No address space for Linux Kernel threads - multithreading

Why the Linux kernel threads do not have an address space. For any task to execute, it should have a memory region right? Where do the text and data of kernel threads go?

Kernel threads do have an address space. It's just that they all share the same one. This does not prevent them from each having a different stack.
Text and data are laid out in the kernel address space (the one that is shared by all the threads), depending on how and when it was allocated, and what it's used for.
The Linux MM site has a lot of documentation about this aspect of Linux. Head over there.

I don't know the precise answer, because I'm not a Linux architect.
But in general, so-called kernel threads do have an address space: it is the address space which contains the kernel. It might not need to be explicitly represented for each kernel thread, since it is shared among many threads.
I'd expect any real thread implementation to have a machine context block containing register values (and stack pointer, etc.), and a pointer to a the address space in which the thread is supposed to run. Then a scheduler, starting a ready thread, can easily determine whether the memory management unit is set up to enable access to the address space (and if not, set it up) to enable the thread to run in its desired space.

Related

Where are the stacks for the other threads located in a process virtual address space?

The following image shows where the sections of a process are laid out in the process's virtual address space (in Linux):
You can see that there is only one stack section (since this process only has one thread I assume).
But what if this process has another thread, where will the stack for this second thread be located? will it be located immediately below the first stack?
Stack space for a new thread is created by the parent thread with mmap(MAP_ANONYMOUS|MAP_STACK). So they're in the "memory map segment", as your diagram labels it. It can end up anywhere that a large malloc() could go. (glibc malloc(3) uses mmap(MAP_ANONYMOUS) for large allocations.)
(MAP_STACK is currently a no-op, and exists in case some future architecture needs special handling).
You pass a pointer to the new thread's stack space to the clone(2) system call which actually creates the thread. (Try using strace -f on a multi-threaded process sometime). See also this blog post about creating a thread using raw Linux syscalls.
See this answer on a related question for some more details about mmaping stacks. e.g. MAP_GROWSDOWN doesn't prevent another mmap() from picking the address right below the thread stack, so you can't depend on it to dynamically grow a small stack the way you can for the main thread's stack (where the kernel reserves the address space even though it's not mapped yet).
So even though mmap(MAP_GROWSDOWN) was designed for allocating stacks, it's so bad that Ulrich Drepper proposed removing it in 2.6.29.
Also, note that your memory-map diagram is for a 32-bit kernel. A 64-bit kernel doesn't have to reserve any user virtual-address space for mapping kernel memory, so a 32-bit process running on an amd64 kernel can use the full 4GB of virtual address space. (Except for the low 64k by default (sysctl vm.mmap_min_addr = 65536), so NULL-pointer dereference does actually fault. And the top page is also reserved as error codes, not valid pointers.)
Related:
See Relation between stack limit and threads for more about stack-size for pthreads. getrlimit(RLIMIT_STACK) is the main thread's stack size. Linux pthreads uses RLIMIT_STACK as the stack size for new threads, too.

why kernel needs virtual addressing?

In Linux each process has its virtual address space (e.g. 4 GB in case of 32 bit system, wherein 3GB is reserved for process and 1 GB for kernel). This virtual addressing mechanism helps isolating the address space of each process. This is understandable in case of process since there are many processes. But since we have 1 kernel only so why do we need virtual addressing for kernel?
The reason the kernel is "virtual" is not to deal with paging as such, it is becuase the processor can only run in one mode at a time. So once you turn on paged memory mapping (Bit 31 in CR0 on x86), the processor is expecting ALL memory accesses to go through the page-mapping mechanism. So, since we do want to access the kernel even after we have enabled paging (virtual memory), it needs to exist somewhere in the virtual space.
The "reserving" of memory is more about "easy way to determine if an address is kernel or user-space" than anything else. It would be perfectly possible to put a little bit of kernel at address 12345-34121, another bit of kernel at 101900-102400 and some other bit of kernel at 40000000-40001000. But it would make life difficult for every aspect of the kernel and userspace - there would be gaps/holes to deal with [there already are such holes/gapes, but having more wouldn't exactly help things]. By setting a fixed limit for "userspace is from here to here, kernel is from end of userspace to X", it makes life much easier in that respect. We can just say kernel = 0; if (address > max_userspace) kernel=1; in some code.
Of course, the kerneln only takes up as much PHYSICAL memory as it will actually use - so the common thinking that "it's a waste to take up a whole gigabyte for the kernel" is wrong - the kernel itself is a few (a dozen or so for a very "big" kernel) megabytes. The modules loaded can easily add up to several more megabytes, and graphics drivers from ATI and nVidia easily another few megabytes just for the kernel moduel for that itself. The kernel also uses some bits of memory to store "kernel data", such as tasks, queues, semaphores, files and other "stuff" the kernel has to deal with. A few megabytes is used for this as well.
Virtual Memory Management is that feature of Linux which enables Multi-tasking in system without any limitation on no. of task or amount of memory used by each task. The Linux Memory Manager Subsystem (along with MMU hardware) facilitates VMM support, where memory or mem-mapped device are accessed through virtual addresses. Within Linux everything, both kernel and user components, works with virtual address except when dealing with real hardware. That's when the Memory Manager takes its place, does virtual-to-physical address translation and points to physical mem/dev location.
A process is an abstract entity, defined by kernel to which system resources are allocated in order to execute a program. In Linux Process Management the kernel is an integrated part of a process memory map. A process has two main regions, like two faces of one coin:
User Space view - contains user program sections (Code, Data, Stack, Heap, etc...) used by process
Kernel Space view - contains kernel data structures that maintain information (PID. States, FD, Resource Usage, etc...) about the process
Every process in Linux system has a unique and separate User Space Region. This feature of Linux VMM isolates each process program sections from one and other. But all processes in the system shares the common Kernel Space Region. When a process needs service from the kernel it must execute the kernel code in this region, or in other words kernel is performing on behalf of user process request.

