I am new to multithreading paradigm. While learning concurrency, every source I found says:
"The difference between mutex and binary semaphore is the ownership
i.e. a mutex can be signaled for release by only the thread who
created it while a semaphore can be signaled any thread"
Considering a scenario where thread A has acquired a binary semaphore lock on resource x and processing it. If any thread can call a release signal for lock on x, doesn't this open a possibility of any thread calling a release on the lock amidst the time when thread A was using x.
Isn't there a scope of inconsistency in this or am I missing something?
Of course, if threads are arbitrarily acquiring or releasing a semaphore, the result would be disastrous and the fact, that implementations do not prevent this, does not imply that this is a useful scenario.
However, there might be real use cases if the involved threads use another mechanism to coordinate themselves while using the semaphore to hold these threads off which do not participate in these coordination.
Imagine you expand a use case where one thread acquires the semaphore for performing a task to parallel execution of said task. After acquiring the semaphore, several worker threads are spawned, each of them working on a different part of the data, thus naturally working non-interfering. Then the last worker thread releases the semaphore which elides the need for another communication between the initiating thread and the worker threads. Of course, this requires the worker thread to detect whether it is the last one, but a simple atomic integer holding the number of active workers would be sufficient.
Related
I'm motivated with this citation from "Concepts in Programming Languages" by John C. Mitchell:
"Atomicity prevents individual statements of one wait procedure from being
interleaved with individual statements of another wait on the same semaphore."
Wait and signal operations need to be atomic which is often enforced by some "lower" level mechanism of acquiring lock - disabling interrupts, disabling preemption, test and set ... But, conceptually, how these locks can be in some way "private" for each semaphore instance?
In other words, is it allowed for example that one thread acquires lock at the beginning and later be preempted in the middle of executing wait operation on one semaphore, and after that another thread acquires lock at the beginning of wait operation on some other semaphore and enters in the body of its wait operation, so that two thread are in the wait operations on different semaphores at the same time? Or, shortly, whether the wait operations on two different semaphores mutually exclusive?
My point is, if thread acquires lock in wait operation on one semaphore s1, is it allowed for another thread to acquire lock at the same time in wait operation on another semaphore s2? I'm emphasizing that these are two different semaphore instances, not the same one.
For example:
class Semaphore {
...
public:
void wait();
...
}
void Semaphore::wait(){
lock();
//POINT OF CONTINUATION FOR THREAD 2!//
if(--val<0){
//POINT OF PREEMPTION FOR THREAD 1!//
block();
}
unlock();
}
Semaphore s1;
Semaphore s2:
...
So...
Is it allowed at some point of execution that one thread be preempted while executing wait operation on semaphore s1 at //POINT OF PREEMPTION FOR THREAD 1!// , and control transfers to another thread which executes wait operation of semaphore s2 at //POINT OF CONTINUATION FOR THREAD 2!//...
...or...
Is it allowed for instructions of wait operation from one semaphore to be interleaved with instruction of wait operation from another semaphore?
..or...
Is it allowed for more than one threads to be in wait operations on different semaphores at the same time?
Sorry for my wordiness but I really struggle to clarify my question. Thanks in advance.
Yes, it's allowed. One of the reasons you would use two different locks, rather than using the same lock for everything, is to avoid unnecessary dependencies like this.
Is it allowed for instructions of wait operation from one semaphore to be interleaved with instruction of wait operation from another semaphore?
Absolutely.
Is it allowed for more than one threads to be in wait operations on different semaphores at the same time?
Absolutely.
Prohibiting any of these things would hurt performance significantly for no benefit. Contention is the enemy of multi-threaded performance.
I read one of the differences between semaphore and mutex is in case of mutex the process/thread (which ever is having the lock) can only release the lock. But in the case of the semaphore any other process can release the semaphore. My doubt arises when a process that does not have the semaphore with it can release the semaphore. What is the use of having a semaphore?
Let's say I have two processes A and B. Assume process A is having a semaphore with it and executing some critical task. Now let us say process B sends a signal to release the semaphore. In this scenario, will process A release the semaphore even if it is executing some critical task?
You are making half-sense. It is not about ownership. Partner-release in semaphores (and mutexes) is usable, for instance, in my favorite interview question of thread ping-pong. As a matter of fact, I have specifically tried to partner-release a mutex on 3 implementations available to me at a time (Linux/Solaris/AIX) and partner-release did work for mutexes as expected - i.e. mutex was successsfully released and threads blocking on it resumed execution. However, this is, of course, prohibited by Posix.
I think you might be confused on the whole set of differences between a semaphore and a mutex. A mutex provides mutual exclusion. A semaphore counts until it reaches a level where it starts excluding. A semaphore that counted to one would give similar semantics to a mutex though.
