I was reading about page faulting and from what I read, the MMU consults the page table to translate virtual addresses into physical addresses. It is the responsibility of the OS (via the page fault handler) to fill up these page table entries.
What confuses me is how does the page fault handler obtain the physical addresses in the first place? In the diagrams and notes I saw, the CPU seems to use virtual addresses and the MMU transparently translates them to physical addresses. Does the CPU specially work with physical address rather than virtual addresses for page fault handling?
If there is an access to some 4K page that is not present in memory and the page fault handler successfully locates the corresponding 4K page on disk, how does it acquire a 4K page of physical memory and figure out the physical address of the 4K page of physical memory?
Part of the OS`s responsibility is to keep track of a list of physical pages. You can look on OSDev to see how this is done - usually by querying BIOS/UEFI-exposed functions, which give you (usually non-contiguous) lists of free memory.
UEFI in particular exposes GetMemoryMap at boot time to get an array of memory descriptors.
Given a maintained list of available physical pages - when the OS handles a page fault, it has access to the faulting virtual address, and it can decide what to do. If it needs to allocate a new page, then it will consults its list of free pages, and choose an available physical page to map into the virtual address space. On x86 this mapping is done by modifying the page table and loading it into the cr3 register.
Once the page is mapped, it can be written to using virtual addresses.
The operating system works with virtual address at the fundamental level, but it does have to manage physical addresses as well. In an OSes memory management subsystem, there is a part called the physical memory manager. Basically, at startup, it read a table given to it by the firmware which tells it which memory regions are free. It sets up free list to contain all free pages in this map. When a page fault occurs, it pulls a page off of this list, maps it into the PTE, grabs another page for to create a page table if no page table exists for this address (note that it will keep doing that step depending on how many levels haven't been mapped yet), flushes the TLB for this address, and then it carries on.
Note that most physical memory allocators are much more compex then this, but, fundamentally, that is the algorithm.
I have some C code running in RTL mode, I use CPU I6400 the C code is just a simple code to read and write from some subsystem, for example,
I tried to write in this address : 0x001e400000 (physical address) so when CPU executes this address I got a TLB exception because this address is a mapped area, after many research I fond that I need to translate this address from Virtual address to physical address, I replace this address by 0xffffffffbe400000 (Kseg1), now im able to write some value in this address is done, But when I tried to read from this address(W/R) or another address I got an exception on KSEG2,
Have you an idea why the writing step pass correctly but the reading step generate a exception?
The MMU is used to translate virtual address to physical address for a running process with the help of page table corresponding to that process.
Lets take a scenario when the virtual address space is equal to physical address space. Do we really require MMU in that case as we won't we having a situation where same virtual address space maps to different physical addresses?
Lets say
Virtual Address
| 20 bits (V) | 12 bits(PO) |
Physical Address
| 20 bits(PPN) | 12 bits(PO) |
where V = Virtual Page
PO = Page offset
PPN = Physical Page Number
Do we really require Page Table for every process?
What problems might appear when there are more than one process?
Please neglect cache memory for simplification.
The MMU does much more than mapping a virtual address space to a physical address space of different size. The most important point of an MMU is memory protection, which is relevant even if both address spaces have the same size:
The MMU handles pages (e.g. 4 kB of size) of virtual memory that are mapped to pages of physical memory.
In most systems, there is not only a single virtual address space, but one for every process. Under MMU control, every process can only access pages as allowed by the operating system (which programs the MMU). Most pages of different processes are isolated from each other, so that e.g. one process cannot crash another process by writing into its memory.
Mapping of virtual to physical pages under OS control allows address space randomization, so that reading across a virtual page boundary results in reading random data instead of certain data (a protection against e.g. buffer overflow attacks).
Moreover, even if there is a single process, pages can be treated as read-write, read-only, execute only, and access forbidden. This allows to restrict a process to access its own pages in a permitted way, e.g. it can make it impossible to execute "data" stored.
Some more information can be found here.
