I want to learn and fill gaps in my knowledge with the help of this question.
So, a user is running a thread (kernel-level) and it now calls yield (a system call I presume).
The scheduler must now save the context of the current thread in the TCB (which is stored in the kernel somewhere) and choose another thread to run and loads its context and jump to its CS:EIP.
To narrow things down, I am working on Linux running on top of x86 architecture. Now, I want to get into the details:
So, first we have a system call:
1) The wrapper function for yield will push the system call arguments onto the stack. Push the return address and raise an interrupt with the system call number pushed onto some register (say EAX).
2) The interrupt changes the CPU mode from user to kernel and jumps to the interrupt vector table and from there to the actual system call in the kernel.
3) I guess the scheduler gets called now and now it must save the current state in the TCB. Here is my dilemma. Since, the scheduler will use the kernel stack and not the user stack for performing its operation (which means the SS and SP have to be changed) how does it store the state of the user without modifying any registers in the process. I have read on forums that there are special hardware instructions for saving state but then how does the scheduler get access to them and who runs these instructions and when?
4) The scheduler now stores the state into the TCB and loads another TCB.
5) When the scheduler runs the original thread, the control gets back to the wrapper function which clears the stack and the thread resumes.
Side questions: Does the scheduler run as a kernel-only thread (i.e. a thread which can run only kernel code)? Is there a separate kernel stack for each kernel-thread or each process?
At a high level, there are two separate mechanisms to understand. The first is the kernel entry/exit mechanism: this switches a single running thread from running usermode code to running kernel code in the context of that thread, and back again. The second is the context switch mechanism itself, which switches in kernel mode from running in the context of one thread to another.
So, when Thread A calls sched_yield() and is replaced by Thread B, what happens is:
Thread A enters the kernel, changing from user mode to kernel mode;
Thread A in the kernel context-switches to Thread B in the kernel;
Thread B exits the kernel, changing from kernel mode back to user mode.
Each user thread has both a user-mode stack and a kernel-mode stack. When a thread enters the kernel, the current value of the user-mode stack (SS:ESP) and instruction pointer (CS:EIP) are saved to the thread's kernel-mode stack, and the CPU switches to the kernel-mode stack - with the int $80 syscall mechanism, this is done by the CPU itself. The remaining register values and flags are then also saved to the kernel stack.
When a thread returns from the kernel to user-mode, the register values and flags are popped from the kernel-mode stack, then the user-mode stack and instruction pointer values are restored from the saved values on the kernel-mode stack.
When a thread context-switches, it calls into the scheduler (the scheduler does not run as a separate thread - it always runs in the context of the current thread). The scheduler code selects a process to run next, and calls the switch_to() function. This function essentially just switches the kernel stacks - it saves the current value of the stack pointer into the TCB for the current thread (called struct task_struct in Linux), and loads a previously-saved stack pointer from the TCB for the next thread. At this point it also saves and restores some other thread state that isn't usually used by the kernel - things like floating point/SSE registers. If the threads being switched don't share the same virtual memory space (ie. they're in different processes), the page tables are also switched.
So you can see that the core user-mode state of a thread isn't saved and restored at context-switch time - it's saved and restored to the thread's kernel stack when you enter and leave the kernel. The context-switch code doesn't have to worry about clobbering the user-mode register values - those are already safely saved away in the kernel stack by that point.
What you missed during step 2 is that the stack gets switched from a thread's user-level stack (where you pushed args) to a thread's protected-level stack. The current context of the thread interrupted by the syscall is actually saved on this protected stack. Inside the ISR and just before entering the kernel, this protected-stack is again switched to the kernel stack you are talking about. Once inside the kernel, kernel functions such as scheduler's functions eventually use the kernel-stack. Later on, a thread gets elected by the scheduler and the system returns to the ISR, it switchs back from the kernel stack to the newly elected (or the former if no higher priority thread is active) thread's protected-level stack, wich eventually contains the new thread context. Therefore the context is restored from this stack by code automatically (depending on the underlying architecture). Finally, a special instruction restores the latest touchy resgisters such as the stack pointer and the instruction pointer. Back in the userland...
To sum-up, a thread has (generally) two stacks, and the kernel itself has one. The kernel stack gets wiped at the end of each kernel entering. It's interesting to point out that since 2.6, the kernel itself gets threaded for some processing, therefore a kernel-thread has its own protected-level stack beside the general kernel-stack.
Some ressources:
3.3.3 Performing the Process Switch of Understanding the Linux Kernel, O'Reilly
5.12.1 Exception- or Interrupt-Handler Procedures of the Intel's manual 3A (sysprogramming). Chapter number may vary from edition to other, thus a lookup on "Stack Usage on Transfers to Interrupt and Exception-Handling Routines" should get you to the good one.
Hope this help!