A good example would be a television set. Only so many people can watch the same television set, so protecting it with a semaphore would make sense. Anyone can stop watching the television. The remote control for the television can only be operated by one person at a time though, so you could protect it with a mutex.
Some reading...
https://en.wikipedia.org/wiki/Mutual_exclusion
https://en.wikipedia.org/wiki/Semaphore_%28programming%29
"Let's say I have two processes A and B. Assume process A is having a semaphore with it and executing some critical task. Now let us say process B sends a signal to release the semaphore. In this scenario, will process A release the semaphore even if it is executing some critical task?"
One key point to note here is the role of OS kernel. Process B can't send a signal to Process A 'to release the semaphore'. What it can do is request the kernel to give it access to the resource. Process A had requested the kernel and the kernel granted it access to the resource.
Now process A, after it finishes its job, will let the kernel know that it is done with the resource and then kernel grants access to B.
"My doubt arises when a process that does not have the semaphore with it can release the semaphore. What is the use of having a semaphore?"
The key difference between a mutex and a semaphore is, a semaphore serializes access to multiple instances of a resource. Mutex does the same when there is one instance of the resource.
A count is maintained by kernel in case of semaphore and mutex is a special case where the count is 1.
Consider the processes as customers waiting in line at a bank.
The use of semaphore is analogous to the case where there are multiple tellers serving the customers. Usage of mutex is analogous to the case where there is just one teller.
Say there are processes A, B and C that need concurrent access to a resource (lock, file or a data structure in memory, etc.). Further suppose there are 2 instances of the resource. So at most two processes can be granted access at a time.
Process A requests access to an instance of the resource following the required semantics. This request to the kernel involves data structures to identify the resource and maximum number of instances as 2. kernel creates the semaphore with a count of 2, grants A access to the resource and decrements the count to 1, because now only one other process can get access.
Now process B requests access to the resource by following the same semantics. Kernel grants it access and decrements the count to 0.
Now process C requests access, but kernel keeps it in waiting state, because count is 0 and no more than 2 processes can get concurrent access.
Process A is done with the resource and lets kernel know. Kernel notices this and grants access to process C that has been waiting.
In case of mutex, kernel grants access to the resource only one process at a time.
A normal binary semaphore is basically used for synchronization. However, the mutex is for exclusive access to a resource. A mutex is a special variant of semaphore that allows only one locker at a time and with more stringency on ownership than a normal semaphore such as the mutex should be released only by the thread that acquired it. Also, please note that in case of pthreads, fast mutex may not check for this error related to ownership, whereas the error checking mutex shall return error.
For the query related to 2 process A and B, the Process A shall intimate via kernel that it is done with its critical work so that the resource can be made available for waiting processes like B.
You could find some related information in this link too :
When should we use mutex and when should we use semaphore
There is no such thing as "having" a semaphore. Semaphores don't have ownership like mutexes do. The code you describe would simply be buggy. Mutexes won't work if your code is buggy either.
Consider the most classic example of a semaphore -- allowing one train at a time on a section of track. You could implement this with a mutex if the train is a thread. The train would lock the track mutex before going on the track and unlock it after leaving the track.
But what if the train itself is multi-threaded? Which thread should own the track?
And what if the signalling devices are the threads, not the train? Here, the signalling device that detects the train entering the track has to lock the track while the signalling device that detects the train leaving the track has to unlock it.
Mutexes are suitable for cases where there is something that is owned by a particular thread for a short period of time. That thread can "own" the mutex. Semaphores are useful for cases where there is no thread to own anything or nothing for the thread to own.
For example, suppose you are using atomic spinlock on an integer flag to ensure only one thread modifies the wait-queue that the mutex maintains at any given time. When a thread tries to lock the mutex, we want it to enqueue itself and set the flag to zero before it blocks itself and the unlocker to dequeue a thread from the queue and set it to runnable.
Consider only two threads to be present, one locking and the other releasing the mutex at the same time. if the locker was preempted after it added himself to the queue and set the flag to zero (but not blocked itself yet) and the unlocker then tried to dequeue and make the thread runnable, it wouldn't be useful since the thread hasn't blocked itself yet. So the make-runnable call would be waste but more importantly, the locker thread would then block itself after that and would remain blocked forever.
How is this atomicity achieved to ensure correctness? A similar scenario can be imagined in condition variables with the release of mutex and blocking itself.
I have an interesting problem related to Java thread live lock. Here it goes.
There are four global locks - L1,L2,L3,L4
There are four threads - T1, T2, T3, T4
T1 requires locks L1,L2,L3
T2 requires locks L2
T3 required locks L3,L4
T4 requires locks L1,L2
So, the pattern of the problem is - Any of the threads can run and acquire the locks in any order. If any of the thread detects that a lock which it needs is not available, it release all other locks it had previously acquired waits for a fixed time before retrying again. The cycle repeats giving rise to a live lock condition.
