I'm new to RTOS (uCOS II) and learning it by reading the book written by uCOS author. I have a doubt and I'm unable to find the answer to it.
In uCOS the task with highest priority is given CPU as per the scheduling algorithm. So, if I create write a uCOS application by creating two tasks One with High priority ( Prio = 1 for ex) and the other with low priority ( for ex Prio = 9).
If for example the highest priority task is waiting for an event, then the scheduler should start executing the next higher priority task ? If thats correct then what part of the code switches High priority with low priority ?
The three arch dependent codes are :
1. Interrupt level context switch
2. Start highest priority task ready to run
3. Task level context switch
In case 1 after serving the interrupt the scheduler returns to the highest priority task. In case 2, its called when we start the OS by OSStart()
In case 3, When ever a higher priority task is made ready and its called by timer interrupt
Now, where exactly or how exactly will the scheduler assigns CPU to a lower priority task given the high priority task is in wait ??
Thanks
Another way to consider your question is to ask yourself how did the high priority task get into the waiting state. The answer to both questions is that the high priority task calls an RTOS routine such as GetEvent(). (I don't know whether that is a real uCOS-II routine -- I'm just generalizing.). The RTOS routine puts the high priority task into the waiting state (i.e. blocked) and then the RTOS scheduler finds the next highest priority task that is ready to run and switches to that task's context. The RTOS will have several blocking functions that allow for a task context switch. For example when you read from a queue or mailbox or when you wait for a semaphore or mutex.
The scheduler runs whenever a scheduling event occurs. In your example, that occurs when the high priority task calls the event wait. In general OS calls that may block or yield cause the scheduler to run. The scheduler also runs on exit from ISRs including the IS timer ISR.
In general, when the scheduler performs a context switch, it copies the current processor core registers to the task's control block, and copies the stored register values for the task being switched to into the processor registers, with the stack pointer and program-counter copies last. The change to the program-counter causes execution to continue in the new task with the task's own stack, in the state it was when it last blocked or was preemted. Preemption can occur when a scheduling event occurs in an ISR that causes a higher priority task to become ready.
The thing about uC/OS-II is that it is described in intricate detail in Jean Labrosse's book. The general principles of RTOS with examples using uC/OS-II are described in
this online course by Jack Ganssle.
Interrupt level context switch is used for preemptive, for example, you have an low priority task running, and high priority need to run (OSTimeDly timeout, for example), in this situation, Interrupt level context switch will pause low priority task, then switch to high priority one.
For high to low priority switch, it need high one give up CPU resource by calling OS_Sched
Why do RTOSes not have any implementation to prevent priority inversion for semaphore even though it exists for mutex.
Semaphores do not need to prevent priority inversion?
The same situation happens both on uC/OS and GreenHills RTOS.
Thanks in advance.
Priority inversion occurs when a low-priority task owns a semaphore,
and a high-priority task is forced to wait on the semaphore until the
low-priority task releases it. If, prior to releasing the semaphore,
the low priority task is preempted by one or more mid-priority tasks,
then unbounded priority inversion has occurred because the delay of
the high-priority task is no longer predictable. This defeats Deadline
Monotonic Analysis (DMA) because it is not possible to predict if the
high-priority task will meet its deadline.
Sharing a critical resource between high and low priority tasks is not
a desirable design practice. It is better to share a resource only
among equal priority tasks or to limit resource accesses to a single
resource server task. Examples are a print server task and a file
server task. We have long advocated this practice. However, with the
layering of increasingly diverse and complicated middleware onto
RTOSs, it is becoming impractical to enforce such simple strategies.
Hence, in the interest of safety, it is best to implement some method
of preventing unbounded priority inversion.
Check full link at http://www.smxrtos.com/articles/techppr/mutex.htm
Regards,
Otacon
Vxworks states that mutual exculsion semaphores : Cannot be given inside ISR, while the condition is vaild for binary and counting semaphore.
I am not able to understand the reason out the same.
Thanks,
Zaks.
Remember that a Mutex must first be acquired/taken then released/given.
In addition, the task that acquires the mutex owns it. This prevents another task from releasing a mutex it doesn't own.
With that being the case, it becomes clear that since an ISR cannot acquire a mutex (or any semaphore for that matter - it's a blocking operation), then it follows that it can't give the mutex.
It is quite possible for an ISR to do give a Binary or Counting semaphore to signal a task that something happens. But mutexes are always a take/give pair.
To clarify a point. In VxWorks, the ISR context is not the same as the context of a task!
The following scenario is invalid:
Task A ISR
semTake(mutex)
....
semGive(mutex)
Task A owns the mutex. When the ISR runs, it executes in a totaly different context. Most current processors have a separate ISR stack. Since Task A owns the mutex, how could the ISR give it up? In fact, what guarantee do you have that the ISR will fire while A has the mutex.
Even assuming you "could" give a mutex in an ISR, how would you handle the following scenario:
Task A Task B ISR
semTake(mutex)
...
<context switch happens>
<B runs>
semGive(mutex)
Task A gets switched out due to a call unrelated to the mutex, and Task B runs. The ISR now executes while B was running. Would it still be valid for the ISR to be given?
