Here is a scenario. Let’s say that a kernel task is running on a uniprocessor system with preemption disabled. The task acquires a spin lock. Now it is executing it’s critical section. At this time, what if the time slice available for this task expires and it has to schedule out?
Does the spin_lock have a mechanism to prevent this?
Can it be scheduled out? If yes, then what happens to the critical section?
Can it be interrupted by an IRQ? (Assuming that preemption is disabled)
Is this scenario feasible? In other words, could this scenario happen?
From the kernel code, I understand that the spin_lock is basically a nop on a uniprocessor with preemption disabled. To be accurate, all it does is barrier()
I understand why it is a nop (as it is a uniprocessor and no other task could be manipulating the data at that instant) but I still don’t understand how it could be uninterrupted(due to IRQs or scheduling).
What am I missing here? Pointers to the Linux kernel code which indicates about this could be really helpful.
My basic assumptions:
32 bit Linux kernel
Actually spin_lock() disables preemption by calling preempt_disable() before it tries to acquire the lock, so scenario #1, #2, #3 could never happen.
From recent source code, spin_lock() eventually calls __raw_spin_lock(), which calls preempt_disable() before calling spin_acquire() to acquire the lock. spin_lock_irqsave() which is commonly used in interrupt context has similar context.
Regarding #3, if the variable is shared between process/interrupt context, you should always use spin_lock_irq()/spin_lock_irqsave() instead of spin_lock() to avoid deadlock scenario.
The mechanism that handles time slices expiring is a timer interrupt. The interrupt will set the TIF_NEEDS_RESCHED flag for the process. When returning from the timer's interrupt context back to your critical section, a check will be made whether or not to preempt the process due to the TIF_NEEDS_RESCHED flag. Since preemption is disabled, nothing will happen and it will return to your critical section.
When your critical section is over, the release of the lock will call preempt_enable() to reenable preemption. At that moment another check is done as to whether or not to preempt. Since the TIF_NEEDS_RESCHED flag is set and preemption is now enabled, the process will be preempted.
Spin locks disable preemption.
No, because preemption is disabled.
Yes. There are spin lock versions that disable IRQs to prevent this.
No because spin locks disable preemption.
Spinlocks don't exist on unitprocessor systems anyway because they don't make sense. If a a thread that doesn't own the lock attempts to acquire it, that means that the thread that does own it is currently asleep (only one cpu). So there's no reason to spin wait for something that's asleep. For this reason spinlocks are optimized away in these cases to just a preemption disable so that no other thread can touch the critical section.
What would happen if I use semaphore and mutex locks in interrupt context?
Normally semaphore is used in synchronization mechanism. What would happen if I use this one in an interrupt context?
I am working on a project on gpio pins and when interrupt happens, I have to send one signal in ISR. I am using spinlocks.
What would happen if I use semaphore and mutext in ISR?
Waiting in mutexes and semaphores are implemented using switching current task state to TASK_INTERRUPTIBLE/TASK_UNINTERRUPTIBLE and similar with futher call to schedule().
Calling schedule() with current task state differed from TASK_RUNNING leads to switching to another process. And if current refers to interrupt context, you will never return back to it, because scheduling can switch only to the process.
So, when you lock contended(that is, currently locked) semaphore/mutex in interrupt context, you just lost current execution "thread".
If you lock semaphore/mutex which is uncontended(currently not locked), execution will be correct except warning in system log about improper semaphore/mutex usage.
First of all sorry for a little bit ambiguity in Question... What I want to understand is the below scenario
Suppose porcess is running, it holds one lock, Now after acquiring the lock HW interrupt is generated, So How kernel will handle this situation, will it wait for lock ? if yes, what if the interrupt handler need to access that lock or the shared data protected by that lock in process ?
The Linux kernel has a few functions for acquiring spinlocks, to deal with issues like the one you're raising here. In particular, there is spin_lock_irq(), which disables interrupts (on the CPU the process is running on) and acquires the spinlock. This can be used when the code knows interrupts are enabled before the spinlock is acquired; in case the function might be called in different contexts, there is also spin_lock_irqsave(), which stashes away the current state of interrupts before disabling them, so that they can be reenabled by spin_unlock_irqrestore().
