Does a code that runs on kernel mode uses Physical Addresses or Virtual Addresses?
If it is virtual, which translation tables does it use? are they updated differently with comparison to user-mode tables?
Thanks
The answer is that kernel mode code uses a mixture of virtual, logical, and physical addressing. How much of each is used depends upon the underlying hardware and operating system.
There has to be some physical addressing in the kernel. Otherwise, there would be the chicken and egg problem of how does one set up logical addressing without knowing physical addresses to map to.
The ideal would be for the system to only use physical addresses for system registers.
The logical/virtual address distinction is important as well. It is possible for some parts of the kernel address space to be pageable (virtual memory). Others parts could be nonpageable (logical memory only).
Kernel mode address translation is very hardware dependent. Some systems make user mode page tables pageable themselves. In such systems, there is two levels of address translation (virtual addresses for the page table and virtual addresses using the page table). The underlying hardware defines specific addresses ranges as kernel addresses. Other systems allow the system (kernel) address space to be defined entirely in software. In such systems, kernel and user address translation is pretty much the same.
Related
For 32bit OS. linux kernel occupies the high logical address 3~4GB (and 2~4GB for windows).
I wonder why the designer choose the high address, instead of 0~1GB for linux, and because the kernel is actually in the phiscal low address, the choice of low logical address will lead to a more harmonious logical/physical relationship. why not?
Before PIC (Position-independent code) was popular, there are lots of static linked programs can only be loaded at specified address, likely 0x400000
To be able to compatible with these programs, the kernel must not obtain the address space. So the kernel is located at high 1G address space.
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.
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).
In nutshell, as I understand memory management, processor produces virtual addresses. These addresses are translated to corresponding physical addresses using per-process address table by MMU (with TLBs and page-faults in-between, as and when needed).
My question is does processor always produces Virtual addresses? In terms of Address-spaces(user/kernel), Processor modes (user/kernel) and contexts (process/system) when all times does processor produce physical addresses?
Memory typically knows nothing about virtual addresses or segments, which are CPU concepts, it is just memory, a collection of addressable and readable/writable bits. The processor talks to memory using physical addresses. Many simple processors (especially old ones or for special embedded uses) have no MMUs, virtual addresses or privileged modes. Those that have MMUs and virtual addresses normally start either with those disabled or they at first use fixed mapping because otherwise nothing would be able to work if there's no mapping at all.
So, physical addresses are always in use, while for virtual addresses it depends on the CPU and the software in use.
Processor is unaware about whether it is physical address or virtual address , it is the job of respective MMU to do the translation.
Processor has to place the address on it address bus , so now the path depends whether MMU is enabled or disabled. if MMU is enabled it will follow the path of MMU translation and respective physical address will get placed on address bus and if MMU is disabled the same address generated by respective instruction will be placed on address bus.
so it is responsibility of programmer if MMU is disabled all address access should be physical address otherwise it will be a exception or abort in system
Generally the CPU ONLY knows "virtual addresses". Ie, when you do any assembly programming, any "load regs, *(memory ptr)" or such-like operation, the addresses are virtual addresses.
The following diagram illustrates well the concept:
http://slideplayer.com/slide/4394245/ (page 7)
Any addresses coming out of CPU, is always virtual. But if the processor has a MMU, then the MMU will intercept the addresses and convert it to physical addresses (through Page Table mechanism) before putting it on the memory bus. So if you sniff the memory bus, you will see physical addresses.
References:
https://www.quora.com/What-is-the-return-address-of-kmalloc-Physical-or-Virtual
Yes when a system works on VM then instructions/programs produce only VA even at the time of booting, addresses generated are not the same as physical despite page tables are still not in existence.
But processor need physical addresses to access memory. So there is mechanism to produce physical addresses from virtual addresses which is called address translation done through page tables. Yes whatever you see in kernel mode/user mode, register values all are virtual addresses. but when processor actually perform computation, processor always uses physical addresses
My question is does processor always produces Virtual addresses? In terms of Address-spaces(user/kernel), Processor modes (user/kernel) and contexts (process/system) when all times does processor produce physical addresses?
An x86 CPU operates with 3 different "kinds" of addresses:
physical: the actual addresses that are used to select the byte in memory. Used for things like segment base addresses, interrupt descriptor table, global descriptor table.
logical: these are addresses that you use most of the time when paging is disabled. They're converted to physical addresses by adding the respective segments (physical) base address.
virtual: the addresses used by most instructions when paging is enabled. These are translated into logical addresses using page directories and tables.
Which kind of address is used isn't dependent on the current privilege level (CPL; "system mode", "user mode" or between). It depends on the state of the processor (paging enabled or not) and the actual instruction (lidt for example). Though I guess it's safe to assume that there won't be physical addressing in "user mode" (CPL > 0) since instructions using physical addresses are usually privileged instructions.
In x86 arch, linux kernel 2.6.x, 32bit system
I understand that virtual address 0xC0000000 ~ 0xFFFFFFFF
is reserved for kernel.
and this virtual address can be translated to physical address by
subtracting 0xC0000000.
however, I think even the result is same, MMU will translate
the kernel virtual address(such as 0xC0851000) to physical address by walking through page table.
such as
CR3 -> page directory -> page table -> PFN.
am I correct or wrong?, please correct me if I'm wrong.
I need to develop hardware based kernel monitor in x86, linux 32bit system.
so I need to know this problem
please help.
For kernel logical addresses, you are correct. Kernel virtual addresses, like memory allocated by vmalloc, do not necessarily have a one-to-one mapping to physical addresses that characterize the logical address space, however.
Just bear in mind that kernel logical addresses aren't always translated to physical by subtracting an offset (that's true in x86 but not, say, AVR32).
"and this virtual address can be translated to physical address by
subtracting 0xC0000000"
since page tables for the kernel virtual addresses are configured that way, people have come up with a shortcut you mentioned.