What is the difference between user and kernel modes in operating systems?

Operating System

Operating System Problem Overview

What are the differences between User Mode and Kernel Mode, why and how do you activate either of them, and what are their use cases?

Operating System Solutions

Solution 1 - Operating System

> 1. Kernel Mode > > In Kernel mode, the executing code has complete and unrestricted > access to the underlying hardware. It > can execute any CPU instruction and > reference any memory address. Kernel > mode is generally reserved for the > lowest-level, most trusted functions > of the operating system. Crashes in > kernel mode are catastrophic; they > will halt the entire PC. > > 2. User Mode > > In User mode, the executing code has no ability to directly access > hardware or reference memory. Code > running in user mode must delegate to > system APIs to access hardware or > memory. Due to the protection afforded > by this sort of isolation, crashes in > user mode are always recoverable. Most > of the code running on your computer > will execute in user mode.

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Understanding User and Kernel Mode

Solution 2 - Operating System

These are two different modes in which your computer can operate. Prior to this, when computers were like a big room, if something crashes – it halts the whole computer. So computer architects decide to change it. Modern microprocessors implement in hardware at least 2 different states.

User mode:

  • mode where all user programs execute. It does not have access to RAM and hardware. The reason for this is because if all programs ran in kernel mode, they would be able to overwrite each other’s memory. If it needs to access any of these features – it makes a call to the underlying API. Each process started by windows except of system process runs in user mode.

Kernel mode:

  • mode where all kernel programs execute (different drivers). It has access to every resource and underlying hardware. Any CPU instruction can be executed and every memory address can be accessed. This mode is reserved for drivers which operate on the lowest level

How the switch occurs.

The switch from user mode to kernel mode is not done automatically by CPU. CPU is interrupted by interrupts (timers, keyboard, I/O). When interrupt occurs, CPU stops executing the current running program, switch to kernel mode, executes interrupt handler. This handler saves the state of CPU, performs its operations, restore the state and returns to user mode.





Solution 3 - Operating System

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: https://stackoverflow.com/questions/6710040/cpu-privilege-rings-why-rings-1-and-2-arent-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: https://stackoverflow.com/questions/18431261/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: https://stackoverflow.com/questions/2711044/why-doesnt-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
compile and run:
    as -o hello_world.o hello_world.S
    ld -o hello_world.out hello_world.o

 [GitHub upstream](https://github.com/cirosantilli/x86-assembly-cheat/blob/d221ebec3a87ad811bbab48c4073e0bb2cc0df58/x86-64/gas/linux/hello_world.S).

 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](https://github.com/cirosantilli/x86-bare-metal-examples/blob/18772b1403133b2328d5ad44791445f9859de320/interrupt.S).

 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: https://stackoverflow.com/questions/7415515/how-to-access-the-control-registers-cr0-cr2-cr3-from-a-program-getting-segmenta/7419306#7419306

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:


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:


    .global _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
        .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:

enter image description here

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:

enter image description here

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: https://stackoverflow.com/questions/31787617/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: https://stackoverflow.com/questions/42824706/qemu-system-aarch64-entering-el1-when-emulating-a53-power-up

Code snippets tested on Ubuntu 18.10.

Solution 4 - Operating System

A processor in a computer running Windows has two different modes: user mode and kernel mode. The processor switches between the two modes depending on what type of code is running on the processor. Applications run in user mode, and core operating system components run in kernel mode. While many drivers run in kernel mode, some drivers may run in user mode.

When you start a user-mode application, Windows creates a process for the application. The process provides the application with a private virtual address space and a private handle table. Because an application's virtual address space is private, one application cannot alter data that belongs to another application. Each application runs in isolation, and if an application crashes, the crash is limited to that one application. Other applications and the operating system are not affected by the crash.

In addition to being private, the virtual address space of a user-mode application is limited. A processor running in user mode cannot access virtual addresses that are reserved for the operating system. Limiting the virtual address space of a user-mode application prevents the application from altering, and possibly damaging, critical operating system data.

All code that runs in kernel mode shares a single virtual address space. This means that a kernel-mode driver is not isolated from other drivers and the operating system itself. If a kernel-mode driver accidentally writes to the wrong virtual address, data that belongs to the operating system or another driver could be compromised. If a kernel-mode driver crashes, the entire operating system crashes.

If you are a Windows user once go through this link you will get more.

Communication between user mode and kernel mode

Solution 5 - Operating System

I'm going to take a stab in the dark and guess you're talking about Windows. In a nutshell, kernel mode has full access to hardware, but user mode doesn't. For instance, many if not most device drivers are written in kernel mode because they need to control finer details of their hardware.

See also this wikibook.

Solution 6 - Operating System

Other answers already explained the difference between user and kernel mode. If you really want to get into detail you should get a copy of Windows Internals, an excellent book written by Mark Russinovich and David Solomon describing the architecture and inside details of the various Windows operating systems.

Solution 7 - Operating System


Basically the difference between kernel and user modes is not OS dependent and is achieved only by restricting some instructions to be run only in kernel mode by means of hardware design. All other purposes like memory protection can be done only by that restriction.


It means that the processor lives in either the kernel mode or in the user mode. Using some mechanisms the architecture can guarantee that whenever it is switched to the kernel mode the OS code is fetched to be run.


Having this hardware infrastructure these could be achieved in common OSes:

  • Protecting user programs from accessing whole the memory, to not let programs overwrite the OS for example,
  • preventing user programs from performing sensitive instructions such as those that change CPU memory pointer bounds, to not let programs break their memory bounds for example.


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QuestionAlexView Question on Stackoverflow
Solution 1 - Operating SystemrahulView Answer on Stackoverflow
Solution 2 - Operating SystemSalvador DaliView Answer on Stackoverflow
Solution 3 - Operating SystemCiro Santilli Путлер Капут 六四事View Answer on Stackoverflow
Solution 4 - Operating SystemSangeen KhanView Answer on Stackoverflow
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