In any case, if a lock is used in both process and interrupt context (which is a good and very common design if there is data that needs to be shared between the contexts), then process context must disable interrupts (locally on the CPU it's running on) when acquiring the spinlock to avoid deadlocks. In fact, lockdep ("CONFIG_PROVE_LOCKING") will verify this and warn if a spinlock is used in a way that is susceptible to the "interrupt while process context holds a lock" deadlock.
Let me explain some basic properties of interrupt handler or bottom half.
A handler can’t transfer data to or from user space, because it doesn’t execute in the context of a process.
Handlers also cannot do anything that would sleep, such as calling wait_event, allocating memory with anything other than GFP_ATOMIC, or locking a semaphore
handlers cannot call schedule.
What i am trying to say is that Interrupt handler runs in atomic context. They can not sleep as they cannot be rescheduled. interrupts do not have a backing process context
The above is by design. You can do whatever you want in code, just be prepared for the consequences
Let us assume that you acquire a lock in interrupt handler(bad design).
When an interrupt occur the process saves its register on stack and start ISR. now after acquiring a lock you would be in a deadlock as their is no way ISR know what the process was doing.
The process will not be able to resume execution until it is done it with ISR
In a preemptive kernel the ISR and the process can be preempt but for a non-preemptive kernel you are dead.
I've recently read section 5.5.2 (Spinlocks and Atomic Context) of LDDv3 book:
Avoiding sleep while holding a lock can be more difficult; many kernel functions can sleep, and this behavior is not always well documented. Copying data to or from user space is an obvious example: the required user-space page may need to be swapped in from the disk before the copy can proceed, and that operation clearly requires a sleep. Just about any operation that must allocate memory can sleep; kmalloc can decide to give up the processor, and wait for more memory to become available unless it is explicitly told not to. Sleeps can happen in surprising places; writing code that will execute under a spinlock requires paying attention to every function that you call.
It's clear to me that spinlocks must always be held for the minimum time possible and I think that it's relatively easy to write correct spinlock-using code from scratch.
Suppose, however, that we have a big project where spinlocks are widely used.
How can we make sure that functions called from critical sections protected by spinlocks will never sleep?
Thanks in advance!
What about enabling "Sleep-inside-spinlock checking" for your kernel ? It is usually found under Kernel Debugging when you run make config. You might also try to duplicate its behavior in your code.
One thing I noticed on a lot of projects is people seem to misuse spinlocks, they get used instead of the other locking primitives that should have be used.
A linux spinlock only exists in multiprocessor builds (in single process builds the spinlock preprocessor defines are empty) spinlocks are for short duration locks on a multi processor platform.
If code fails to aquire a spinlock it just spins the processor until the lock is free. So either another process running on a different processor must free the lock or possibly it could be freed by an interrupt handler but the wait event mechanism is much better way of waiting on an interrupt.
The irqsave spinlock primitive is a tidy way of disabling/ enabling interrupts so a driver can lock out an interrupt handler but this should only be held for long enough for the process to update some variables shared with an interrupt handler, if you disable interupts you are not going to be scheduled.
If you need to lock out an interrupt handler use a spinlock with irqsave.
For general kernel locking you should be using mutex/semaphore api which will sleep on the lock if they need to.
To lock against code running in other processes use muxtex/semaphore
To lock against code running in an interrupt context use irq save/restore or spinlock_irq save/restore
To lock against code running on other processors then use spinlocks and avoid holding the lock for long.
I hope this helps
I am reading following article by Robert Love
http://www.linuxjournal.com/article/6916
that says
"...Let's discuss the fact that work queues run in process context. This is in contrast to the other bottom-half mechanisms, which all run in interrupt context. Code running in interrupt context is unable to sleep, or block, because interrupt context does not have a backing process with which to reschedule. Therefore, because interrupt handlers are not associated with a process, there is nothing for the scheduler to put to sleep and, more importantly, nothing for the scheduler to wake up..."
