On the surface, this appears to be a silly question. Some patience please.. :-)
Am structuring this qs into 2 parts:
Part 1:
I fully understand that platform RAM is mapped into the kernel segment; esp on 64-bit systems this will work well. So each kernel virtual address is indeed just an offset from physical memory (DRAM).
Also, it's my understanding that as Linux is a modern virtual memory OS, (pretty much) all addresses are treated as virtual addresses and must "go" via hardware - the TLB/MMU - at runtime and then get translated by the TLB/MMU via kernel paging tables. Again, easy to understand for user-mode processes.
HOWEVER, what about kernel virtual addresses? For efficiency, would it not be simpler to direct-map these (and an identity mapping is indeed setup from PAGE_OFFSET onwards). But still, at runtime, the kernel virtual address must go via the TLB/MMU and get translated right??? Is this actually the case? Or is kernel virtual addr translation just an offset calculation?? (But how can that be, as we must go via hardware TLB/MMU?). As a simple example, lets consider:
char *kptr = kmalloc(1024, GFP_KERNEL);
Now kptr is a kernel virtual address.
I understand that virt_to_phys() can perform the offset calculation and return the physical DRAM address.
But, here's the Actual Question: it can't be done in this manner via software - that would be pathetically slow! So, back to my earlier point: it would have to be translated via hardware (TLB/MMU).
Is this actually the case??
Part 2:
Okay, lets say this is the case, and we do use paging in the kernel to do this, we must of course setup kernel paging tables; I understand it's rooted at swapper_pg_dir.
(I also understand that vmalloc() unlike kmalloc() is a special case- it's a pure virtual region that gets backed by physical frames only on page fault).
If (in Part 1) we do conclude that kernel virtual address translation is done via kernel paging tables, then how exactly does the kernel paging table (swapper_pg_dir) get "attached" or "mapped" to a user-mode process?? This should happen in the context-switch code? How? Where?
Eg.
On an x86_64, 2 processes A and B are alive, 1 cpu.
A is running, so it's higher-canonical addr
0xFFFF8000 00000000 through 0xFFFFFFFF FFFFFFFF "map" to the kernel segment, and it's lower-canonical addr
0x0 through 0x00007FFF FFFFFFFF map to it's private userspace.
Now, if we context-switch A->B, process B's lower-canonical region is unique But
it must "map" to the same kernel of course!
How exactly does this happen? How do we "auto" refer to the kernel paging table when
in kernel mode? Or is that a wrong statement?
Thanks for your patience, would really appreciate a well thought out answer!
First a bit of background.
This is an area where there is a lot of potential variation between
architectures, however the original poster has indicated he is mainly
interested in x86 and ARM, which share several characteristics:
no hardware segments or similar partitioning of the virtual address space (when used by Linux)
hardware page table walk
multiple page sizes
physically tagged caches (at least on modern ARMs)
So if we restrict ourselves to those systems it keeps things simpler.
Once the MMU is enabled, it is never normally turned off. So all CPU
addresses are virtual, and will be translated to physical addresses
using the MMU. The MMU will first look up the virtual address in the
TLB, and only if it doesn't find it in the TLB will it refer to the
page table - the TLB is a cache of the page table - and so we can
ignore the TLB for this discussion.
The page table
describes the entire virtual 32 or 64 bit address space, and includes
information like:
whether the virtual address is valid
which mode(s) the processor must be in for it to be valid
special attributes for things like memory mapped hardware registers
and the physical address to use
Linux divides the virtual address space into two: the lower portion is
used for user processes, and there is a different virtual to physical
mapping for each process. The upper portion is used for the kernel,
and the mapping is the same even when switching between different user
processes. This keep things simple, as an address is unambiguously in
user or kernel space, the page table doesn't need to be changed when
entering or leaving the kernel, and the kernel can simply dereference
pointers into user space for the
current user process. Typically on 32bit processors the split is 3G
user/1G kernel, although this can vary. Pages for the kernel portion
of the address space will be marked as accessible only when the processor
is in kernel mode to prevent them being accessible to user processes.
The portion of the kernel address space which is identity mapped to RAM
(kernel logical addresses) will be mapped using big pages when possible,
which may allow the page table to be smaller but more importantly
reduces the number of TLB misses.
