问题
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?
回答1:
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.
回答2:
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!
回答3:
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)
来源：https://stackoverflow.com/questions/12630214/context-switch-internals