I don't get it. AFAIK, scheduler in the kernel is O(1), that is implemented through the bitmap. So what stops the scehduler from putting interrupt context to sleep and taking next schedulable process and passing it the control?
So what stops the scehduler from putting interrupt context to sleep and taking next schedulable process and passing it the control?
The problem is that the interrupt context is not a process, and therefore cannot be put to sleep.
When an interrupt occurs, the processor saves the registers onto the stack and jumps to the start of the interrupt service routine. This means that when the interrupt handler is running, it is running in the context of the process that was executing when the interrupt occurred. The interrupt is executing on that process's stack, and when the interrupt handler completes, that process will resume executing.
If you tried to sleep or block inside an interrupt handler, you would wind up not only stopping the interrupt handler, but also the process it interrupted. This could be dangerous, as the interrupt handler has no way of knowing what the interrupted process was doing, or even if it is safe for that process to be suspended.
A simple scenario where things could go wrong would be a deadlock between the interrupt handler and the process it interrupts.
Process1 enters kernel mode.
Process1 acquires LockA.
Interrupt occurs.
ISR starts executing using Process1's stack.
ISR tries to acquire LockA.
ISR calls sleep to wait for LockA to be released.
At this point, you have a deadlock. Process1 can't resume execution until the ISR is done with its stack. But the ISR is blocked waiting for Process1 to release LockA.
I think it's a design idea.
Sure, you can design a system that you can sleep in interrupt, but except to make to the system hard to comprehend and complicated(many many situation you have to take into account), that's does not help anything. So from a design view, declare interrupt handler as can not sleep is very clear and easy to implement.
From Robert Love (a kernel hacker):
http://permalink.gmane.org/gmane.linux.kernel.kernelnewbies/1791
You cannot sleep in an interrupt handler because interrupts do not have
a backing process context, and thus there is nothing to reschedule back
into. In other words, interrupt handlers are not associated with a task,
so there is nothing to "put to sleep" and (more importantly) "nothing to
wake up". They must run atomically.
This is not unlike other operating systems. In most operating systems,
interrupts are not threaded. Bottom halves often are, however.
The reason the page fault handler can sleep is that it is invoked only
by code that is running in process context. Because the kernel's own
memory is not pagable, only user-space memory accesses can result in a
page fault. Thus, only a few certain places (such as calls to
copy_{to,from}_user()) can cause a page fault within the kernel. Those
places must all be made by code that can sleep (i.e., process context,
no locks, et cetera).
Because the thread switching infrastructure is unusable at that point. When servicing an interrupt, only stuff of higher priority can execute - See the Intel Software Developer's Manual on interrupt, task and processor priority. If you did allow another thread to execute (which you imply in your question that it would be easy to do), you wouldn't be able to let it do anything - if it caused a page fault, you'd have to use services in the kernel that are unusable while the interrupt is being serviced (see below for why).
Typically, your only goal in an interrupt routine is to get the device to stop interrupting and queue something at a lower interrupt level (in unix this is typically a non-interrupt level, but for Windows, it's dispatch, apc or passive level) to do the heavy lifting where you have access to more features of the kernel/os. See - Implementing a handler.
It's a property of how O/S's have to work, not something inherent in Linux. An interrupt routine can execute at any point so the state of what you interrupted is inconsistent. If you interrupted the thread scheduling code, its state is inconsistent so you can't be sure you can "sleep" and switch threads. Even if you protect the thread switching code from being interrupted, thread switching is a very high level feature of the O/S and if you protected everything it relies on, an interrupt becomes more of a suggestion than the imperative implied by its name.
So what stops the scehduler from putting interrupt context to sleep and taking next schedulable process and passing it the control?
Scheduling happens on timer interrupts. The basic rule is that only one interrupt can be open at a time, so if you go to sleep in the "got data from device X" interrupt, the timer interrupt cannot run to schedule it out.
Interrupts also happen many times and overlap. If you put the "got data" interrupt to sleep, and then get more data, what happens? It's confusing (and fragile) enough that the catch-all rule is: no sleeping in interrupts. You will do it wrong.
