I was reading this on a page that:
Because of hardware limitations, the kernel cannot treat all pages as identical. Some pages, because of their physical address in memory, cannot be used for certain tasks. Because of this limitation, the kernel divides pages into different zones.
I was wondering about those hardware limitation. Can somebody please explain me those hardware limitation and give an example. As well, is there any software guide from intel explaining this?
Also, I read that virtual memory is divided into two parts 1GB for kernel space and 3GB for user space. Why do we give 1GB space in the virtual space of all processes to kernel? How is it mapped to actual physical pages? Can somebody please point me to a clean text explaining this?
Thanks in advance.
The hardware limitations mostly concern old devies. For example, you have the ZONE_DMA, which is from 0 - 16MB. This is e.g. needed for older ISA Devices, which are not capable of adressing above the 16MB limit. Then you have the ZONE_NORMAL, where most of the kernel operations take place and is adressed permanently into the kernels adress space.
The 1GB and 3GB split is simple. You have virtual adresses here, so for your application, the memory adress always starts at 0x00000000, reserved are the 1st GB of this for kernel stuff. Why this is done is pretty simple: You have the kernel mode and the user mode. In kernel mode you are allowed to use system calls. If you would not have the kernel memory mapped to your virtual adress space, you would have to do a context switch to trap you into kernel mode (context switch: Save current context to memory, load another context from memory -> time consuming). But as kernel-mode operations can take place in the same virtual adress space, you dont need to switch the context to, for example, allocate new memory or do any other system call.
for your second question about 1GB kernel mapping in user space for a processor
kernel is mapped of course for time saving by not having switch. 1 GB is for kernel functionality so that if kernel maps new memory for its functionality kernel can do that. any book on Unix can give you details
What is the difference between the kernel space and the user space? Do kernel space, kernel threads, kernel processes and kernel stack mean the same thing? Also, why do we need this differentiation?
The really simplified answer is that the kernel runs in kernel space, and normal programs run in user space. User space is basically a form of sand-boxing -- it restricts user programs so they can't mess with memory (and other resources) owned by other programs or by the OS kernel. This limits (but usually doesn't entirely eliminate) their ability to do bad things like crashing the machine.
The kernel is the core of the operating system. It normally has full access to all memory and machine hardware (and everything else on the machine). To keep the machine as stable as possible, you normally want only the most trusted, well-tested code to run in kernel mode/kernel space.
The stack is just another part of memory, so naturally it's segregated right along with the rest of memory.
The Random Access Memory (RAM) can be logically divided into two distinct regions namely - the kernel space and the user space.(The Physical Addresses of the RAM are not actually divided only the Virtual Addresses, all this implemented by the MMU)
The kernel runs in the part of memory entitled to it. This part of memory cannot be accessed directly by the processes of the normal users, while the kernel can access all parts of the memory. To access some part of the kernel, the user processes have to use the predefined system calls i.e. open, read, write etc. Also, the C library functions like printf call the system call write in turn.
The system calls act as an interface between the user processes and the kernel processes. The access rights are placed on the kernel space in order to stop the users from messing with the kernel unknowingly.
So, when a system call occurs, a software interrupt is sent to the kernel. The CPU may hand over the control temporarily to the associated interrupt handler routine. The kernel process which was halted by the interrupt resumes after the interrupt handler routine finishes its job.
CPU rings are the most clear distinction
In x86 protected mode, the CPU is always in one of 4 rings. The Linux kernel only uses 0 and 3:
0 for kernel
3 for users
This is the most hard and fast definition of kernel vs userland.
Why Linux does not use rings 1 and 2: CPU Privilege Rings: Why rings 1 and 2 aren't used?
How is the current ring determined?
The current ring is selected by a combination of:
global descriptor table: a in-memory table of GDT entries, and each entry has a field Privl which encodes the ring.
The LGDT instruction sets the address to the current descriptor table.
See also: http://wiki.osdev.org/Global_Descriptor_Table
the segment registers CS, DS, etc., which point to the index of an entry in the GDT.
