Required reading: proc.c (focus on scheduler() and sched()), swtch.S. Also process creation: sys_fork() and copyproc().
Big picture: more programs than CPUs. How to share the limited number of CPUs among the programs?
Idea: have the kernel transparently switch the CPU(s) among programs, giving them a "virtual CPU" abstraction.
Terminology: I'm going to call a switch from one program to another a context switch.
The kernel also exposes system calls for handling processes; here are the main XV6 process system calls:
Terminology: if a process has entered the kernel, the kernel stack and CPU registers of the resulting thread of control is called the processes kernel thread.
In this arrangment, the kernel doesn't directly switch CPUs among processes. Instead, the kernel has a number of mechanisms which combine to switch among processes:
Processes can be in a number of states (see proc.h): UNUSED, EMBRYO, SLEEPING, RUNNABLE, RUNNING, ZOMBIE. Diagram of possible states (skip ZOMBIE for now) and transitions between them.
allocproc, userinit.
sys_fork. why save pid in a temp variable? [1883]
main(){ fork(); while(1) ; }
Run bochs with a single CPU (bochs -q -f uni, then load-symbols "kernel.sym"). Break in yield(), then print stack using x/16x $esp on kernel stack of one of the processes, The return EIP should be trap().
Why does yield() need to acquire proc_table_lock? [1973] When will the lock be released?
Why state = RUNNABLE? What was state's previous value?
yield() calls sched() which calls swtch() b "swtch", s, c, print stack. Look at definition of struct context: it's a set of saved registers. swtch() will save into one context, restore from another. What are the two contexts? x/8x the context addrs.
Where do we think swtch() will return to?
Why doesn't swtch() save EAX in struct context?
At end of swtch(), what stack are we running on?
Where does print-stack show swtch() will return to? Why did this happen?
scheduler() is in the middle of a loop that checks to see whether each process would like to run (is RUNNABLE).
If scheduler() finds no RUNNABLE process, it finishes the loop, releases the lock, but then immediately re-acquires the lock at the start of the outer loop. Why release just to re-acquire?
That is, suppose for a while there are no RUNNABLE processes. What will the CPUs spend their time doing?
Why do we need to disable interrupts while holding locks in the kernel? For example, why do we disable interrupts when grabbing a lock, e.g. in scheduler? or tickslock in sys_sleep, trap?
If we always had enough idle processes, we could avoid having to release the lock.
On the other hand, what would happen if we never enabled interrupts in scheduler() and kept spinning with no processes to run?
How does the scheduler avoid picking the same process that just called yield (if there's another process that can run)?
scheduler() has its own stack (in cpus[x]). Why? Why not just
call it directly from sched()? I.e. why not have scheduler()
run on the stack of the last process that gave up the CPU?
Because the scheduler() on a multiprocessor system may need
to release the ptable lock. And the parent of an exiting process
might deallocate it's child stack. So if the scheduler was
running on the child's stack at that time, we have a problem.
Hence, scheduler needs a separate stack. Could idle processes
make this possible? yes, because in that case the scheduler
does not need to release the lock (in some sense, we are using
the idle process's stack).
- On an SMP, scheduler may have to release ptable lock
- If a ptable lock is released, it is possible that the old process's stack
gets deallocated (if it had just called exit)
- Hence the scheduler must run on a different stack from old process's stack.
What policy does scheduler() use to give CPU time to processes?
We know scheduler() is going to run the other cpu-bound process, by calling usegment() and swtch() to restore other process's registers and stack. Why does it set state to RUNNING? (That is, what's the functional difference between RUNNABLE and RUNNING?)
Do we need to call usegment in scheduler? Can we call it just before returning to user-space instead? What does usegment do exactly (set stack-top, user segments)?
Break in swtch() again, and check that we are indeed switching to the other process's stack, and step through to the return from its call to yield() from trap().
Break in swtch() a few times to see that we're bouncing between two processes (based on their context address values).
How do we garbage-collect processes in xv6? Can't deallocate the stack we're running on right now. When does a process get deallocated? Back to process state diagram, ZOMBIE state. Why does exit pass children to init, rather than to its own parent?
How do we kill a process? Why not turn it into a ZOMBIE directly?
How would a context switch between threads differ from a context switch between processes?
Added complication with kernel threads: concurrent syscalls?
We could also implement threads in user-space (maybe even pre-emptive with a periodic interrupt like SIGALRM). What's the difference? What do we gain from kernel threads?
We've seen the complete story of how xv6 switches between user processes. This is the concrete realization of the virtual CPU abstraction. The core mechanism is swtch(), which saves and restores thread register state and thus switches stacks (ESP) and changes the control flow (EIP). We also saw a bit of what it takes to integrate thread switching with the concurrency brought by multiple CPUs and by interrupts.