So, to solve this problem, I have two solutions in mind
1) Let each thread wait for a random period of time before retrying.
OR,
2) Let each thread acquire all the locks in a particular order ( even if a thread does not require all the
locks)
I am not convinced that these are the only two options available to me. Please advise.
Have all the threads enter a single mutex-protected state-machine whenever they require and release their set of locks. The threads should expose methods that return the set of locks they require to continue and also to signal/wait for a private semaphore signal. The SM should contain a bool for each lock and a 'Waiting' queue/array/vector/list/whatever container to store waiting threads.
If a thread enters the SM mutex to get locks and can immediately get its lock set, it can reset its bool set, exit the mutex and continue on.
If a thread enters the SM mutex and cannot immediately get its lock set, it should add itself to 'Waiting', exit the mutex and wait on its private semaphore.
If a thread enters the SM mutex to release its locks, it sets the lock bools to 'return' its locks and iterates 'Waiting' in an attempt to find a thread that can now run with the set of locks available. If it finds one, it resets the bools appropriately, removes the thread it found from 'Waiting' and signals the 'found' thread semaphore. It then exits the mutex.
You can twiddle with the algorithm that you use to match up the available set lock bools with waiting threads as you wish. Maybe you should release the thread that requires the largest set of matches, or perhaps you would like to 'rotate' the 'Waiting' container elements to reduce starvation. Up to you.
A solution like this requires no polling, (with its performance-sapping CPU use and latency), and no continual aquire/release of multiple locks.
It's much easier to develop such a scheme with an OO design. The methods/member functions to signal/wait the semaphore and return the set of locks needed can usually be stuffed somewhere in the thread class inheritance chain.
Unless there is a good reason (performance wise) not to do so,
I would unify all locks to one lock object.
This is similar to solution 2 you suggested, only more simple in my opinion.
And by the way, not only is this solution more simple and less bug proned,
The performance might be better than solution 1 you suggested.
Personally, I have never heard of Option 1, but I am by no means an expert on multithreading. After thinking about it, it sounds like it will work fine.
However, the standard way to deal with threads and resource locking is somewhat related to Option 2. To prevent deadlocks, resources need to always be acquired in the same order. For example, if you always lock the resources in the same order, you won't have any issues.
Go with 2a) Let each thread acquire all of the locks that it needs (NOT all of the locks) in a particular order; if a thread encounters a lock that isn't available then it releases all of its locks
As long as threads acquire their locks in the same order you can't have deadlock; however, you can still have starvation (a thread might run into a situation where it keeps releasing all of its locks without making forward progress). To ensure that progress is made you can assign priorities to threads (0 = lowest priority, MAX_INT = highest priority) - increase a thread's priority when it has to release its locks, and reduce it to 0 when it acquires all of its locks. Put your waiting threads in a queue, and don't start a lower-priority thread if it needs the same resources as a higher-priority thread - this way you guarantee that the higher-priority threads will eventually acquire all of their locks. Don't implement this thread queue unless you're actually having problems with thread starvation, though, because it's probably less efficient than just letting all of your threads run at once.
You can also simplify things by implementing omer schleifer's condense-all-locks-to-one solution; however, unless threads other than the four you've mentioned are contending for these resources (in which case you'll still need to lock the resources from the external threads), you can more efficiently implement this by removing all locks and putting your threads in a circular queue (so your threads just keep running in the same order).
What are the major differences between a Monitor and a Semaphore?
A Monitor is an object designed to be accessed from multiple threads. The member functions or methods of a monitor object will enforce mutual exclusion, so only one thread may be performing any action on the object at a given time. If one thread is currently executing a member function of the object then any other thread that tries to call a member function of that object will have to wait until the first has finished.
A Semaphore is a lower-level object. You might well use a semaphore to implement a monitor. A semaphore essentially is just a counter. When the counter is positive, if a thread tries to acquire the semaphore then it is allowed, and the counter is decremented. When a thread is done then it releases the semaphore, and increments the counter.
If the counter is already zero when a thread tries to acquire the semaphore then it has to wait until another thread releases the semaphore. If multiple threads are waiting when a thread releases a semaphore then one of them gets it. The thread that releases a semaphore need not be the same thread that acquired it.
A monitor is like a public toilet. Only one person can enter at a time. They lock the door to prevent anyone else coming in, do their stuff, and then unlock it when they leave.
A semaphore is like a bike hire place. They have a certain number of bikes. If you try and hire a bike and they have one free then you can take it, otherwise you must wait. When someone returns their bike then someone else can take it. If you have a bike then you can give it to someone else to return --- the bike hire place doesn't care who returns it, as long as they get their bike back.