For example, CS = 0 means the first entry of the GDT is currently active for the executing code.
What can each ring do?
The CPU chip is physically built so that:
ring 0 can do anything
ring 3 cannot run several instructions and write to several registers, most notably:
cannot change its own ring! Otherwise, it could set itself to ring 0 and rings would be useless.
In other words, cannot modify the current segment descriptor, which determines the current ring.
cannot modify the page tables: How does x86 paging work?
In other words, cannot modify the CR3 register, and paging itself prevents modification of the page tables.
This prevents one process from seeing the memory of other processes for security / ease of programming reasons.
cannot register interrupt handlers. Those are configured by writing to memory locations, which is also prevented by paging.
Handlers run in ring 0, and would break the security model.
In other words, cannot use the LGDT and LIDT instructions.
cannot do IO instructions like in and out, and thus have arbitrary hardware accesses.
Otherwise, for example, file permissions would be useless if any program could directly read from disk.
More precisely thanks to Michael Petch: it is actually possible for the OS to allow IO instructions on ring 3, this is actually controlled by the Task state segment.
What is not possible is for ring 3 to give itself permission to do so if it didn't have it in the first place.
Linux always disallows it. See also: Why doesn't Linux use the hardware context switch via the TSS?
How do programs and operating systems transition between rings?
when the CPU is turned on, it starts running the initial program in ring 0 (well kind of, but it is a good approximation). You can think this initial program as being the kernel (but it is normally a bootloader that then calls the kernel still in ring 0).
when a userland process wants the kernel to do something for it like write to a file, it uses an instruction that generates an interrupt such as int 0x80 or syscall to signal the kernel. x86-64 Linux syscall hello world example:
.data
hello_world:
.ascii "hello world\n"
hello_world_len = . - hello_world
.text
.global _start
_start:
/* write */
mov $1, %rax
mov $1, %rdi
mov $hello_world, %rsi
mov $hello_world_len, %rdx
syscall
/* exit */
mov $60, %rax
mov $0, %rdi
syscall
compile and run:
as -o hello_world.o hello_world.S
ld -o hello_world.out hello_world.o
./hello_world.out
GitHub upstream.
When this happens, the CPU calls an interrupt callback handler which the kernel registered at boot time. Here is a concrete baremetal example that registers a handler and uses it.
This handler runs in ring 0, which decides if the kernel will allow this action, do the action, and restart the userland program in ring 3. x86_64
when the exec system call is used (or when the kernel will start /init), the kernel prepares the registers and memory of the new userland process, then it jumps to the entry point and switches the CPU to ring 3
If the program tries to do something naughty like write to a forbidden register or memory address (because of paging), the CPU also calls some kernel callback handler in ring 0.
But since the userland was naughty, the kernel might kill the process this time, or give it a warning with a signal.
When the kernel boots, it setups a hardware clock with some fixed frequency, which generates interrupts periodically.
This hardware clock generates interrupts that run ring 0, and allow it to schedule which userland processes to wake up.
This way, scheduling can happen even if the processes are not making any system calls.
What is the point of having multiple rings?
There are two major advantages of separating kernel and userland:
it is easier to make programs as you are more certain one won't interfere with the other. E.g., one userland process does not have to worry about overwriting the memory of another program because of paging, nor about putting hardware in an invalid state for another process.
it is more secure. E.g. file permissions and memory separation could prevent a hacking app from reading your bank data. This supposes, of course, that you trust the kernel.
How to play around with it?
I've created a bare metal setup that should be a good way to manipulate rings directly: https://github.com/cirosantilli/x86-bare-metal-examples
I didn't have the patience to make a userland example unfortunately, but I did go as far as paging setup, so userland should be feasible. I'd love to see a pull request.
Alternatively, Linux kernel modules run in ring 0, so you can use them to try out privileged operations, e.g. read the control registers: How to access the control registers cr0,cr2,cr3 from a program? Getting segmentation fault
Here is a convenient QEMU + Buildroot setup to try it out without killing your host.