Direct access user memory from kernel on Linux

I've got a user-mode process and kernel module. Now I want to read certain regions of usermode process from kernel, but there's one catch: no copying of usermode memory and simple access by VA.
So what we have: task_struct for target process, other related structs (like mm_struct, vma_struct) and virtual address like 0x0070abcd that I want to read or rather map somehow to my kernel module.
I can get page list using get_user_pages for desired memory regions, but what next? Should I map pages somehow into kernel and then try to read them as continuous memory region or there are better solutions?
The problem is that "looking" at user-space requires locking a ton of stuff. So it's better that you do a short copy than leave everything locked for arbitrary amounts of time. Your user-space process may not be VM-mapped into the current CPU. In fact, it may be entirely swapped out to disk, running on another CPU, in the middle of it's own kernel call, etc.
Linux Kernel: copy_from_user - struct with pointers

Running two processes in Unix/Linux

When the kernel creates two processes whose code section is same, does the kernel actually copy the code to the virtual address space of both processes? In other words, if I create two processes of the same program, in memory, do we have two copies of the program or just one copy?
Obviously, it may depend on implementation but I'm asking in traditional Unix OS.
Does the kernel actually copy the code to the virtual address space of both processes?
The text segment will be mapped (rather than copied) into the virtual address space of each process, but will be referring to the same physical space (so the kernel will only have one copy of the text in memory).
The data and bss segments will also be mapped into the virtual address space of each process, but these will be created per process. At process initiation, the data from the data and bss segments from the executable will be mapped/copied into the process's virtual memory; if it was not copied ab initio then as soon as the processes start writing to the data the process will be given its own private copy.
Clearly, shared memory and mmap'd memory are handled after the process starts. Shared memory is always shared between processes; that's its raison d'ĂȘtre. What happens with mmap depends on the flags used, but it is often shared too.
Modern operating systems will use Copy-on-Write to avoid duplicating pages until they are actually updated. Note that on many systems (including Linux) this can lead to overcommit, where the OS doesn't actually have enough RAM to cope with all the copying required should every process decide to modify un-duplicated pages.

For arm Linux, could threads in user space access virtual address of Kernel space?

Virtual memory is split two parts. In tradition, 0~3GB is for user space and 3GB~4GB for kernel space.
My question:
Could the thread in user space access memory of kernel space?
For ARM datasheet, the access attribution is in the charge of domain access control register. But in kernel source code,the domain value in page table entry of user space virtual memory is same as kernel space's page table entry.
In fact, your application might access page 0xFFFF0000, as it contains the swi-handler and a couple of other userspace-helpers. So no, the 3/1 split is nothing magical, it's just very easy for the kernel to manage.
Usually the kernel will setup all memory above 3GB to be only accessible by the kernel-domain itself. If a driver needs to share memory between user and kernel-space it will usually provide an mmap interface, which then creates an aliased mapping, so you have two virtual addresses for the same physical address. This only works reliably on VIPT-Cache systems or with a LOT of careful explicit cache flushing. If you don't want this you CAN hack the kernel to make a chunk of memory ABOVE the 3G-split accessible to userspace. But then all userspace applications will share this memory. I've done this once for a special application on a armv5-system.
Userspace code getting Kernel memory? The only kernel that ever allowed that was DOS and its archaic friends.
But back to the question, look at this example C code:
char c=42;
*c=42;
We take one byte (a char) and assign it the numeric value 42. We then dereference this non-pointer, which will probably try to access the 42nd byte of virtual memory, which is almost definitely not your memory, and, for the sake of this example, Kernel memory. guess what happens when you run this (if you manage to hold the compiler at gunpoint):
Segmentation fault
Linux has memory protection like any modern operating system. If you try to access the memory of another process, your process will be terminated before it can do anything (other things I'm not so sure about happen with debuggers though). Even if that memory was that of another Userland process, you would still get terminated. I'm almost sure that root programs can't access other programs memory, or Kernel memory. The only way to access Kernel memory is to be part of the Kernel, or indirectly through the kernel's cooperation.

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