This is a rather basic question regarding the memory ordering guarantees inside the Rust async ecosystems. However, I don't seem to find a clear answer anywhere.
C++ memory ordering specifies the memory ordering from a thread-based perspective:
Release-Acquire ordering
If an atomic store in thread A is tagged memory_order_release and an atomic load in thread B from the same variable is tagged memory_order_acquire, all memory writes (non-atomic and relaxed atomic) that happened-before the atomic store from the point of view of thread A, become visible side-effects in thread B. That is, once the atomic load is completed, thread B is guaranteed to see everything thread A wrote to memory.
Various Rust async executors supports moving tasks across thread at .await points. A natural scenario would be:
Task A was at Thread 1 and performed a Relaxed store at variable x.
Task A hit a .await point
Task A was moved to Thread 2 and awaked. And then performed a Release store at variable y.
Task B was at Thread 3 and performed an Acquire load at variable y.
The question is: Is the Relaxed store from Thread 1 guaranteed to be visible in Task B after the Acquire load? E.g. if Task B at Thread 3 performed an Relaxed load at variable x afterwards, is it guaranteed that Task A's side effect will be visible?
This should depend on the implementation of synchronization mechanisms inside each async executor's task scheduler. Anyone familiar with them is appreciated.
Comment from Tokio maintainer:
When the task is moved from one thread to another, there is a synchronizes-with relation between the things before and after the .await. Depending on the exact circumstances, the synchronization comes from a mutex or an atomic using acquire & release semantics.
In C++ memory spec, synchronizse-with guarantees inter-thread happens-before. Therefore at least in Tokio, it should be safe to ignore memory ordering impact from a .await point.
I'm very new to Dart and stilling learning it. As I understand, Dart executes code in different isolates. An isolate could start up another isolate to execute some long-running code. For each isolate, there is a thread and some memory allocated for it. These isolates are isolated like a bunch of little VMs.
I also read from the Dart document that Dart is a single threaded language. But, think about it, each isolate has its own thread. If isolate A has thread t1 and isolate B has thread t2, t1 and t2 are not the same thread, right?
If t1 and t2 are the same thread, then t1 and t2 can't execute code at the same time, which is ridiculous. So, t1 and t2 must be different thread.
If so, why we say Dart is a single-threaded language?
Yes and no.
"Yes" in the sense that you don't have to worry about locks or mutexes.
"No" in the sense that you list.
Dart tries to offer some of the benefits of multi-threading with isolates while avoiding all of the issues with shared memory multi-threading.
I was just wondering, suppose PC is having multi cores. There are three threads running in three different cores. Thread(T1) has acquired spin lock(S) in core(C1) and acquired lock by T1, same time T2 and T3 threads running in core C2 and C3 try to acquire lock and waiting for release of lock. once T1 thread releases lock which thread will acquire lock either T2 or T3? I am considering same priority of T2 and T3,and also waiting in different cores same time.
The linux kernel uses MCS spin locks. The gist is that waiters end up adding themselves to a queue. However, if there are 2 threads doing this, there are no guarantees as to who is going to succeed first.
Similarly for more simplistic spin locks where the code just tries to flip the "taken" bit, there are no guarantees whatsoever. However, certain hardware characteristics can make it so certain cores have an easier time than others (if they share the same socket).
You want to read https://www.kernel.org/pub/linux/kernel/people/paulmck/perfbook/perfbook.html
I repeat: if 2 different threads compete for a lock, there is no guaranteed order in which they will take it and looking for one is wrong in the first place.
This is an interview question.
How to detect and find out if a program is in deadlock? Are there some tools that can be used to do that on Linux/Unix systems?
My idea:
If a program makes no progress and its status is running, it is deadlock. But, other reasons can also cause this problem. Open source tools are valgrind (halgrind) can do that. Right?
If you suspect a deadlock, do a ps aux | grep <exe name>, if in output, the PROCESS STATE CODE is D (Uninterruptible sleep) means it is a deadlock.
Because as #daijo explained, say you have two threads T1 & T2 and two critical sections each protected by semaphores S1 & S2 then if T1 acquires S1 and T2 acquires S2 and after that they try to acquire the other lock before relinquishing the one already held by them, this will lead to a deadlock and on doing a ps aux | grep <exe name>, the process state code will be D (ie Uninterruptible sleep).