Virtual Memory is a quite complex topic for me. I am trying to understand it. Here is my understanding for a 32-bit system. Example RAM is just 2GB. I have tried reading many links, and I am not confident at the moment. I would like you people to help me in clearing up my concepts. Please acknowledge my points, and also please answer for what you feel is wrong. I have also a confused section in my points. So, here starts the summary.
Every process thinks it is only running. It can access the 4GB of memory - virtual address space.
When a process access a virtual address it is translated to physical address via MMU.
This MMU is a part of a CPU - a hardware.
When the MMU cannot translate the address to a physical one, it raises a page fault.
On page fault, the kernel is notified. The kernel check the VM area struct. If it can find it - may be on disk. It will do some page-in /page-out. And get this memory on the RAM.
Now MMU will again try and will succeed this time.
In case the kernel cannot find the address, it will raise a signal. For example, invalid access will raise a SIGSEGV.
Confused points.
Does Page table is maintained in Kernel? This VM area struct has a page table ?
How MMU cannot find the address in physical RAM. Let's say it translates to some wrong address in RAM. Still the code will execute, but it will be a bad address. How MMU ensures that it is reading a right data? Does it consult Kernel VM area everytime?
Is the Mapping table - virtual to physical is inside a MMU. I have read it that is maintained by an individual process. If it is inside a process, why I can't see it.
Or if it is MMU, how MMU generates the address - is it that Segment + 12-bit shift -> Page frame number, and then the addition of offset (bits -1 to 10) -> gives a physical address.
Does it mean that for a 32-bit architecture, with this calculation in my mind. I can determine the physical address from a virtual address.
cat /proc/pid_value/maps. This shows me the current mapping of the vmarea. Basically, it reads the Vmarea struct and prints it. That means that this is important. I am not able to fit this piece in the complete picture. When the program is executed does the vmarea struct is generated. Is VMAREA comes only into the picture when the MMU cannnot translate the address i.e. Page fault? When I print the vmarea it displays the address range , permission and mapped to file descriptor, and offset. I am sure this file descriptor is the one in the hard-disk and the offset is for that file.
The high-mem concept is that kernel cannot directly access the Memory region greater than 1 GB(approx). Thus, it needs a page table to indirectly map it. Thus, it will temporarily load some page table to map the address. Does HIGH MEM will come into the picture everytime. Because Userspace can directly translate the address via MMU. On what scenario, does kernel really want to access the High MEM. I believe the kernel drivers will mostly be using kmalloc. This is a direct memory + offset address. In this case no mapping is really required. So, the question is on what scenario a kernel needs to access the High Mem.
Does the processor specifically comes with the MMU support. Those who doesn't have MMU support cannot run LInux?
Does Page table is maintained in Kernel? This VM area struct has a page table ?
Yes. Not exactly: each process has a mm_struct, which contains a list of vm_area_struct's (which represent abstract, processor-independent memory regions, aka mappings), and a field called pgd, which is a pointer to the processor-specific page table (which contains the current state of each page: valid, readable, writable, dirty, ...).
The page table doesn't need to be complete, the OS can generate each part of it from the VMAs.
How MMU cannot find the address in physical RAM. Let's say it translates to some wrong address in RAM. Still the code will execute, but it will be a bad address. How MMU ensures that it is reading a right data? Does it consult Kernel VM area everytime?
The translation fails, e.g. because the page was marked as invalid, or a write access was attempted against a readonly page.
Is the Mapping table - virtual to physical is inside a MMU. I have read it that is maintained by an individual process. If it is inside a process, why I can't see it.
Or if it is MMU, how MMU generates the address - is it that Segment + 12-bit shift -> Page frame number, and then the addition of offset (bits -1 to 10) -> gives a physical address.
Does it mean that for a 32-bit architecture, with this calculation in my mind. I can determine the physical address from a virtual address.
There are two kinds of MMUs in common use. One of them only has a TLB (Translation Lookaside Buffer), which is a cache of the page table. When the TLB doesn't have a translation for an attempted access, a TLB miss is generated, the OS does a page table walk, and puts the translation in the TLB.
The other kind of MMU does the page table walk in hardware.
