createmutex questions - c++

Let's say I call
h=CreateMutex(NULL,FALSE,"full");
y=WaitForSingleObject(h,INFINITE);
//Read from a queue (critical section)
ReleaseMutex(h);
What issues can arise that can lead to an access violation reading a location?
For example is it possible for multiple threads to enter that critical section as the same time?

Although you're storing the results of those functions in variables, you're not reading them to determine whether the functions succeeded. Perhaps you didn't create or open the given mutex, so h is 0. Or perhaps instead of acquiring ownership of the mutex, the the wait failed. In either case, you should call GetLastError to find out why, and then don't execute the protected section of code.
It's possible for a mutex to be abandoned. That means that the thread that previously owned the mutex was terminated before it released ownership of the mutex. (Only a mutex can be abandoned; critical sections and semaphores don't have thread affinity the way mutex objects do.) If that happens, you'll still be granted ownership of the mutex, but you can't really trust the validity of the data that the mutex is supposed to be protecting because the previous owner might not have left things in a stable state before it terminated.
If you call the functions correctly and check for errors, there's no way for multiple threads to enter a critical section simultaneously. That's the whole purpose of synchronization objects.

Related

std::condition_variable memory writes visibility

Does std::condition_variable::notify_one() or std::condition_variable::notify_all() guarantee that non-atomic memory writes in the current thread prior to the call will be visible in notified threads?
Other threads do:
{
std::unique_lock lock(mutex);
cv.wait(lock, []() { return values[threadIndex] != 0; });
// May a thread here see a zero value and therefore start to wait again?
}
Main thread does:
fillData(values); // All values are zero and all threads wait() before calling this.
cv.notify_all(); // Do need some memory fence or lock before this
// to ensure that new non-zero values will be visible
// in other threads immediately after waking up?
Doesn't notify_all() store some atomic value therefore enforcing memory ordering? I did not clarified it.
UPD: according to Superlokkus' answer and an answer here: we have to acquire a lock to ensure memory writes visibility in other threads (memory propagation), otherwise threads in my case may read zero values.
Also I missed this quote here about condition_variable, which specifically answers my question. Even an atomic variable has to be modified under a lock in a case when the modification must become visible immediately.
Even if the shared variable is atomic, it must be modified under the
mutex in order to correctly publish the modification to the waiting
thread.
I guess you are mixing up memory ordering of so called atomic values and the mechanisms of classic lock based synchronization.
When you have a datum which is shared between threads, lets say an int for example, one thread can not simply read it while the other thread might be write to it meanwhile. Otherwise we would have a data race.
To get around this for long time we used classic lock based synchronization:
The threads share at least a mutex and the int. To read or to write any thread has to hold the lock first, meaning they wait on the mutex. Mutexes are build so that they are fine that this can happen concurrently. If a thread wins gettting the mutex it can change or read the int and then should unlock it, so others can read/write too. Using a conditional variable like you used is just to make the pattern "readers wait for a change of a value by a writer" more efficient, they get woken up by the cv instead of periodically waiting on the lock, reading, and unlocking, which would be called busy waiting.
So because you hold the lock in any after waiting on the mutex or in you case, correctly (mutex is still needed) waiting on the conditional variable, you can change the int. And readers will read the new value after the writer was able to wrote it, never the old. UPDATE: However one thing if have to add, which might also be the cause of confusion: Conditional variables are subject for so called spurious wakeups. Meaning even though you write did not have notified any thread, a read thread might still wake up, with the mutex locked. So you have to check if you writer actually waked you up, which is usually done by the writer by changing another datum just to notify this, or if its suitable by using the same datum you already wanted to share. The lambda parameter overload of std::condition_variable::wait was just made to make the checking and going back to sleep code looking a bit prettier. Based on your question now I don't know if you want to use you values for this job.
However at snippet for the "main" thread is incorrect or incomplete:
You are not synchronizing on the mutex in order to change values.
You have to hold the lock for that, but notifying can be done without the lock.
std::unique_lock lock(mutex);
fillData(values);
lock.unlock();
cv.notify_all();
But these mutex based patters have some drawbacks and are slow, only one thread at a time can do something. This is were so called atomics, like std::atomic<int> came into play. They can be written and read at the same time without an mutex by multiple threads concurrently. Memory ordering is only a thing to consider there and an optimization for cases where you uses several of them in a meaningful way or you don't need the "after the write, I never see the old value" guarantee. However with it's default memory ordering memory_order_seq_cst you would also be fine.