The downside of kernel modules is that other kthreads are running and could interfere with your experiments. But in theory you can take over all interrupt handlers with your kernel module and own the system, that would be an interesting project actually.
Negative rings
While negative rings are not actually referenced in the Intel manual, there are actually CPU modes which have further capabilities than ring 0 itself, and so are a good fit for the "negative ring" name.
One example is the hypervisor mode used in virtualization.
For further details see:
https://security.stackexchange.com/questions/129098/what-is-protection-ring-1
https://security.stackexchange.com/questions/216527/ring-3-exploits-and-existence-of-other-rings
ARM
In ARM, the rings are called Exception Levels instead, but the main ideas remain the same.
There exist 4 exception levels in ARMv8, commonly used as:
EL0: userland
EL1: kernel ("supervisor" in ARM terminology).
Entered with the svc instruction (SuperVisor Call), previously known as swi before unified assembly, which is the instruction used to make Linux system calls. Hello world ARMv8 example:
hello.S
.text
.global _start
_start:
/* write */
mov x0, 1
ldr x1, =msg
ldr x2, =len
mov x8, 64
svc 0
/* exit */
mov x0, 0
mov x8, 93
svc 0
msg:
.ascii "hello syscall v8\n"
len = . - msg
GitHub upstream.
Test it out with QEMU on Ubuntu 16.04:
sudo apt-get install qemu-user gcc-arm-linux-gnueabihf
arm-linux-gnueabihf-as -o hello.o hello.S
arm-linux-gnueabihf-ld -o hello hello.o
qemu-arm hello
Here is a concrete baremetal example that registers an SVC handler and does an SVC call.
EL2: hypervisors, for example Xen.
Entered with the hvc instruction (HyperVisor Call).
A hypervisor is to an OS, what an OS is to userland.
For example, Xen allows you to run multiple OSes such as Linux or Windows on the same system at the same time, and it isolates the OSes from one another for security and ease of debug, just like Linux does for userland programs.
Hypervisors are a key part of today's cloud infrastructure: they allow multiple servers to run on a single hardware, keeping hardware usage always close to 100% and saving a lot of money.
AWS for example used Xen until 2017 when its move to KVM made the news.
EL3: yet another level. TODO example.
Entered with the smc instruction (Secure Mode Call)
The ARMv8 Architecture Reference Model DDI 0487C.a - Chapter D1 - The AArch64 System Level Programmer's Model - Figure D1-1 illustrates this beautifully:
The ARM situation changed a bit with the advent of ARMv8.1 Virtualization Host Extensions (VHE). This extension allows the kernel to run in EL2 efficiently:
VHE was created because in-Linux-kernel virtualization solutions such as KVM have gained ground over Xen (see e.g. AWS' move to KVM mentioned above), because most clients only need Linux VMs, and as you can imagine, being all in a single project, KVM is simpler and potentially more efficient than Xen. So now the host Linux kernel acts as the hypervisor in those cases.
Note how ARM, maybe due to the benefit of hindsight, has a better naming convention for the privilege levels than x86, without the need for negative levels: 0 being the lower and 3 highest. Higher levels tend to be created more often than lower ones.
The current EL can be queried with the MRS instruction: what is the current execution mode/exception level, etc?
ARM does not require all exception levels to be present to allow for implementations that don't need the feature to save chip area. ARMv8 "Exception levels" says:
An implementation might not include all of the Exception levels. All implementations must include EL0 and EL1.
EL2 and EL3 are optional.
QEMU for example defaults to EL1, but EL2 and EL3 can be enabled with command line options: qemu-system-aarch64 entering el1 when emulating a53 power up
Code snippets tested on Ubuntu 18.10.
Kernel space & virtual space are concepts of virtual memory....it doesn't mean Ram(your actual memory) is divided into kernel & User space.
Each process is given virtual memory which is divided into kernel & user space.
So saying
"The random access memory (RAM) can be divided into two distinct regions namely - the kernel space and the user space." is wrong.
