IO memory address(for eg a DMA controller registers, Memory > controller register, etc). When MMU is enabled all the address > generated by the cpu are virtual address. Hence io memory should have > a valid virtual to physical memory mapping. > 3. Vmalloc address space. A dynamic kernel memory allocation > mechanism, which only guarantees continuity in virtual address space. > 4. Persistent kernel map. (if kernel want to use HIGH memory, it maps > high memory to this portion of virtual address). > 5. vector table. > > Let me give a rough calculation for a better understanding. Lets say a > system with configuration as follows, > 2GB of physical RAM, 40 MB of physical io address space, 240 MB of > vmalloc address space, 32 MB for persistent kernel map > The maximum RAM which can be mapped as low mem = 1GB - (40 MB + 240 MB > + 32MB) = 712MB. > > Rest of RAM 1336MB( 1GB - 712MB) will fall as HIGH_MEM. > > Now how system uses HIGH memory. Major user of HIGH mem is user space > code. Kernel directly maps high mem to user space virtual address. > Hight mem is also used by kernel though PK mappings. Even vmalloc > allocation can also fall from HIGH mem region. > > Now if we decides to use 1:3 user space to kernel space split, high > memory is not required. Because we have plenty of kernel virtual > address(3GB) and can easily map 2GB of RAM in to it. > > HTH. > > Thanks, > Arun > > > For a RAM of 896 MiB - 4096 MiB, the book says: > > "In this case, the RAM cannot be mapped entirely into the kernel linear > > address space. The best Linux can do during the initialization phase is > to > > map a RAM window of size 896 MB into the kernel linear address space." > > > > Why is there a need to map the whole RAM into the kernel space (the > usage of > > the word "entirely") ? Shouldn't it be only LOW_MEM ? Or am I confusing > the > > two things here ? > > > > > > I believe all doubts are pointing to the concepts of LOW_MEM and > HIGH_MEM, > > but I'm still not being able to wrap my head around them. > > > > Thanks, > > Sunny > > > > On Thu, Mar 12, 2015 at 11:49 PM, Jeff Haran > wrote: > >> > >> -Original Message- > >> From: kernelnewbies-boun...@kernelnewbies.org > >> [mailto:kernelnewbies-boun...@kernelnewbies.org] On Behalf Of Arun KS > >> Sent: Thursday, March 12, 2015 11:03 AM > >> To: Sunny Shah > >> Cc: kernelnewbies > >> Subject: Re: Understanding the mapping of physical memory to kernel > >> address space > >> > >> Hello Sunny, > >> > >> On Thu, Mar 12, 2015 at 10:32 PM, Sunny Shah > >> wrote: > >> > Hello, > >> > > >> > This is my first mail on this list, so please let me know if I'm > erring. > >> > > >> > I'm reading Bovet and Cesati's "Understanding the Linux Kernel", > >> > specifically the chapter "Memory Addressing", sub-section "Kernel Page > >> > Tables". Here they describe how Linux initializes its page tables for > >> > various RAM sizes and how much of the physical address space is mapped > >> > onto the kernel virtual address space. > >> > > >> > I have several questions from my reading: > >> > > >> > My understanding is that the 32 bit
Re: Understanding the mapping of physical memory to kernel address space
Thank you guys! I have two more questions from your replies: - I thought I had understood HIGH_MEM and LOW_MEM, but it appears I was wrong. Does the concept of high memory/low memory correspond to physical address space or virtual address space? Also, does LOW_MEM always have to be 1 GiB maximum? - For a RAM of 896 MiB - 4096 MiB, the book says: "In this case, the RAM cannot be mapped entirely into the kernel linear address space. The best Linux can do during the initialization phase is to map a RAM window of size 896 MB into the kernel linear address space." Why is there a need to map the whole RAM into the kernel space (the usage of the word "entirely") ? Shouldn't it be only LOW_MEM ? Or am I confusing the two things here ? I believe all doubts are pointing to the concepts of LOW_MEM and HIGH_MEM, but I'm still not being able to wrap my head around them. Thanks, Sunny On Thu, Mar 12, 2015 at 11:49 PM, Jeff Haran wrote: > -Original Message- > From: kernelnewbies-boun...@kernelnewbies.org [mailto: > kernelnewbies-boun...@kernelnewbies.org] On Behalf Of Arun KS > Sent: Thursday, March 12, 2015 11:03 AM > To: Sunny Shah > Cc: kernelnewbies > Subject: Re: Understanding the mapping of physical memory to kernel > address space > > Hello Sunny, > > On Thu, Mar 12, 2015 at 10:32 PM, Sunny Shah > wrote: > > Hello, > > > > This is my first mail on this list, so please let me know if I'm erring. > > > > I'm reading Bovet and Cesati's "Understanding the Linux Kernel", > > specifically the chapter "Memory Addressing", sub-section "Kernel Page > > Tables". Here they describe how Linux initializes its page tables for > > various RAM sizes and how much of the physical address space is mapped > > onto the kernel virtual address space. > > > > I have several questions from my reading: > > > > My understanding is that the 32 bit virtual address space of a process > > is split into 2 parts - the first 3 GiB for the user space and the > > remaining 1GiB for the kernel (with the same kernel mapping being used > > for all processes. However, although the kernel is mapped into the > > higher portion of the address space, it resides in the lower 1 GiB of > RAM. Is this correct? > Yes. Incase of 3:1 mapping, kernel virtual address starts at 0xc000. > You can also have 2:2 mappings aswell. It is a configurable option > > Just an FYI, I've seen 1:3 mapping too. We had to do that with the kernels > we built > when I was at one company because we needed 3GB of virtual address space > to map all > of the memory mapped registers on their ASICs. > > There's lots of options here. > > Jeff Haran > > > ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
Understanding the mapping of physical memory to kernel address space
Hello, This is my first mail on this list, so please let me know if I'm erring. I'm reading Bovet and Cesati's "Understanding the Linux Kernel", specifically the chapter "Memory Addressing", sub-section "Kernel Page Tables". Here they describe how Linux initializes its page tables for various RAM sizes and how much of the physical address space is mapped onto the kernel virtual address space. I have several questions from my reading: - My understanding is that the 32 bit virtual address space of a process is split into 2 parts - the first 3 GiB for the user space and the remaining 1GiB for the kernel (with the same kernel mapping being used for all processes. However, although the kernel is mapped into the higher portion of the address space, it resides in the lower 1 GiB of RAM. Is this correct? - There is a frequent mention of "mapping RAM to the kernel address space". Is the whole RAM mapped to the kernel space? What would happen in the case of a machine having only 1 GiB of RAM? - Related to the above, what exactly is max_low_pfn? It is mentioned it is the "Page frame number of the last page frame directly mapped by the kernel (low memory)". What would it's value be for a machine with 1 GiB RAM? For a machine with 4 GiB RAM, would it be the number of the frame covering the last portion of kernel space (896 MiB)? Sorry if my questions are a bit vague. I'm still new to this and having difficulty relating everything. Thanks, Sunny ___ Kernelnewbies mailing list Kernelnewbies@kernelnewbies.org http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