Boost, mutex concept

I am new to multi-threading programming, and confused about how Mutex works. In the Boost::Thread manual, it states:
Mutexes guarantee that only one thread can lock a given mutex. If a code section is surrounded by a mutex locking and unlocking, it's guaranteed that only a thread at a time executes that section of code. When that thread unlocks the mutex, other threads can enter to that code region:
My understanding is that Mutex is used to protect a section of code from being executed by multiple threads at the same time, NOT protect the memory address of a variable. It's hard for me to grasp the concept, what happen if I have 2 different functions trying to write to the same memory address.
Is there something like this in Boost library:
lock a memory address of a variable, e.g., double x, lock (x); So
that other threads with a different function can not write to x.
do something with x, e.g., x = x + rand();
unlock (x)
Thanks.
The mutex itself only ensures that only one thread of execution can lock the mutex at any given time. It's up to you to ensure that modification of the associated variable happens only while the mutex is locked.
C++ does give you a way to do that a little more easily than in something like C. In C, it's pretty much up to you to write the code correctly, ensuring that anywhere you modify the variable, you first lock the mutex (and, of course, unlock it when you're done).
In C++, it's pretty easy to encapsulate it all into a class with some operator overloading:
class protected_int {
int value; // this is the value we're going to share between threads
mutex m;
public:
operator int() { return value; } // we'll assume no lock needed to read
protected_int &operator=(int new_value) {
lock(m);
value = new_value;
unlock(m);
return *this;
}
};
Obviously I'm simplifying that a lot (to the point that it's probably useless as it stands), but hopefully you get the idea, which is that most of the code just treats the protected_int object as if it were a normal variable.
When you do that, however, the mutex is automatically locked every time you assign a value to it, and unlocked immediately thereafter. Of course, that's pretty much the simplest possible case -- in many cases, you need to do something like lock the mutex, modify two (or more) variables in unison, then unlock. Regardless of the complexity, however, the idea remains that you centralize all the code that does the modification in one place, so you don't have to worry about locking the mutex in the rest of the code. Where you do have two or more variables together like that, you generally will have to lock the mutex to read, not just to write -- otherwise you can easily get an incorrect value where one of the variables has been modified but the other hasn't.
No, there is nothing in boost(or elsewhere) that will lock memory like that.
You have to protect the code that access the memory you want protected.
what happen if I have 2 different functions trying to write to the same
memory address.
Assuming you mean 2 functions executing in different threads, both functions should lock the same mutex, so only one of the threads can write to the variable at a given time.
Any other code that accesses (either reads or writes) the same variable will also have to lock the same mutex, failure to do so will result in indeterministic behavior.
It is possible to do non-blocking atomic operations on certain types using Boost.Atomic. These operations are non-blocking and generally much faster than a mutex. For example, to add something atomically you can do:
boost::atomic<int> n = 10;
n.fetch_add(5, boost:memory_order_acq_rel);
This code atomically adds 5 to n.
In order to protect a memory address shared by multiple threads in two different functions, both functions have to use the same mutex ... otherwise you will run into a scenario where threads in either function can indiscriminately access the same "protected" memory region.
So boost::mutex works just fine for the scenario you describe, but you just have to make sure that for a given resource you're protecting, all paths to that resource lock the exact same instance of the boost::mutex object.
I think the detail you're missing is that a "code section" is an arbitrary section of code. It can be two functions, half a function, a single line, or whatever.
So the portions of your 2 different functions that hold the same mutex when they access the shared data, are "a code section surrounded by a mutex locking and unlocking" so therefore "it's guaranteed that only a thread at a time executes that section of code".
Also, this is explaining one property of mutexes. It is not claiming this is the only property they have.
Your understanding is correct with respect to mutexes. They protect the section of code between the locking and unlocking.
As per what happens when two threads write to the same location of memory, they are serialized. One thread writes its value, the other thread writes to it. The problem with this is that you don't know which thread will write first (or last), so the code is not deterministic.
Finally, to protect a variable itself, you can find a near concept in atomic variables. Atomic variables are variables that are protected by either the compiler or the hardware, and can be modified atomically. That is, the three phases you comment (read, modify, write) happen atomically. Take a look at Boost atomic_count.