Following explanation actually explains how wait() and signal() of monitor differ from P and V of semaphore.
The wait() and signal() operations on condition variables in a monitor are similar to P and V operations on counting semaphores.
A wait statement can block a process's execution, while a signal statement can cause another process to be unblocked. However, there are some differences between them. When a process executes a P operation, it does not necessarily block that process because the counting semaphore may be greater than zero. In contrast, when a wait statement is executed, it always blocks the process. When a task executes a V operation on a semaphore, it either unblocks a task waiting on that semaphore or increments the semaphore counter if there is no task to unlock. On the other hand, if a process executes a signal statement when there is no other process to unblock, there is no effect on the condition variable. Another difference between semaphores and monitors is that users awaken by a V operation can resume execution without delay. Contrarily, users awaken by a signal operation are restarted only when the monitor is unlocked. In addition, a monitor solution is more structured than the one with semaphores because the data and procedures are encapsulated in a single module and that the mutual exclusion is provided automatically by the implementation.
Link: here for further reading. Hope it helps.
Semaphore allows multiple threads (up to a set number) to access a shared object. Monitors allow mutually exclusive access to a shared object.
Monitor
Semaphore
One Line Answer:
Monitor: controls only ONE thread at a time can execute in the monitor. (need to acquire lock to execute the single thread)
Semaphore: a lock that protects a shared resource. (need to acquire the lock to access resource)
A semaphore is a signaling mechanism used to coordinate between threads. Example: One thread is downloading files from the internet and another thread is analyzing the files. This is a classic producer/consumer scenario. The producer calls signal() on the semaphore when a file is downloaded. The consumer calls wait() on the same semaphore in order to be blocked until the signal indicates a file is ready. If the semaphore is already signaled when the consumer calls wait, the call does not block. Multiple threads can wait on a semaphore, but each signal will only unblock a single thread.
A counting semaphore keeps track of the number of signals. E.g. if the producer signals three times in a row, wait() can be called three times without blocking. A binary semaphore does not count but just have the "waiting" and "signalled" states.
A mutex (mutual exclusion lock) is a lock which is owned by a single thread. Only the thread which have acquired the lock can realease it again. Other threads which try to acquire the lock will be blocked until the current owner thread releases it. A mutex lock does not in itself lock anything - it is really just a flag. But code can check for ownership of a mutex lock to ensure that only one thread at a time can access some object or resource.
A monitor is a higher-level construct which uses an underlying mutex lock to ensure thread-safe access to some object. Unfortunately the word "monitor" is used in a few different meanings depending on context and platform and context, but in Java for example, a monitor is a mutex lock which is implicitly associated with an object, and which can be invoked with the synchronized keyword. The synchronized keyword can be applied to a class, method or block and ensures only one thread can execute the code at a time.
Semaphore :
Using a counter or flag to control access some shared resources in a concurrent system, implies use of Semaphore.
Example:
A counter to allow only 50 Passengers to acquire the 50 seats (Shared resource) of any Theatre/Bus/Train/Fun ride/Classroom. And to allow a new Passenger only if someone vacates a seat.
A binary flag indicating the free/occupied status of any Bathroom.
Traffic lights are good example of flags. They control flow by regulating passage of vehicles on Roads (Shared resource)
Flags only reveal the current state of Resource, no count or any other information on the waiting or running objects on the resource.
Monitor :
A Monitor synchronizes access to an Object by communicating with threads interested in the object, asking them to acquire access or wait for some condition to become true.
Example:
A Father may acts as a monitor for her daughter, allowing her to date only one guy at a time.
A school teacher using baton to allow only one child to speak in the class.
Lastly a technical one, transactions (via threads) on an Account object synchronized to maintain integrity.
When a semaphore is used to guard a critical region, there is no direct relationship between the semaphore and the data being protected. This is part of the reason why semaphores may be dispersed around the code, and why it is easy to forget to call wait or notify, in which case the result will be, respectively, to violate mutual exclusion or to lock the resource permanently.
In contrast, niehter of these bad things can happen with a monitor. A monitor is tired directly to the data (it encapsulates the data) and, because the monitor operations are atomic actions, it is impossible to write code that can access the data without calling the entry protocol. The exit protocol is called automatically when the monitor operation is completed.
A monitor has a built-in mechanism for condition synchronisation in the form of condition variable before proceeding. If the condition is not satisfied, the process has to wait until it is notified of a change in the condition. When a process is waiting for condition synchronisation, the monitor implementation takes care of the mutual exclusion issue, and allows another process to gain access to the monitor.
Taken from The Open University M362 Unit 3 "Interacting process" course material.