In any case, the OS maintains a page table per process, this maps Virtual Page Numbers to Physical Frame Numbers. This mapping can change at any moment, when a page is paged-in, the physical frame it is mapped to depends on the availability of free memory.
cat /proc/pid_value/maps. This shows me the current mapping of the vmarea. Basically, it reads the Vmarea struct and prints it. That means that this is important. I am not able to fit this piece in the complete picture. When the program is executed does the vmarea struct is generated. Is VMAREA comes only into the picture when the MMU cannnot translate the address i.e. Page fault? When I print the vmarea it displays the address range , permission and mapped to file descriptor, and offset. I am sure this file descriptor is the one in the hard-disk and the offset is for that file.
To a first approximation, yes. Beyond that, there are many reasons why the kernel may decide to fiddle with a process' memory, e.g: if there is memory pressure it may decide to page out some rarely used pages from some random process. User space can also manipulate the mappings via mmap(), execve() and other system calls.
The high-mem concept is that kernel cannot directly access the Memory region greater than 1 GB(approx). Thus, it needs a page table to indirectly map it. Thus, it will temporarily load some page table to map the address. Does HIGH MEM will come into the picture everytime. Because Userspace can directly translate the address via MMU. On what scenario, does kernel really want to access the High MEM. I believe the kernel drivers will mostly be using kmalloc. This is a direct memory + offset address. In this case no mapping is really required. So, the question is on what scenario a kernel needs to access the High Mem.
Totally unrelated to the other questions. In summary, high memory is a hack to be able to access lots of memory in a limited address space computer.
Basically, the kernel has a limited address space reserved to it (on x86, a typical user/kernel split is 3Gb/1Gb [processes can run in user space or kernel space. A process runs in kernel space when a syscall is invoked. To avoid having to switch the page table on every context-switch, on x86 typically the address space is split between user-space and kernel-space]). So the kernel can directly access up to ~1Gb of memory. To access more physical memory, there is some indirection involved, which is what high memory is all about.
Does the processor specifically comes with the MMU support. Those who doesn't have MMU support cannot run Linux?
Laptop/desktop processors come with an MMU. x86 supports paging since the 386.
Linux, specially the variant called µCLinux, supports processors without MMUs (!MMU). Many embedded systems (ADSL routers, ...) use processors without an MMU. There are some important restrictions, among them:
Some syscalls don't work at all: e.g fork().
Some syscalls work with restrictions and non-POSIX conforming behavior: e.g mmap()
The executable file format is different: e.g bFLT or ELF-FDPIC instead of ELF.
The stack cannot grow, and its size has to be set at link-time.
When a program is loaded first the kernel will setup a kernel VM-Area for that process is it? This Kernel VM Area actually holds where the program sections are there in the memory/HDD. Then the entire story of updating CR3 register, and page walkthrough or TLB comes into the picture right? So, whenever there is a pagefault - Kernel will update the page table by looking at Kernel virtual memory area is it? But they say Kernel VM area keeps updating. How this is possible, since cat /proc/pid_value/map will keep updating.The map won't be constant from start to end. SO, the real information is available in the Kernel VM area struct is it? This is the acutal information where the section of program lies, it could be HDD or physical memory -- RAM? So, this is filled during process loading is it, the first job? Kernel does the page in page out on page fault, and will update the Kernel VM area is it? So, it should also know the entire program location on the HDD for page-in / page out right? Please correct me here. This is in continuation to my first question of the previous comment.
When the kernel loads a program, it will setup several VMAs (mappings), according to the segments in the executable file (which on ELF files you can see with readelf --segments), which will be text/code segment, data segment, etc... During the lifetime of the program, additional mappings may be created by the dynamic/runtime linkers, by the memory allocator (malloc(), which may also extend the data segment via brk()), or directly by the program via mmap(),shm_open(), etc..
The VMAs contain the necessary information to generate the page table, e.g. they tell whether that memory is backed by a file or by swap (anonymous memory). So, yes, the kernel will update the page table by looking at the VMAs. The kernel will page in memory in response to page faults, and will page out memory in response to memory pressure.