Disallowing an interrupt handler to block is a design choice. When some data is on the device, the interrupt handler intercepts the current process, prepares the transfer of the data and enables the interrupt; before the handler enables the current interrupt, the device has to hang. We want keep our I/O busy and our system responsive, then we had better not block the interrupt handler.
I don't think the "unstable states" are an essential reason. Processes, no matter they are in user-mode or kernel-mode, should be aware that they may be interrupted by interrupts. If some kernel-mode data structure will be accessed by both interrupt handler and the current process, and race condition exists, then the current process should disable local interrupts, and moreover for multi-processor architectures, spinlocks should be used to during the critical sections.
I also don't think if the interrupt handler were blocked, it cannot be waken up. When we say "block", basically it means that the blocked process is waiting for some event/resource, so it links itself into some wait-queue for that event/resource. Whenever the resource is released, the releasing process is responsible for waking up the waiting process(es).
However, the really annoying thing is that the blocked process can do nothing during the blocking time; it did nothing wrong for this punishment, which is unfair. And nobody could surely predict the blocking time, so the innocent process has to wait for unclear reason and for unlimited time.
Even if you could put an ISR to sleep, you wouldn't want to do it. You want your ISRs to be as fast as possible to reduce the risk of missing subsequent interrupts.
The linux kernel has two ways to allocate interrupt stack. One is on the kernel stack of the interrupted process, the other is a dedicated interrupt stack per CPU. If the interrupt context is saved on the dedicated interrupt stack per CPU, then indeed the interrupt context is completely not associated with any process. The "current" macro will produce an invalid pointer to current running process, since the "current" macro with some architecture are computed with the stack pointer. The stack pointer in the interrupt context may point to the dedicated interrupt stack, not the kernel stack of some process.
By nature, the question is whether in interrupt handler you can get a valid "current" (address to the current process task_structure), if yes, it's possible to modify the content there accordingly to make it into "sleep" state, which can be back by scheduler later if the state get changed somehow. The answer may be hardware-dependent.
But in ARM, it's impossible since 'current' is irrelevant to process under interrupt mode. See the code below:
#linux/arch/arm/include/asm/thread_info.h
94 static inline struct thread_info *current_thread_info(void)
95 {
96 register unsigned long sp asm ("sp");
97 return (struct thread_info *)(sp & ~(THREAD_SIZE - 1));
98 }
sp in USER mode and SVC mode are the "same" ("same" here not mean they're equal, instead, user mode's sp point to user space stack, while svc mode's sp r13_svc point to the kernel stack, where the user process's task_structure was updated at previous task switch, When a system call occurs, the process enter kernel space again, when the sp (sp_svc) is still not changed, these 2 sp are associated with each other, in this sense, they're 'same'), So under SVC mode, kernel code can get the valid 'current'. But under other privileged modes, say interrupt mode, sp is 'different', point to dedicated address defined in cpu_init(). The 'current' calculated under these mode will be irrelevant to the interrupted process, accessing it will result in unexpected behaviors. That's why it's always said that system call can sleep but interrupt handler can't, system call works on process context but interrupt not.
High-level interrupt handlers mask the operations of all lower-priority interrupts, including those of the system timer interrupt. Consequently, the interrupt handler must avoid involving itself in an activity that might cause it to sleep. If the handler sleeps, then the system may hang because the timer is masked and incapable of scheduling the sleeping thread.
Does this make sense?
If a higher-level interrupt routine gets to the point where the next thing it must do has to happen after a period of time, then it needs to put a request into the timer queue, asking that another interrupt routine be run (at lower priority level) some time later.
When that interrupt routine runs, it would then raise priority level back to the level of the original interrupt routine, and continue execution. This has the same effect as a sleep.
It is just a design/implementation choices in Linux OS. The advantage of this design is simple, but it may not be good for real time OS requirements.
Other OSes have other designs/implementations.
For example, in Solaris, the interrupts could have different priorities, that allows most of devices interrupts are invoked in interrupt threads. The interrupt threads allows sleep because each of interrupt threads has separate stack in the context of the thread.
The interrupt threads design is good for real time threads which should have higher priorities than interrupts.