When the kernel starts it creates a single page table for itself
(swapper_pg_dir) which just describes the kernel portion of the
virtual address space and with no mappings for the user portion of the
address space. Then every time a user process is created a new page
table will be generated for that process, the portion which describes
kernel memory will be the same in each of these page tables. This could be
done by copying all of the relevant portion of swapper_pg_dir, but
because page tables are normally a tree structures, the kernel is
frequently able to graft the portion of the tree which describes the
kernel address space from swapper_pg_dir into the page tables for each
user process by just copying a few entries in the upper layer of the
page table structure. As well as being more efficient in memory (and possibly
cache) usage, it makes it easier to keep the mappings consistent. This
is one of the reasons why the split between kernel and user virtual
address spaces can only occur at certain addresses.
To see how this is done for a particular architecture look at the
implementation of pgd_alloc(). For example ARM
(arch/arm/mm/pgd.c) uses:
pgd_t *pgd_alloc(struct mm_struct *mm)
{
...
init_pgd = pgd_offset_k(0);
memcpy(new_pgd + USER_PTRS_PER_PGD, init_pgd + USER_PTRS_PER_PGD,
(PTRS_PER_PGD - USER_PTRS_PER_PGD) * sizeof(pgd_t));
...
}
or
x86 (arch/x86/mm/pgtable.c) pgd_alloc() calls pgd_ctor():
static void pgd_ctor(struct mm_struct *mm, pgd_t *pgd)
{
/* If the pgd points to a shared pagetable level (either the
ptes in non-PAE, or shared PMD in PAE), then just copy the
references from swapper_pg_dir. */
...
clone_pgd_range(pgd + KERNEL_PGD_BOUNDARY,
swapper_pg_dir + KERNEL_PGD_BOUNDARY,
KERNEL_PGD_PTRS);
...
}
So, back to the original questions:
Part 1: Are kernel virtual addresses really translated by the TLB/MMU?
Yes.
Part 2: How is swapper_pg_dir "attached" to a user mode process.
All page tables (whether swapper_pg_dir or those for user processes)
have the same mappings for the portion used for kernel virtual
addresses. So as the kernel context switches between user processes,
changing the current page table, the mappings for the kernel portion
of the address space remain the same.
The kernel address space is mapped to a section of each process for example on 3:1 mapping after address 0xC0000000. If the user code try to access this address space it will generate a page fault and it is guarded by the kernel.
The kernel address space is divided into 2 parts, the logical address space and the virtual address space. It is defined by the constant VMALLOC_START. The CPU is using the MMU all the time, in user space and in kernel space (can't switch on/off).
The kernel virtual address space is mapped the same way as user space mapping. The logical address space is continuous and it is simple to translate it to physical so it can be done on demand using the MMU fault exception. That is the kernel is trying to access an address, the MMU generate fault , the fault handler map the page using macros __pa , __va and change the CPU pc register back to the previous instruction before the fault happened, now everything is ok. This process is actually platform dependent and in some hardware architectures it mapped the same way as user (because the kernel doesn't use a lot of memory).
Related
what is kernel mapping? What are permanent mapping and temporary mapping. What is a window in this context? I went through code and explanation of this but could not understand this
I'm assuming you're talking about memory mapping in linux kernel.
Memory mapping is a process of mapping kernel address space directly to users process's address space.
Types of addresses :
User virtual address : These are the regular addresses seen by user-space programs
Physical addresses : The addresses used between the processor and the system’s memory.
Bus addresses : The addresses used between peripheral buses and memory. Often, they are the same as the physical addresses used by the processor, but that is not necessarily the case.
Kernel logical addresses : These make up the normal address space of the kernel.
Kernel virtual addresses : Kernel virtual addresses are similar to logical addresses in that they are a mapping from a kernel-space address to a physical address.
High and Low Memory :
Low memory : Memory for which logical addresses exist in kernel space. On almost every system you will likely encounter, all memory is low memory.
High memory : Memory for which logical addresses do not exist, because it is beyond the address range set aside for kernel virtual addresses.This means the kernel needs to start using temporary mappings of the pieces of physical memory that it wants to access.
Kernel splits virtual address into two part user address space and kernel address space. The kernel’s code and data structures must fit into that space, but the biggest consumer of kernel address space is virtual mappings for physical memory. Thus kernel needs its own virtual address for any memory it must touch directly. So, the maximum amount of physical memory that could be handled by the kernel was the amount that could be mapped into the kernel’s portion of the virtual address space, minus the space used by kernel code.
