![]() ![]() 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 the CPU is turned on, it starts running the initial program in ring 0 (well kind of, but it is a good approximation). How do programs and operating systems transition between rings? See also: Why doesn't Linux use the hardware context switch via the TSS? 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. 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. Otherwise, for example, file permissions would be useless if any program could directly read from disk. In other words, cannot use the LGDT and LIDT instructions.Ĭannot do IO instructions like in and out, and thus have arbitrary hardware accesses. ![]() Handlers run in ring 0, and would break the security model. Those are configured by writing to memory locations, which is also prevented by paging. This prevents one process from seeing the memory of other processes for security / ease of programming reasons.Ĭannot register interrupt handlers. In other words, cannot modify the CR3 register, and paging itself prevents modification of the page tables. In other words, cannot modify the current segment descriptor, which determines the current ring.Ĭannot modify the page tables: How does x86 paging work? Ring 3 cannot run several instructions and write to several registers, most notably:Ĭannot change its own ring! Otherwise, it could set itself to ring 0 and rings would be useless. The CPU chip is physically built so that: The segment registers CS, DS, etc., which point to the index of an entry in the GDT.įor example, CS = 0 means the first entry of the GDT is currently active for the executing code. The LGDT instruction sets the address to the current descriptor table. Global descriptor table: a in-memory table of GDT entries, and each entry has a field Privl which encodes the ring. The current ring is selected by a combination of: Why Linux does not use rings 1 and 2: CPU Privilege Rings: Why rings 1 and 2 aren't used? This is the most hard and fast definition of kernel vs userland. In x86 protected mode, the CPU is always in one of 4 rings. This handler saves the state of CPU, performs its operations, restore the state and returns to user mode. When interrupt occurs, CPU stops executing the current running program, switch to kernel mode, executes interrupt handler. CPU is interrupted by interrupts (timers, keyboard, I/O). The switch from user mode to kernel mode is not done automatically by CPU. Is reserved for drivers which operate on the lowest level Any CPU instructionĬan be executed and every memory address can be accessed. It hasĪccess to every resource and underlying hardware. mode where all kernel programs execute (different drivers).Each process started by windows except of system It needs to access any of these features – it makes a call to the Kernel mode, they would be able to overwrite each other’s memory. The reason for this is because if all programs ran in ![]() It does not have access to RAMĪnd hardware. Modern microprocessors implement in hardware at least 2 different states. So computer architects decide to change it. Prior to this, when computers were like a big room, if something crashes – it halts the whole computer. These are two different modes in which your computer can operate. ![]()
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