Kernel itself have no stack at all. The same is true for the process. It also have no stack. Threads are only system citizens which are considered as execution units. Due to this only threads can be scheduled and only threads have stacks. But there is one point which kernel mode code exploits heavily - every moment of time system works in the context of the currently active thread. Due to this kernel itself can reuse the stack of the currently active stack. Note that only one of them can execute at the same moment of time either kernel code or user code. Due to this when kernel is invoked it just reuse thread stack and perform a cleanup before returning control back to the interrupted activities in the thread. The same mechanism works for interrupt handlers. The same mechanism is exploited by signal handlers.
In its turn thread stack is divided into two isolated parts, one of which called user stack (because it is used when thread executes in user mode), and second one is called kernel stack (because it is used when thread executes in kernel mode). Once thread crosses the border between user and kernel mode, CPU automatically switches it from one stack to another. Both stack are tracked by kernel and CPU differently. For the kernel stack, CPU permanently keeps in mind pointer to the top of the kernel stack of the thread. It is easy, because this address is constant for the thread. Each time when thread enters the kernel it found empty kernel stack and each time when it returns to the user mode it cleans kernel stack. In the same time CPU doesn't keep in mind pointer to the top of the user stack, when thread runs in the kernel mode. Instead during entering to the kernel, CPU creates special "interrupt" stack frame on the top of the kernel stack and stores the value of the user mode stack pointer in that frame. When thread exits the kernel, CPU restores the value of ESP from previously created "interrupt" stack frame, immediately before its cleanup. (on legacy x86 the pair of instructions int/iret handle enter and exit from kernel mode)
During entering to the kernel mode, immediately after CPU will have created "interrupt" stack frame, kernel pushes content of the rest of CPU registers to the kernel stack. Note that is saves values only for those registers, which can be used by kernel code. For example kernel doesn't save content of SSE registers just because it will never touch them. Similarly just before asking CPU to return control back to the user mode, kernel pops previously saved content back to the registers.
Note that in such systems as Windows and Linux there is a notion of system thread (frequently called kernel thread, I know it is confusing). System threads a kind of special threads, because they execute only in kernel mode and due to this have no user part of the stack. Kernel employs them for auxiliary housekeeping tasks.
Thread switch is performed only in kernel mode. That mean that both threads outgoing and incoming run in kernel mode, both uses their own kernel stacks, and both have kernel stacks have "interrupt" frames with pointers to the top of the user stacks. Key point of the thread switch is a switch between kernel stacks of threads, as simple as:
pushad; // save context of outgoing thread on the top of the kernel stack of outgoing thread
; here kernel uses kernel stack of outgoing thread
mov [TCB_of_outgoing_thread], ESP;
mov ESP , [TCB_of_incoming_thread]
; here kernel uses kernel stack of incoming thread
popad; // save context of incoming thread from the top of the kernel stack of incoming thread
Note that there is only one function in the kernel that performs thread switch. Due to this each time when kernel has stacks switched it can find a context of incoming thread on the top of the stack. Just because every time before stack switch kernel pushes context of outgoing thread to its stack.
Note also that every time after stack switch and before returning back to the user mode, kernel reloads the mind of CPU by new value of the top of kernel stack. Making this it assures that when new active thread will try to enter kernel in future it will be switched by CPU to its own kernel stack.
Note also that not all registers are saved on the stack during thread switch, some registers like FPU/MMX/SSE are saved in specially dedicated area in TCB of outgoing thread. Kernel employs different strategy here for two reasons. First of all not every thread in the system uses them. Pushing their content to and and popping it from the stack for every thread is inefficient. And second one there are special instructions for "fast" saving and loading of their content. And these instructions doesn't use stack.
Note also that in fact kernel part of the thread stack has fixed size and is allocated as part of TCB. (true for Linux and I believe for Windows too)
Related
I have a little misunderstanding about the difference between the task's TCB and the task stack in an RTOS, doesn't the TCB already carry all needed info about context switching such as registers values and so?
Thanks.
C does not support threading or multitasking directly and is not thread aware, however a C implementation typically requires a stack for local variables, function parameters and function return addresses.
In a pre-emptive multi-threading environment, each thread must have an independent stack so that the non-thread aware C code has a distinct execution environment.
As you state the TCB contains "info about context switching such as registers values", which may be true in some implementations, in others the TCB may contain only the the value of the task's stack-pointer - with all other registers pushed to the thread's own stack on the context switch. During a context switch, the stack-pointer will be restored, then the context stored on the stack restored. The last register to be popped-off the thread's stack will be the program-counter causing an immediate jump to the same location with the same same stack (and therefore same local variables and call stack), that it had when it was pre-empted.
Note that implementations may differ in various ways, but the above is a generalised description, rather than a description of FreeRTOS specifically.
The details of a FreeRTOS context switch using an AVR target as an example and a context switch caused by a OS tick is given at: https://www.freertos.org/implementation/a00018.html. There the TCB retains only the task's stack-pointer, the program counter is placed on the stack (automatically by the interrupt in this case example).
You might examine the struct tskTaskControlBlock in your specific implementation.
As I understood before, each process has its own address space called vitual address space or program memory,
and every process has a location called stack which is used to store local variables and parameters of a function.
Also, when an exception occurs the processor (say an ARM cortex-A) switches to privileged mode and then branches to the exception handler.