& regarding "kernel space vs user space" thing
When a process is created and its virtual memory is divided into user-space and a kernel-space , where user space region contains data, code, stack, heap of the process & kernel-space contains things such as the page table for the process, kernel data structures and kernel code etc.
To run kernel space code, control must shift to kernel mode(using 0x80 software interrupt for system calls) & kernel stack is basically shared among all processes currently executing in kernel space.
Kernel space and user space is the separation of the privileged operating system functions and the restricted user applications. The separation is necessary to prevent user applications from ransacking your computer. It would be a bad thing if any old user program could start writing random data to your hard drive or read memory from another user program's memory space.
User space programs cannot access system resources directly so access is handled on the program's behalf by the operating system kernel. The user space programs typically make such requests of the operating system through system calls.
Kernel threads, processes, stack do not mean the same thing. They are analogous constructs for kernel space as their counterparts in user space.
Each process has its own 4GB of virtual memory which maps to the physical memory through page tables. The virtual memory is mostly split in two parts: 3 GB for the use of the process and 1 GB for the use of the Kernel. Most of the variables you create lie in the first part of the address space. That part is called user space. The last part is where the kernel resides and is common for all the processes. This is called Kernel space and most of this space is mapped to the starting locations of physical memory where the kernel image is loaded at boot time.
The maximum size of address space depends on the length of the address register on the CPU.
On systems with 32-bit address registers, the maximum size of address space is 232 bytes, or 4 GiB.
Similarly, on 64-bit systems, 264 bytes can be addressed.
Such address space is called virtual memory or virtual address space. It is not actually related to physical RAM size.
On Linux platforms, virtual address space is divided into kernel space and user space.
An architecture-specific constant called task size limit, or TASK_SIZE, marks the position where the split occurs:
the address range from 0 up to TASK_SIZE-1 is allotted to user space;
the remainder from TASK_SIZE up to 232-1 (or 264-1) is allotted to kernel space.
On a particular 32-bit system for example, 3 GiB could be occupied for user space and 1 GiB for kernel space.
Each application/program in a Unix-like operating system is a process; each of those has a unique identifier called Process Identifier (or simply Process ID, i.e. PID). Linux provides two mechanisms for creating a process: 1. the fork() system call, or 2. the exec() call.
A kernel thread is a lightweight process and also a program under execution.
A single process may consist of several threads sharing the same data and resources but taking different paths through the program code. Linux provides a clone() system call to generate threads.
Example uses of kernel threads are: data synchronization of RAM, helping the scheduler to distribute processes among CPUs, etc.
Briefly : Kernel runs in Kernel Space, the kernel space has full access to all memory and resources, you can say the memory divide into two parts, part for kernel , and part for user own process, (user space) runs normal programs, user space cannot access directly to kernel space so it request from kernel to use resources. by syscall (predefined system call in glibc)
there is a statement that simplify the different "User Space is Just a test load for the Kernel " ...
To be very clear : processor architecture allow CPU to operate in two mode, Kernel Mode and User Mode, the Hardware instruction allow switching from one mode to the other.
memory can be marked as being part of user space or kernel space.
When CPU running in User Mode, the CPU can access only memory that is being in user space, while cpu attempts to access memory in Kernel space the result is a "hardware exception", when CPU running in Kernel mode, the CPU can access directly to both kernel space and user space ...
The kernel space means a memory space can only be touched by kernel. On 32bit linux it is 1G(from 0xC0000000 to 0xffffffff as virtual memory address).Every process created by kernel is also a kernel thread, So for one process, there are two stacks: one stack in user space for this process and another in kernel space for kernel thread.
the kernel stack occupied 2 pages(8k in 32bit linux), include a task_struct(about 1k) and the real stack(about 7k). The latter is used to store some auto variables or function call params or function address in kernel functions. Here is the code(Processor.h (linux\include\asm-i386)):
#define THREAD_SIZE (2*PAGE_SIZE)
#define alloc_task_struct() ((struct task_struct *) __get_free_pages(GFP_KERNEL,1))
#define free_task_struct(p) free_pages((unsigned long) (p), 1)
__get_free_pages(GFP_KERNEL,1)) means alloc memory as 2^1=2 pages.
But the process stack is another thing, its address is just bellow 0xC0000000(32bit linux), the size of it can be quite bigger, used for the user space function calls.