Using x86 no PAE as an example:
On x86 with no PAE, a linear address can be split into 3 parts: the top 10 bits point to an entry in the page directory, the middle 10 bits point to an entry in the page table pointed to by the aforementioned page directory entry. The page table entry may contain a valid physical frame number: the top 22 bits of a physical address. The bottom 12 bits of the virtual address is an offset into the page that goes untranslated into the physical address.
Each time the kernel schedules a different process, the CR3 register is written to with a pointer to the page directory for the current process. Then, each time a memory access is made, the MMU tries to look for a translation cached in the TLB, if it doesn't find one, it looks for one doing a page table walk starting from CR3. If it still doesn't find one, a GPF fault is raised, the CPU switches to Ring 0 (kernel mode), and the kernel tries to find one in the VMAs.
Also, I believe this reading from CR, page directory->page-table->Page frame number-memory address this all done by MMU. Am I correct?
On x86, yes, the MMU does the page table walk. On other systems (e.g: MIPS), the MMU is little more than the TLB, and on TLB miss exceptions the kernel does the page table walk by software.
Though this is not going to be the best answer, iw ould like to share my thoughts on confused points.
1. Does Page table is maintained...
Yes. kernel maintains the page tables. In fact it maintains nested page tables. And top of the page tables is stored in top_pmd. pmd i suppose it is page mapping directory. You can traverse through all the page tables using this structure.
2. How MMU cannot find the address in physical RAM.....
I am not sure i understood the question. But in case because of some problem, the instruction is faulted or out of its instruction area is being accessed, you generally get undefined instruction exception resulting in undefined exception abort. If you look at the crash dumps, you can see it in the kernel log.
3. Is the Mapping table - virtual to physical is inside a MMU...
Yes. MMU is SW+HW. HW is like TLB and all. The mapping tables are stored here. For instructions, that is for code section i always converted the physical-virtual address and always they matched. And almost all the times it matches for Data sections as well.
4. cat /proc/pid_value/maps. This shows me the current mapping of the vmarea....
This is more used for analyzing the virtual addresses of user space stacks. As you know virtually all the user space programs can have 4 GB of virtual address. So unlike kernel if i say 0xc0100234. You cannot directly go and point to the istruction. So you need this mapping and the virtual address to point the instruction based on the data you have.
5. The high-mem concept is that kernel cannot directly access the Memory...
High-mem corresponds to user space memory(some one correct me if i am wrong). When kernel wants to read some data from a address at user space you will be accessing the HIGHMEM.
6. Does the processor specifically comes with the MMU support. Those who doesn't have MMU support cannot run LInux?
MMU as i mentioned is HW + SW. So mostly it would be coming with the chipset. and the SW would be generally architecture dependent. You can disable MMU from kernel config and build. I have never tried it though. Mostly these days allthe chipsets have it. But small boards i think they disable MMU. I am not entirely sure though.
As all these are conceptual questions, i may be lacking some knowledge and be wrong at places. If so others please correct me.
I have a doubt when each process has its own separate page table then why is a system wide page table required? Also if a page table is such that it maps virtual address to a physical address then I think two process may map to same physical address because all process have same virtual address space. Is this true?
About the second part, which is mapping virtual address to the same physical address, for library code and different instances of an application code, this is indeed what is done. The code is given read only access and the same virtual address is mapped to the same physical address. In this way, there is no need to have multiple copies of the same code in the physical memory, all this assuming ASLR is not enabled.
Now concerning the data part, Modern OSs like Linux, use demand paging, that is a page is only brought to the physical memory when it is accessed (read or write). At that point, the kernel can make sure to assign a unique physical address for that page. I don't know what is the purpose of system wide page table though.
A system-wide page table would be used by the kernel, which in most systems is always mapped in to memory. (Typically a 32-bit system will allocate the lower 2-3 GB of virtual address space to the user process and the upper 1-2 GB to the kernel.) Making the kernel mappings common across all processes means that you don't have to worry about making sure kernel code you're about to run is mapped when you enter a system call from userland.