Regardless of all this, the simple fact is that a mutex is always used in a pair of Get/Set. I fail to see a use case where you would have an isolated semGive.
Is there a specific situation you have in mind that would require a semGive from an ISR context?
Mutexes should not be used from an interrupt because:
They include a priority inheritance mechanism which only makes sense if the mutex is given and taken from a task, not an interrupt.
An interrupt cannot block to wait for a resource that is guarded by a mutex to become available
https://www.freertos.org/Real-time-embedded-RTOS-mutexes.html
I've heard the phrase 'priority inversion' in reference to development of operating systems.
What exactly is priority inversion?
What is the problem it's meant to solve, and how does it solve it?
Imagine three (3) tasks of different priority: tLow, tMed and tHigh. tLow and tHigh access the same critical resource at different times; tMed does its own thing.
tLow is running, tMed and tHigh are presently blocked (but not in critical section).
tLow comes along and enters the critical section.
tHigh unblocks and since it is the highest priority task in the system, it runs.
tHigh then attempts to enter the critical resource but blocks as tLow is in there.
tMed unblocks and since it is now the highest priority task in the system, it runs.
tHigh can not run until tLow gives up the resource. tLow can not run until tMed blocks or ends. The priority of the tasks has been inverted; tHigh though it has the highest priority is at the bottom of the execution chain.
To "solve" priority inversion, the priority of tLow must be bumped up to be at least as high as tHigh. Some may bump its priority to the highest possible priority level. Just as important as bumping up the priority level of tLow, is dropping the priority level of tLow at the appropriate time(s). Different systems will take different approaches.
When to drop the priority of tLow ...
No other tasks are blocked on any of the resources that tLow has. This may be due to timeouts or the releasing of resources.
No other tasks contributing to the raising the priority level of tLow are blocked on the resources that tLow has. This may be due to timeouts or the releasing of resources.
When there is a change in which tasks are waiting for the resource(s), drop the priority of tLow to match the priority of the highest priority level task blocked on its resource(s).
Method #2 is an improvement over method #1 in that it shortens the length of time that tLow has had its priority level bumped. Note that its priority level stays bumped at tHigh's priority level during this period.
Method #3 allows the priority level of tLow to step down in increments if necessary instead of in one all-or-nothing step.
Different systems will implement different methods depending upon what factors they consider important.
memory footprint
complexity
real time responsiveness
developer knowledge
Hope this helps.
Priority inversion is a problem, not a solution. The typical example is a low priority process acquiring a resource that a high priority process needs, and then being preempted by a medium priority process, so the high priority process is blocked on the resource while the medium priority one finishes (effectively being executed with a lower priority).
A rather famous example was the problem experienced by the Mars Pathfinder rover: http://www.cs.duke.edu/~carla/mars.html, it's a pretty interesting read.
Suppose an application has three threads:
Thread 1 has high priority.
Thread 2 has medium priority.
Thread 3 has low priority.
Let's assume that Thread 1 and Thread 3 share the same critical section code
Thread 1 and thread 2 are sleeping or blocked at the beginning of the example. Thread 3 runs and enters a critical section.
At that moment, thread 2 starts running, preempting thread 3 because thread 2 has a higher priority. So, thread 3 continues to own a critical section.
Later, thread 1 starts running, preempting thread 2. Thread 1 tries to enter the critical section that thread 3 owns, but because it is owned by another thread, thread 1 blocks, waiting for the critical section.
At that point, thread 2 starts running because it has a higher priority than thread 3 and thread 1 is not running. Thread 3 never releases the critical section that thread 1 is waiting for because thread 2 continues to run.
Therefore, the highest-priority thread in the system, thread 1, becomes blocked waiting for lower-priority threads to run.
It is the problem rather than the solution.
It describes the situation that when low-priority threads obtain locks during their work, high-priority threads will have to wait for them to finish (which might take especially long since they are low-priority). The inversion here is that the high-priority thread cannot continue until the low-priority thread does, so in effect it also has low priority now.
A common solution is to have the low-priority threads temporarily inherit the high priority of everyone who is waiting on locks they hold.
[ Assume, Low process = LP, Medium Process = MP, High process = HP ]
LP is executing a critical section. While entering the critical section, LP must have acquired a lock on some object, say OBJ.
LP is now inside the critical section.
Meanwhile, HP is created. Because of higher priority, CPU does a context switch, and HP is now executing (not the same critical section, but some other code). At some point during HP's execution, it needs a lock on the same OBJ (may or may not be on the same critical section), but the lock on OBJ is still held by LP, since it was pre-empted while executing the critical section. LP cannot relinquish now because the process is in READY state, not RUNNING. Now HP is moved to BLOCKED / WAITING state.
Now, MP comes in, and executes its own code. MP does not need a lock on OBJ, so it keeps executing normally. HP waits for LP to release lock, and LP waits for MP to finish executing so that LP can come back to RUNNING state (.. and execute and release lock). Only after LP has released lock can HP come back to READY (and then go to RUNNING by pre-empting the low priority tasks.)