Tools:
Valgrind, Lockdep (linux kernel utility)
Check this link on types of deadlocks and how to avoid them :
http://cmdlinelinux.blogspot.com/2014/01/linux-kernel-deadlocks-and-how-to-avoid.html
Edit: ps aux output D "could" mean process is in deadlock, from this redhat doc:
Uninterruptible Sleep State
An Uninterruptible sleep state is one
that won't handle a signal right away. It will wake only as a result
of a waited-upon resource becoming available or after a time-out
occurs during that wait (if the time-out is specified when the process
is put to sleep).
I would suggest you look at Helgrind: a thread error detector.
The simplest example of such a problem is as follows.
Imagine some shared resource R, which, for whatever reason, is guarded by two locks, L1 and L2, which must both be held when R is accessed.
Suppose a thread acquires L1, then L2, and proceeds to access R. The implication of this is that all threads in the program must acquire the two locks in the order first L1 then L2. Not doing so risks deadlock.
The deadlock could happen if two threads -- call them T1 and T2 -- both want to access R. Suppose T1 acquires L1 first, and T2 acquires L2 first. Then T1 tries to acquire L2, and T2 tries to acquire L1, but those locks are both already held. So T1 and T2 become deadlocked."
When should we use mutex and when should we use semaphore ?
Here is how I remember when to use what -
Semaphore:
Use a semaphore when you (thread) want to sleep till some other thread tells you to wake up. Semaphore 'down' happens in one thread (producer) and semaphore 'up' (for same semaphore) happens in another thread (consumer)
e.g.: In producer-consumer problem, producer wants to sleep till at least one buffer slot is empty - only the consumer thread can tell when a buffer slot is empty.
Mutex:
Use a mutex when you (thread) want to execute code that should not be executed by any other thread at the same time. Mutex 'down' happens in one thread and mutex 'up' must happen in the same thread later on.
e.g.: If you are deleting a node from a global linked list, you do not want another thread to muck around with pointers while you are deleting the node. When you acquire a mutex and are busy deleting a node, if another thread tries to acquire the same mutex, it will be put to sleep till you release the mutex.
Spinlock:
Use a spinlock when you really want to use a mutex but your thread is not allowed to sleep.
e.g.: An interrupt handler within OS kernel must never sleep. If it does the system will freeze / crash. If you need to insert a node to globally shared linked list from the interrupt handler, acquire a spinlock - insert node - release spinlock.
A mutex is a mutual exclusion object, similar to a semaphore but that only allows one locker at a time and whose ownership restrictions may be more stringent than a semaphore.
It can be thought of as equivalent to a normal counting semaphore (with a count of one) and the requirement that it can only be released by the same thread that locked it(a).
A semaphore, on the other hand, has an arbitrary count and can be locked by that many lockers concurrently. And it may not have a requirement that it be released by the same thread that claimed it (but, if not, you have to carefully track who currently has responsibility for it, much like allocated memory).
So, if you have a number of instances of a resource (say three tape drives), you could use a semaphore with a count of 3. Note that this doesn't tell you which of those tape drives you have, just that you have a certain number.
Also with semaphores, it's possible for a single locker to lock multiple instances of a resource, such as for a tape-to-tape copy. If you have one resource (say a memory location that you don't want to corrupt), a mutex is more suitable.
Equivalent operations are:
Counting semaphore Mutual exclusion semaphore
-------------------------- --------------------------
Claim/decrease (P) Lock
Release/increase (V) Unlock
Aside: in case you've ever wondered at the bizarre letters (P and V) used for claiming and releasing semaphores, it's because the inventor was Dutch. In that language:
Probeer te verlagen: means to try to lower;
Verhogen: means to increase.
(a) ... or it can be thought of as something totally distinct from a semaphore, which may be safer given their almost-always-different uses.
It is very important to understand that a mutex is not a semaphore with count 1!
This is the reason there are things like binary semaphores (which are really semaphores with count 1).
The difference between a Mutex and a Binary-Semaphore is the principle of ownership:
A mutex is acquired by a task and therefore must also be released by the same task.
This makes it possible to fix several problems with binary semaphores (Accidental release, recursive deadlock, and priority inversion).
Caveat: I wrote "makes it possible", if and how these problems are fixed is up to the OS implementation.
Because the mutex has to be released by the same task it is not very good for the synchronization of tasks. But if combined with condition variables you get very powerful building blocks for building all kinds of IPC primitives.
So my recommendation is: if you got cleanly implemented mutexes and condition variables (like with POSIX pthreads) use these.