Temporary mapping : When a mapping must be created but the current context cannot sleep, the kernel provides temporary mappings (also called atomic mappings). The kernel can atomically map a high memory page into one of the reserved mappings (which can hold temporary mappings). Consequently, a temporary mapping can be used in places that cannot sleep, such as interrupt handlers, because obtaining the mapping never blocks.
Ref :
kernel.org/doc/Documentation/vm/highmem.txt
static.lwn.net/images/pdf/LDD3/ch15.pdf
man mmap
notes.shichao.io/lkd/ch12/
A full answer would be very long, for details refers (for example) to Linux Kernel Addressing or Understanding the Linux Kernel (pages 306-). These concepts are related to the way address spaces are organized in Linux. Firstly how kernel space is mapped into user space (kernel mapped onto user space simplifies the switching in between user and kernel mode) and, secondly the way physical memory is mapped onto kernel space (because kernel have to manage physical memory).
Beware that this is of no concern in modern 64bit architectures.
I'm having some doubts about how kernel physical and logical addresses are handled by the MMU. I'll try to explain my question doing an example.
Let's doing the assumption that we are on an ARM architecture.
The system starts with the MMU off, so all the addresses which pass inside the CPU are physical. Before we enable the MMU we make a page table where we say that all our physical addresses are mapped at the virtual addresses physical address + 0xC0000000. After this we turn on the MMU. All of this is clear. But now questions start:
Since we are in a pipeline architecture let's say that the instruction after is a load from the address 0x8000. Now for my knowledge here we should have a page fault since the MMU doesn't find this address anywhere inside the page table, so it's invoked a page fault for handle the situation. But also if we've set the interrupt vector, inside it there's a branch to another physical address, so the MMU doesn't find this address and we fall unavoidably in an endless loop. What am I missing?
You miss that all used virtual addresses should be mapped even if in fact they match the same physical memory regions. Let's elaborate.
Say there is startup code (for primary initialisation and enabling MMU) and other main code (only to work with MMU enabled) with a following layout.
startup - phys: 0x4000-0x7FFFF, virt: 0x4000-0x7FFFF
main - phys: 0x8000-0xBFFF, virt: 0xC0008000-0xC000BFFF
Such a layout is linker job and your job is to feed it with a right script. CPU entry point in this example is supposed to be 0x4000. That pretty much is a hardware specific address.
CPU starts with MMU disabled and 'PC=0x4000', and it continues till say 0x4800 when MMU is enabled.
So before enabling MMU there have to be maps for a 0x4000-0x7FFFF (startup) and 0xC0008000-0xC000BFFF (all other code).
Now MMU is enabled. PC has 0x4804 (assuming 32b CPU). 0x4804 is a virtual address now and it's mapped on 0x4804 physical address. CPU proceeds with 0x4804, 0x4808, 0x480C etc.
At some point you should jump into a main. For ARM that would be something like
ldr r0, =0xC0008000
bx r0
Note, that virtual address of main entry is used. So after branching PC=0xC0008000 which is resolved into 0x8000 in physical memory.
0x4000-0x7FFFF mapping could be removed after that.
I understand that the mmu of the processor uses register cr3 to translate linear addresses into physical ones, provided that cr3 is properly set to the physical address of the page directory. But after the kernel has allocated the page tables, how would it find the physical address of the tables and set cr3 to it?
EDIT: I'm talking about the linux kernel.
I'm going to assume that what's bugging you is this: assuming that (once switching to protected mode) the kernel only ever writes to virtual addresses, then this means that it writes the page tables it creates (e.g. for new processes) into virtual addresses. But since the kernel must put a physical address into cr3, then how can it convert the virtual address of the page tables into a physical one?
The short answer is basically what Margaret said: the page tables are found in kernel address space and the kernel keeps a close track of the virtual->physical mapping there.
To flesh this out a little bit more, Linux differentiates between two types of virtual addresses in the kernel:
Kernel virtual addresses - which can map (conceptually) to any physical address; and
Kernel logical addresses - which are virtual addresses that have a linear mapping to physical addresses
The kernel places the page tables in logical addresses, so you only need to focus on those for this discussion.