According to what I understood, most applications run in non-privileged user mode, and this mode has a special register called stack pointer to hold the address of top of the stack; but this is a single register and can't actually hold the address of top of the stack of several processes at the same time. Would you please explain what actually happens?
As for all registers, it's saved and put away in a data structure associated to the process once the OS decides it's time for another process to run ("context switch"); it's as if it took a snapshot of the current processor state.
When the process is scheduled again, all registers are restored (including the instruction pointer) and execution resumes as if nothing happened.
According to what I understood, most applications run in non-privileged user mode, and this mode has a special register called stack pointer to hold the address of top of the stack
The stack pointer is not specific of user mode, the processor always has (and can use) it, regardless of the mode.
For a (pet) project which is a virtual machine (written in pure C) I am developing a threading mechanism. A few notes to understand much better the problem:
the virtual machine interprets a sequence of bytecode, more or less similar to the x86 instructions
it has a set of registers, stack, IP, etc... which are all grouped into an execution context of the current thread.
each thread has its own execution context, so they do not mess around with other threads' data (however in their local stack they have a part of the global stack which was filled up till the point the thread started its own life with variables from the global context).
the VM has a list of execution contexts, representing each of the threads, and also has a current execution contex.
the code part (the bytes of the byteode) is stored common place
the threading mechanism is implemented as cycling through the execution contexts and always executing the bytecode from the current threads' execution contexts' instruction pointer (IP) (yes, it is a fake multi threaded system for now).
threads are (will be) put in a priority queue, which is always updated if the thread requires a new priority.
when there is (will be) a new thread created a new execution context is created and the VM will populate it with data, then will switch to it and this thread will run till the thread scheduler decides it is time to switch to another thread.
And now comes the question:
Based on what should the thread scheduler decide that ok, it is time to switch to a new thread automatically (not considering thread yields control, thread finished or created)?
I was thinking at the following solutions:
at the completion of each full (CPU level atomic) instruction the thread scheduler will switch to the next thread based on its priority (full instruction: mov ax, 13 so it will always complete it, will not switch after mov ax).
each thread has allocated a specific time slice and upon finishing it after the first completed instruction it will switch to the new thread
What are your suggestions?
Some random thought... Depends on for what reason your VM is created. If it simulates some real or imaginable hardware with cycle-precision or so, you have to follow its specification (I guess you wouldn't ask this question in this case :) ). Otherwise, I'd consider performance of the VM as one of the top priority, and for this reason, second-like solution sounds reasonable since it looks more cache-friendly. But instead of literal time-slice, I'd consider some buffer-size-based limits since it, again is closer to cache-efficiency.
Is the Linux kernel aware of pthreads in the user address space ( which i dont think it is..but i did not find any info abt that). How does the Instruction pointer change when thread switching takes place.. ??
The native NPTL (native posix thread library) used in Linux maps pthreads to "processes that share resources and therefore look like threads" in the kernel. In this way, the kernel's scheduler directly controls the scheduling of pthreads.
A "pthread switch" is done by the exact same code (in the kernel) that handles process switches. Simplified, this would be something like "store previous process state; if the next process uses a different virtual address space then switch virtual address spaces; load next process state;" (where "process state" includes the instruction pointer for the process/thread).
Well the Linux kernel doesn't know about user threads (pthread does in userspace, moreover the kernel doesn't really care about them except it just needs to know what to schedule).
The instruction pointer is changed in the kernel during what's called a context switch. During this switch the kernel essentially asks the scheduler what's next? the scheduler will hand it a task_struct which contains all the information about the thread and the interrupt handler for a context switch will go ahead and set the values on the CPU accordingly (page tables, instruction pointer, etc...) and when that code is done the CPU simply just starts executing from there.
1) The kernel doesn't know about user-level threads. However, NPTL isn't user level
2) This is a really broad question. You should look at an OS book. It will go into depth on that issue and all other involved in a context switch.
I have a simple bootloader, which initializes and prepares SDRAM. Then it loads an application from the Flash and starts it at some address in the RAM. After the application has finished its execution, the system does restart. There is no system stack.
Now, I would like this bootloader receives control back after an application finished its execution. The bootloader (let's call it OS) must also read an application's return code.
How can an application return a value to the calling OS and how the calling OS gets control back? I suppose, it can be done using interrupts - OS has a special resident function joined with some interrupt and every application just calls this interrupt at the end of its own execution. But how can a return code be read by OS if there is no system stack?
Normally you would leave a return code in one or more registers, but since you're in control, you can leave it wherever you like!
When an application is interrupted, the interrupt handling routine needs to save the application's state somewhere, which will probably mean copying from shadow registers to a predefined location in memory.
If an application surrenders control back to the OS (through a software interrupt / sytem call) then you need to define your own calling convention for which registers arguments are placed in, and the event handler needs to follow this before passing control back to the OS. You probably want to make the calling convention match up with that of your c compiler as much as possible, to keep things easy for yourself.
One solution is for the program to write its exit code at a fixed, known location in memory - the "OS" can then read it.