Use semaphores only if they fit exactly to the problem you are trying to solve, don't try to build other primitives (e.g. rw-locks out of semaphores, use mutexes and condition variables for these)
There is a lot of misunderstanding between mutexes and semaphores. The best explanation I found so far is in this 3-Part article:
Mutex vs. Semaphores – Part 1: Semaphores
Mutex vs. Semaphores – Part 2: The Mutex
Mutex vs. Semaphores – Part 3 (final part): Mutual Exclusion Problems
While #opaxdiablo answer is totally correct I would like to point out that the usage scenario of both things is quite different. The mutex is used for protecting parts of code from running concurrently, semaphores are used for one thread to signal another thread to run.
/* Task 1 */
pthread_mutex_lock(mutex_thing);
// Safely use shared resource
pthread_mutex_unlock(mutex_thing);
/* Task 2 */
pthread_mutex_lock(mutex_thing);
// Safely use shared resource
pthread_mutex_unlock(mutex_thing); // unlock mutex
The semaphore scenario is different:
/* Task 1 - Producer */
sema_post(&sem); // Send the signal
/* Task 2 - Consumer */
sema_wait(&sem); // Wait for signal
See http://www.netrino.com/node/202 for further explanations
See "The Toilet Example" - http://pheatt.emporia.edu/courses/2010/cs557f10/hand07/Mutex%20vs_%20Semaphore.htm:
Mutex:
Is a key to a toilet. One person can have the key - occupy the toilet - at the time. When finished, the person gives (frees) the key to the next person in the queue.
Officially: "Mutexes are typically used to serialise access to a section of re-entrant code that cannot be executed concurrently by more than one thread. A mutex object only allows one thread into a controlled section, forcing other threads which attempt to gain access to that section to wait until the first thread has exited from that section."
Ref: Symbian Developer Library
(A mutex is really a semaphore with value 1.)
Semaphore:
Is the number of free identical toilet keys. Example, say we have four toilets with identical locks and keys. The semaphore count - the count of keys - is set to 4 at beginning (all four toilets are free), then the count value is decremented as people are coming in. If all toilets are full, ie. there are no free keys left, the semaphore count is 0. Now, when eq. one person leaves the toilet, semaphore is increased to 1 (one free key), and given to the next person in the queue.
Officially: "A semaphore restricts the number of simultaneous users of a shared resource up to a maximum number. Threads can request access to the resource (decrementing the semaphore), and can signal that they have finished using the resource (incrementing the semaphore)."
Ref: Symbian Developer Library
Mutex is to protect the shared resource.
Semaphore is to dispatch the threads.
Mutex:
Imagine that there are some tickets to sell. We can simulate a case where many people buy the tickets at the same time: each person is a thread to buy tickets. Obviously we need to use the mutex to protect the tickets because it is the shared resource.
Semaphore:
Imagine that we need to do a calculation as below:
c = a + b;
Also, we need a function geta() to calculate a, a function getb() to calculate b and a function getc() to do the calculation c = a + b.
Obviously, we can't do the c = a + b unless geta() and getb() have been finished.
If the three functions are three threads, we need to dispatch the three threads.
int a, b, c;
void geta()
{
a = calculatea();
semaphore_increase();
}
void getb()
{
b = calculateb();
semaphore_increase();
}
void getc()
{
semaphore_decrease();
semaphore_decrease();
c = a + b;
}
t1 = thread_create(geta);
t2 = thread_create(getb);
t3 = thread_create(getc);
thread_join(t3);
With the help of the semaphore, the code above can make sure that t3 won't do its job untill t1 and t2 have done their jobs.
In a word, semaphore is to make threads execute as a logicial order whereas mutex is to protect shared resource.
So they are NOT the same thing even if some people always say that mutex is a special semaphore with the initial value 1. You can say like this too but please notice that they are used in different cases. Don't replace one by the other even if you can do that.
Trying not to sound zany, but can't help myself.
Your question should be what is the difference between mutex and semaphores ?
And to be more precise question should be, 'what is the relationship between mutex and semaphores ?'
(I would have added that question but I'm hundred % sure some overzealous moderator would close it as duplicate without understanding difference between difference and relationship.)
In object terminology we can observe that :
observation.1 Semaphore contains mutex
observation.2 Mutex is not semaphore and semaphore is not mutex.
There are some semaphores that will act as if they are mutex, called binary semaphores, but they are freaking NOT mutex.
There is a special ingredient called Signalling (posix uses condition_variable for that name), required to make a Semaphore out of mutex.
Think of it as a notification-source. If two or more threads are subscribed to same notification-source, then it is possible to send them message to either ONE or to ALL, to wakeup.
There could be one or more counters associated with semaphores, which are guarded by mutex. The simple most scenario for semaphore, there is a single counter which can be either 0 or 1.
This is where confusion pours in like monsoon rain.
A semaphore with a counter that can be 0 or 1 is NOT mutex.