So, effectively it means that until MP finishes, LP cannot execute and hence HP cannot execute. So, it seems like HP is waiting for MP, even though they are not directly related through any OBJ locks. -> Priority Inversion.
A solution to Priority Inversion is Priority Inheritance -
increase the priority of a process (A) to the maximum priority of any
other process waiting for any resource on which A has a resource lock.
Let me make it very simple and clear. (This answer is based on the answers above but presented in crisp way).
Say there is a resource R and 3 processes. L, M, H. where p(L) < p(M) < p(H) (where p(X) is priority of X).
Say
L starts executing first and catch holds on R. (exclusive access to R)
H comes later and also want exclusive access to R and since L is holding it, H has to wait.
M comes after H and it doesn't need R. And since M has got everything it wants to execute it forces L to leave as it has high priority compared to L. But H cannot do this as it has a resource locked by L which it needs for execution.
Now making the problem more clear, actually the M should wait for H to complete as p(H) > p(M) which didn't happen and this itself is the problem. If many processes such as M come along and don't allow the L to execute and release the lock H will never execute. Which can be hazardous in time critical applications
And for solutions refer the above answers :)
Priority inversion is where a lower priority process gets ahold of a resource that a higher priority process needs, preventing the higher priority process from proceeding till the resource is freed.
eg:
FileA needs to be accessed by Proc1 and Proc2.
Proc 1 has a higher priority than Proc2, but Proc2 manages to open FileA first.
Normally Proc1 would run maybe 10 times as often as Proc2, but won't be able to do anything because Proc2 is holding the file.
So what ends up happening is that Proc1 blocks until Proc2 finishes with FileA, essentially their priorities are 'inverted' while Proc2 holds FileA's handle.
As far as 'Solving a problem' goes, priority inversion is a problem in itself if it keeps happening.
The worst case (most operating systems won't let this happen though) is if Proc2 wasn't allowed to run until Proc1 had. This would cause the system to lock as Proc1 would keep getting assigned CPU time, and Proc2 will never get CPU time, so the file will never be released.
Priority inversion occurs as such:
Given processes H, M and L where the names stand for high, medium and low priorities,
only H and L share a common resource.
Say, L acquires the resource first and starts running. Since H also needs that resource, it enters the waiting queue.
M doesn't share the resource and can start to run, hence it does. When L is interrupted by any means, M takes the running state since it has higher priority and it is running on the instant that interrupt happens.
Although H has higher priority than M, since it is on the waiting queue, it cannot acquire the resource, implying a lower priority than even M.
After M finishes, L will again take over CPU causing H to wait the whole time.
Priority Inversion can be avoided if the blocked high priority thread transfers its high priority to the low priority thread that is holding onto the resource.
A scheduling challenge arises when a higher-priority process needs to read or modify kernel data that are currently being accessed by a lower-priority process—or a chain of lower-priority processes. Since kernel data are typically protected with a lock, the higher-priority process will have to wait for a lower-priority one to finish with the resource. The situation becomes more complicated if the lower-priority process is preempted in favor of another process with a higher priority. As an example, assume we have three processes—L, M, and H—whose priorities follow the order L < M < H. Assume that process H requires resource R,which is currently being accessed by process L.Ordinarily,process H would wait for L to finish using resource R. However, now suppose that process M becomes runnable, thereby preempting process L. Indirectly, a process with a lower priority—process M—has affected how long process H must wait for L to relinquish resource R. This problem is known as priority inversion.It occurs only in systems with more than two priorities,so one solution is to have only two priorities.That is insufficient for most general-purpose operating systems, however. Typically these systems solve the problem by implementing a priority-inheritance protocol. According to this protocol, all processes that are accessing resources needed by a higher-priority process inherit the higher priority until they are finished with the resources in question.When they are finished,their priorities revert to their original values. In the example above, a priority-inheritance protocol would allow process L to temporarily inherit the priority of process H,thereby preventing process M from preempting its execution. When process L had finished using resource R,it would relinquish its inherited priority from H and assume its original priority.Because resource R would now be available, process H—not M—would run next.
Reference :ABRAHAM SILBERSCHATZ
Consider a system with two processes,H with high priority and L with low priority. The scheduling rules are such that H is run whenever it is in ready state because of its high priority. At a certain moment, with L in its critical region, H becomes ready to run (e.g., an I/O operation completes). H now begins busy waiting, but since L is never scheduled while H is running, L never gets the chance to leave the critical section. So H loops forever.
This situation is called Priority Inversion. Because higher priority process is waiting on lower priority process.
How many tasks are needed for priority inversion to happen??..As per as my understanding we need atleast 3.....or can we have it only with 2 task??
I actually went through a book : modern operating system by Andrew Tanenbaum . I knew only when 3 task are there in some patern as u all know ..a priority inversion can happen..however ..I found the book says only 2 two task 1-low and the 1-high can also cause the same..so I am confused...
You need one high-priority task which waits for a resource held by a low-priority task, while a mid-priority task is running.
So yes, you need three.
priority inversion can occur with 2 thread also,
Ex:-
Higer priority task waiting on a low priority task which holds a spinlock effectively disabling preemption of the task by higher priority one.