So here is a question come for system call, it is running in kernel space but was called by process in user space, how does it work? Will linux put its params and function address in kernel stack or process stack? Linux's solution: all system call are triggered by software interruption INT 0x80.
Defined in entry.S (linux\arch\i386\kernel), here is some lines for example:
ENTRY(sys_call_table)
.long SYMBOL_NAME(sys_ni_syscall) /* 0 - old "setup()" system call*/
.long SYMBOL_NAME(sys_exit)
.long SYMBOL_NAME(sys_fork)
.long SYMBOL_NAME(sys_read)
.long SYMBOL_NAME(sys_write)
.long SYMBOL_NAME(sys_open) /* 5 */
.long SYMBOL_NAME(sys_close)
By Sunil Yadav, on Quora:
The Linux Kernel refers to everything that runs in Kernel mode and is
made up of several distinct layers. At the lowest layer, the Kernel
interacts with the hardware via the HAL. At the middle level, the
UNIX Kernel is divided into 4 distinct areas. The first of the four
areas handles character devices, raw and cooked TTY and terminal
handling. The second area handles network device drivers, routing
protocols and sockets. The third area handles disk device drivers,
page and buffer caches, file system, virtual memory, file naming and
mapping. The fourth and last area handles process dispatching,
scheduling, creation and termination as well as signal handling.
Above all this we have the top layer of the Kernel which includes
system calls, interrupts and traps. This level serves as the
interface to each of the lower level functions. A programmer uses
the various system calls and interrupts to interact with the features
of the operating system.
IN short kernel space is the portion of memory where linux kernel runs (top 1 GB virtual space in case of linux) and user space is the portion of memory where user application runs( bottom 3 GB of virtual memory in case of Linux. If you wanna know more the see the link given below :)
http://learnlinuxconcepts.blogspot.in/2014/02/kernel-space-and-user-space.html
Kernel Space and User Space are logical spaces.
Most of the modern processors are designed to run in different privileged mode. x86 machines can run in 4 different privileged modes.
And a particular machine instruction can be executed when in/above particular privileged mode.
Because of this design you are giving a system protection or sand-boxing the execution environment.
Kernel is a piece of code, which manages your hardware and provide system abstraction. So it needs to have access for all the machine instruction. And it is most trusted piece of software. So i should be executed with the highest privilege. And Ring level 0 is the most privileged mode. So Ring Level 0 is also called as Kernel Mode.
User Application are piece of software which comes from any third party vendor, and you can't completely trust them. Someone with malicious intent can write a code to crash your system if he had complete access to all the machine instruction. So application should be provided with access to limited set of instructions. And Ring Level 3 is the least privileged mode. So all your application run in that mode. Hence that Ring Level 3 is also called User Mode.
Note: I am not getting Ring Levels 1 and 2. They are basically modes with intermediate privilege. So may be device driver code are executed with this privilege. AFAIK, linux uses only Ring Level 0 and 3 for kernel code execution and user application respectively.
So any operation happening in kernel mode can be considered as kernel space.
And any operation happening in user mode can be considered as user space.
Trying to give a very simplified explanation
Virtual Memory is divided into kernel space and the user space.
Kernel space is that area of virtual memory where kernel processes will run and user space is that area of virtual memory where user processes will be running.
This division is required for memory access protections.
Whenever a bootloader starts a kernel after loading it to a location in RAM, (on an ARM based controller typically)it needs to make sure that the controller is in supervisor mode with FIQ's and IRQ's disabled.
The correct answer is: There is no such thing as kernel space and user space. The processor instruction set has special permissions to set destructive things like the root of the page table map, or access hardware device memory, etc.