Mutex has two states (0,1) and one ownership(task).
Semaphore has a mutex, some counters and a condition variable.
Now, use your imagination, and every combination of usage of counter and when to signal can make one kind-of-Semaphore.
Single counter with value 0 or 1 and signaling when value goes to 1 AND then unlocks one of the guy waiting on the signal == Binary semaphore
Single counter with value 0 to N and signaling when value goes to less than N, and locks/waits when values is N == Counting semaphore
Single counter with value 0 to N and signaling when value goes to N, and locks/waits when values is less than N == Barrier semaphore (well if they dont call it, then they should.)
Now to your question, when to use what. (OR rather correct question version.3 when to use mutex and when to use binary-semaphore, since there is no comparison to non-binary-semaphore.)
Use mutex when
1. you want a customized behavior, that is not provided by binary semaphore, such are spin-lock or fast-lock or recursive-locks.
You can usually customize mutexes with attributes, but customizing semaphore is nothing but writing new semaphore.
2. you want lightweight OR faster primitive
Use semaphores, when what you want is exactly provided by it.
If you dont understand what is being provided by your implementation of binary-semaphore, then IMHO, use mutex.
And lastly read a book rather than relying just on SO.
I think the question should be the difference between mutex and binary semaphore.
Mutex = It is a ownership lock mechanism, only the thread who acquire the lock can release the lock.
binary Semaphore = It is more of a signal mechanism, any other higher priority thread if want can signal and take the lock.
All the above answers are of good quality,but this one's just to memorize.The name Mutex is derived from Mutually Exclusive hence you are motivated to think of a mutex lock as Mutual Exclusion between two as in only one at a time,and if I possessed it you can have it only after I release it.On the other hand such case doesn't exist for Semaphore is just like a traffic signal(which the word Semaphore also means).
As was pointed out, a semaphore with a count of one is the same thing as a 'binary' semaphore which is the same thing as a mutex.
The main things I've seen semaphores with a count greater than one used for is producer/consumer situations in which you have a queue of a certain fixed size.
You have two semaphores then. The first semaphore is initially set to be the number of items in the queue and the second semaphore is set to 0. The producer does a P operation on the first semaphore, adds to the queue. and does a V operation on the second. The consumer does a P operation on the second semaphore, removes from the queue, and then does a V operation on the first.
In this way the producer is blocked whenever it fills the queue, and the consumer is blocked whenever the queue is empty.
A mutex is a special case of a semaphore. A semaphore allows several threads to go into the critical section. When creating a semaphore you define how may threads are allowed in the critical section. Of course your code must be able to handle several accesses to this critical section.
I find the answer of #Peer Stritzinger the correct one.
I wanted to add to his answer the following quote from the book Programming with POSIX Threads by David R Butenhof. On page 52 of chapter 3 the author writes (emphasis mine):
You cannot lock a mutex when the calling thread already has that mutex locked. The result of attempting to do so may be an error return (EDEADLK), or it may be a self-deadlock, where the unfortunate thread waits forever. You cannot unlock a mutex that is unlocked, or that is locked by another thread. Locked mutexes are owned by the thread that locks them. If you need an "unowned" lock, use a semaphore. Section 6.6.6 discusses semaphores)
With this in mind, the following piece of code illustrates the danger of using a semaphore of size 1 as a replacement for a mutex.
sem = Semaphore(1)
counter = 0 // shared variable
----
Thread 1
for (i in 1..100):
sem.lock()
++counter
sem.unlock()
----
Thread 2
for (i in 1..100):
sem.lock()
++counter
sem.unlock()
----
Thread 3
sem.unlock()
thread.sleep(1.sec)
sem.lock()
If only for threads 1 and 2, the final value of counter should be 200. However, if by mistake that semaphore reference was leaked to another thread and called unlock, than you wouldn't get mutual exclusion.
With a mutex, this behaviour would be impossible by definition.
Binary semaphore and Mutex are different. From OS perspective, a binary semaphore and counting semaphore are implemented in the same way and a binary semaphore can have a value 0 or 1.
Mutex -> Can only be used for one and only purpose of mutual exclusion for a critical section of code.
Semaphore -> Can be used to solve variety of problems. A binary semaphore can be used for signalling and also solve mutual exclusion problem. When initialized to 0, it solves signalling problem and when initialized to 1, it solves mutual exclusion problem.
When the number of resources are more and needs to be synchronized, we can use counting semaphore.
In my blog, I have discussed these topics in detail.
https://designpatterns-oo-cplusplus.blogspot.com/2015/07/synchronization-primitives-mutex-and.html