Writing to a mutex'ed shared resourced

I've a C++ list which is being processed by multiple thread.
Each thread creates a pthread_mutex_lock on the list so that other threads cannot "interfere" with the list. As a part of processing, each thread also push_back data on the list.
My question is - is push_back on a mutex-ed list a bad idea? Is the mutex still valid while the thread is pusing more data on the list? Most of the documentation/examples I've seen on pthread_mutex_lock are only doing "reading" so I am curious to know what happens the same thread which acquired lock, writes on the shared resource.
As long as only that particular thread is holding the lock, and no other thread can take this lock, writing should be fine. think of why a problem could happen? it wouldve been a problem if one thread was writing and the other was reading simultaneously. If a ball is yours, you can do anything with it right? things change when they're shared.
The mutex needs to be unique for the entire group of threads (i.e. all threads must use the same mutex). If you create a mutex for each thread, then you are not thread-safe at all, because each thread will wait on its own mutex and not be synchronized with the rest.
And yes an acquired mutex can be used safely to both read and write.

Confusion on CreateMutex

Let says I call CreateMutex.
HANDLE h;
h=CreateMutex(NULL, TRUE, NULL);
waitforsingleobject(h, INFINITE);
////Random Code
ReleaseMutex(h);
Assuming I have multiple threads running, the first thread to reach the function createmutex essentially blocks all the other threads from the //random code section until release mutex is called right?
It doesn't, because you have created an unnamed mutex (the third parameter is the name). Assuming the example code is run in multiple threads, each thread will create a new unnamed mutex, and will promptly get access to the critical section (Random Code), because they are only waiting for their own mutex.
To fix this, either let h be a global handle that all threads have access to, and call CreateMutex once outside the shared code, or provide CreateMutex with a name (third argument). In the latter case, subsequent calls to CreateMutex will return a handle to the existing mutex.
Since you don't specify a name for the mutex each thread will create its own distinct mutex other threads will not be aware of.
A critical section would be a better choice in your scenario.
Well taking into account that you create an unnamed mutex, each thread will create an unnamed mutex of its own and assume ownership upon creation. Since each thread has its own mutex all threads will be able to run in parallel.
So you should create the mutex once for all threads if you want it to be unnamed. From looking at your code you should also check if mutex creation succeeded.
Then in the function that gets called from multiple threads call WaitForSingleObject on the previously created mutex.
Check the returned result if it returned because you got ownersip of the mutex or the wait terminanted because the mutex was abandoned.
Make sure you call ReleaseMutex on the mutex handle if you got ownersip. Probably you should think of using RAII to manage the mutex for many reasons (to make sure the mutex is properly released when exceptions occur or when a return statement is inserted before the ReleaseMutex to name just a few).
Sometime when the mutex is no longer needed make sure you call CloseHandle on the MutexHandle.
For an example illustrating how to use mutexes see here: "Using Mutex Objects"
For a basic multithreading wrap see here: Using Threads
Only if the same mutex reference is shared across all threads/processes. You're not sharing it, you're creating it each time. You'll need to name it so that they all get the same mutex for your code to work.

How to implement a recursive MRSW lock?