Kernel code has the highest level privileges, and user code the lowest. This prevents user code from crashing the system, modifying other programs, etc.
Generally kernel code is kept under a different memory map than user code (just as user spaces are kept in different memory maps than each other). This is where the "kernel space" and "user space" terms come from. But that is not a hard and fast rule. For example, since the x86 indirectly requires its interrupt/trap handlers to be mapped at all times, part (or some OSes all) of the kernel must be mapped into user space. Again, this does not mean that such code has user privileges.
Why is the kernel/user divide necessary? Some designers disagree that it is, in fact, necessary. Microkernel architecture is based on the idea that the highest privileged sections of code should be as small as possible, with all significant operations done in user privileged code. You would need to study why this might be a good idea, it is not a simple concept (and is famous for both having advantages and drawbacks).
This demarcation need architecture support there are some instructions that are accessed in privileged mode.
In pagetables we have access details if user process try to access address which lies in kernel address range then it will give privilege violation fault.
So to enter privileged mode it is required to run instruction like trap which change CPU mode to privilege and give access to instructions as well as memory regions
In Linux there are two space 1st is user space and another one is kernal space. user space consist of only user application which u want to run. as the kernal service there is process management, file management, signal handling, memory management, thread management, and so many services are present there. if u run the application from the user space that appliction interact with only kernal service. and that service is interact with device driver which is present between hardware and kernal.
the main benefit of kernal space and user space seperation is we can acchive a security by the virus.bcaz of all user application present in user space, and service is present in kernal space. thats why linux doesn,t affect from the virus.
I read that the first 3 GBs are reserved for the process and the last GB is for the Kernel. I also read that the kernel is loaded starting from the 2nd MB of the physical address space (depending on the configuration). My question is that is the mapping of that last 1 GB is same for all processes and maps to this physical area of memory?
Another question is, when a process switches to kernel mode (eg, when a sys call occurs), then what page tables are used, the process page tables or the kernel page tables? If kernel page tables are used, then they can't access the memory locations belonging to the process. If that is the case, then there is apparently no use for the kernel virtual memory since all access to kernel code and data will be through the mapping of the last 1 GB of process address space. Please help me clarify this (any useful links will be much appreciated)
It seems, you are talking about 32-bit x86 systems, right?
If I am not mistaken, the kernel can be configured not only for 3Gb/1Gb memory distrubution, there could be other variants (e.g. 2Gb/2Gb). Still, 3Gb/1Gb is probably the most common one on x86-32.
The kernel part of the address space should be inaccessible from the user space. From the kernel's point of view, yes, the mapping of the memory occupied by the kernel itself is always the same. No matter, in the context of which process (or interrupt handler, or whatever else) the kernel currently operates.
As one of the consequences, if you look at the addresses of kernel symbols in /proc/kallsyms from different processes, you will see the same addresses each time. And these are exactly the addresses of the respective kernel functions, variables and others from the kernel's point of view.
So I suppose, the answer to your first question is "yes" but it is probably not very useful for the user-space code as the kernel space memory is not directly accessible from there anyway.
As for the second question, well, if the kernel currently operates in the context of some process, it can actually access the user-space memory of that process. I can't describe it in detail but probably the implementation of kernel functions copy_from_user and copy_to_user could give you some hints. See arch/x86/lib/usercopy_32.c and arch/x86/include/asm/uaccess.h in the kernel sources. It seems, on x86-32, the user-space memory is accessed in these functions directly, using the default memory mappings for the current process context. The 'magic' stuff there is only related to the optimizations and checking the address of the memory area for correctness.
Yes, the mapping of the kernel part of the address space is the same in all processes. Part of it does map that part of the physical memory where the kernel image is loaded, but that's not the bulk of it - the remainder is used to map other physical memory locations for the kernel's runtime working set.
When a process switches to kernel mode, the page tables are not changed. The kernel part of the address space simply becomes accessible because the CPL (Current Privilege Level) is now zero.