Mapping a logical address to its corresponding physical one requires only the subtraction of a constant (see e.g. the __pa macro in the Linux source code).
For example, on x86, physical address 0 corresponds to logical address 0xC0000000, and physical address 0x8000 corresponds to logical address 0xC0008000.
So once the kernel places the page tables in a particular logical address, it can easily calculate which physical address it corresponds to.
For further details, you can read the relevant Linux Device Drivers chapter.
Im currently trying to understand systems programming for Linux and have a hard time understanding how virtual to physical memory mappings work.
What I understand so far is that two processes P1 and P2 can make references to the same virtual adress for example 0xf11001. Now this memory adress is split up into two parts. 0xf11 is the page number and 0x001 is the offset within that page (assuming 4096 page size is used). To find the physical adress the MMU has hardware registeres that maps the pagenumber to a physical adress lets say 0xfff. The last stage is to combine 0xfff with 0x001 to find the physical 0xfff001 adress.
However this understanding makes no sens, the same virtual adresses would still point to the same physical location??? What step is I missing inorder for my understanding to be correct???
You're missing one (crucial) step here. In general, MMU doesn't have hardware registers with mappings, but instead one register (page table base pointer) which points to the physical memory address of the page table (with mappings) for the currently running process (which are unique to every process). On context switch, kernel with change this register's value, so for each running process different mapping will be performed.
Here's nice presentation on this topic: http://www.eecs.harvard.edu/~mdw/course/cs161/notes/vm.pdf
In Linux running on an x86 platform where is the real mode address space mapped to in protected kernel mode? In kernel mode, a thread can access the kernel address space directly. The kernel is in the lower 8MB, The page table is at a certain position, etc (as describe here). But where does the real mode address space go? Can it be accessed directly? For example the BIOS and BIOS addons (See here)?
(My x86-fu is a bit weak. I'll add some tags so that other people can (hopefully) correct me if I'm lying anywhere.)
Physical addresses are the same in real and protected mode. The only difference is in how you get from an address (offset) specified in an instruction to a physical address:
In real mode, the physical address is basically (segment_reg << 4) + offset.
In protected mode, the physical address is translate_via_page_table([segment_reg] + offset).
By [segment_reg] I mean the base address of the segment, looked up in the Global or Local Descriptor Table at the offset in segment_reg. translate_via_page_table() means the address translation done via paging (if enabled).
Looking here, it seems the BIOS ROM appears at physical addresses 0x000F0000-0x000FFFFF. To get at that memory in protected mode with paging, you would have to map it into the virtual address space somewhere by setting up correct page table entries. Assuming 4 KB pages (the usual case), mapping the entire range should require 16 ((0xFFFFF-0xF0000+1)/4096) entries.
To see how the Linux kernel does things, you could look into how e.g. /dev/mem, which allows reading of arbitrary physical addresses, is implemented. The implementation is in drivers/char/mem.c.
The following command (from e.g. this answer) will dump the memory range 0xC0000-0xFFFFF (meaning it includes the video BIOS too, per the memory map linked above):
$ dd if=/dev/mem bs=1k skip=768 count=256 > bios
1024*768 = 0xC0000, and 1024*(768+256) - 1 = 0xFFFFF, which gives the expected physical memory range.
Tracing things a bit, read_mem() in drivers/char/mem.c calls xlate_dev_mem_ptr(), which has an x86-specific implementation in arch/x86/mm/ioremap.c. The ioremap_cache() call in that function seems to be responsible for mapping in the page if needed.
Note that BIOS routines won't work in protected mode by the way. They assume the CPU is running in real mode.
For Linux x86 32 bits, the first 896MB of physical RAM is mapped to a contiguous block of virtual memory starting at virtual address 0xC0000000 to 0xF7FFFFFF. Virtual addresses from 0xF8000000 to 0xFFFFFFFF are assigned dynamically to various parts of the physical memory, so the kernel can have a window of 128MB mapped into any part of physical memory beyond the 896MB limit.
The kernel itself loads at physical address 1MB and up, leaving the first MB free. This first MB is used, for instance, to have DMA buffers that ISA devices needs to be there, because they use the 8237 DMA controller, which can only be mapped to such addresses.
So, reading from virtual memory address 0xC0000000 is actually reading from physical address 0x00000000 (provided the kernel has flagged that page as present)