I need a fully-recursive multiple-reader/single-writer lock (shared mutex) for my project - I don't agree with the notion that if you have complete const-correctness you shouldn't need them (there was some discussion about that on the boost mailing list), in my case the lock should protect a completely transparent cache which would be mutable in any case.
As for the semantics of recursive MRSW locks, I think the only ones that make sense are that acquiring a exclusive lock in addition to a shared one temporarily releases the shared one, to be reacquired when the exclusive one is released.
Has the somewhat strange effect that unlocking can wait but I can live with that - writing rarely happens anyway and recursive locking usually only happens through recursive code paths, in which case the caller has to be prepared that the call might wait in any case. To avoid it one can still simply upgrade the lock instead of using recursive locking.
Acquiring a shared lock on top of an exclusive one should obviously just increases the lock count.
So the question becomes - how should I implement it? The usual approach with a critical section and two semaphores doesn't work here because - as far as I can see - the woken up thread has to handshake, by inserting it's thread id into the lock's owner map.
I suppose it would be doable with two condition variables and a couple of mutexes but the sheer amount of synchronization primitives that would end up using sounds like a bit too much overhead for my taste.
An idea which just sprang into my mind is to utilize TLS to remember the type of lock I'm holding (and possibly the local lock counts). Have to think it through - but I'll still post the question for now.
Target platform is Win32 but that shouldn't really matter. Note that I'm specifically targeting Win2k so anything related to the new MRSW lock primitive in Windows 7 is not relevant for me. :-)
Okay, I solved it.
It can be done with just 2 semaphores, a critical section and almost no more locking than for a regular non-recursive MRSW lock (there is obviously some more CPU-time spent inside the lock because that multimap must be managed) - but it's tricky. The structure I came up with looks like this:
// Protects everything that follows, except mWriterThreadId and mRecursiveUpgrade
CRITICAL_SECTION mLock;
// Semaphore to wait on for a read lock
HANDLE mSemaReader;
// Semaphore to wait on for a write lock
HANDLE mSemaWriter;
// Number of threads waiting for a write lock.
int mWriterWaiting;
// Number of times the writer entered the write lock.
int mWriterActive;
// Number of threads inside a read lock. Note that this does not include
// recursive read locks.
int mReaderActiveThreads;
// Whether or not the current writer obtained the lock by a recursive
// upgrade. Note that this member might be set outside the critical
// section, so it should only be read from by the writer during his
// unlock.
bool mRecursiveUpgrade;
// This member contains the current thread id once for each
// (recursive) read lock held by the current thread in addition to an
// undefined number of other thread ids which may or may not hold a
// read lock, even inside the critical section (!).
std::multiset<unsigned long> mReaderActive;
// If there is no writer this member contains 0.
// If the current thread is the writer this member contains his
// thread-id.
// Otherwise it can contain either of them, even inside the
// critical section (!).
// Also note that it might be set outside the critical section.
unsigned long mWriterThreadId;
Now, the basic idea is this:
Full update of mWriterWaiting and mWriterActive for an unlock is performed by the unlocking thread.
For mWriterThreadId and mReaderActive this is not possible, as the waiting thread needs to insert itself when it was released.
So the rule is, that you may never access those two members except to check whether you are holding a read lock or are the current writer - specifically it may not be used to checker whether or not there are any readers / writers - for that you have to use the (somewhat redundant but necessary for this reason) mReaderActiveThreads and mWriterActive.
I'm currently running some test code (which has been going on deadlock- and crash-free for 30 minutes or so) - when I'm sure that it's stable and I've cleaned up the code somewhat I'll put it on some pastebin and add a link in a comment here (just in case someone else ever needs this).
Well, I did some thinking. Starting from the simple "two semaphores and a critical section" one adds a writer lock count and a owning writer TID to the structure.
Unlock still set most of the new status in the critsec. Readers still normally increase the lock count - recursive locking simply adds a non-existing reader to the counter.
During writers lock() I compare the owning TID, and if the writer already own it the write lock counter is increased.
Setting the new writer TID can't be done by the unlock() - it doesn't know which one will be wakened, but if writers reset it back to zero in their unlock() it's not a problem - the current thread id won't ever be zero and setting it is an atomic operation.
All sounds simple enough - one nasty problem left: A recursive reader-reader lock while a writer is waiting will deadlock. And I don't know how to solve that short of doing a reader-biased lock... somehow I need to know whether or not I already own a reader lock.
Using TLS doesn't sound too great after I realized that the number if available slots might be rather limited...
As far as I understand, you need to provide your writer exclusive access to the data, while readers can operate simultaneously (if this is not what you want, please clarify your question).
I think you need to implement a sort of "inverse semaphore", i.e. a semaphore that will block a thread when positive, and signal all waiting threads when zero. If you do this, you can use two such semaphores for your program. The operation of your threads could then be the following:
Reader:
(1) wait on sem A
(2) increase sem B
(3) read operation
(4) decrease sem B
Writer:
(1) increase sem A
(2) wait on sem B
(3) write operation
(4) decrease sem A
In this way the writer will perform the write operation as soon as all pending readers have finished reading. As soon as your writer finishes, readers can resume their operation without blocking each other.
I am not familiar with Windows mutex/semaphore facilities but I can think of a way to implement such semaphores using the POSIX threads API (combining a mutex, a counter and a conditional variable).