I've been trying to learn how to multithread and came up with the following understanding. I was wondering if I'm correct or far off and, if I'm incorrect in any way, if someone could give me advice.
To create a thread, first you need to utilize a library such as <thread> or any alternative (I'm using boost's multithreading library to get cross-platform capabilities). Afterwards, you can create a thread by declaring it as such (for std::thread)
std::thread thread (foo);
Now, you can use thread.join() or thread.detach(). The former will wait until the thread finishes, and then continue; while, the latter will run the thread alongside whatever you plan to do.
If you want to protect something, say a vector std::vector<double> data, from threads accessing simultaneously, you would use a mutex.
Mutex's would be declared as a global variable so that they may access the thread functions (OR, if you're making a class that will be multithreaded, the mutex can be declared as a private/public variable of the class). Afterwards, you can lock and unlock a thread using a mutex.
Let's take a quick look at this example pseudo code:
std::mutex mtx;
std::vector<double> data;
void threadFunction(){
// Do stuff
// ...
// Want to access a global variable
mtx.lock();
data.push_back(3.23);
mtx.unlock();
// Continue
}
In this code, when the mutex locks down on the thread, it only locks the lines of code between it and mtx.unlock(). Thus, other threads will still continue on their merry way until they try accessing data (Note, we would likely through a mutex in the other threads as well). Then they would stop, wait to use data, lock it, push_back, unlock it and continue. Check here for a good description of mutex's.
That's about it on my understanding of multithreading. So, am I horribly wrong or accurate?
Your comments refer to "locking the whole thread". You can't lock part of a thread.
When you lock a mutex, the current thread takes ownership of the mutex. Conceptually, you can think of it as the thread places its mark on the mutex (stores its threadid in the mutex data structure). If any other thread comes along and attempts to acquire the same mutex instance, it sees that the mutex is already "claimed" by somebody else and it waits until the first thread has released the mutex. When the owning thread later releases the mutex, one of the threads that is waiting for the mutex can wake up, acquire the mutex for themselves, and carry on.
In your code example, there is a potential risk that the mutex might not be released once it is acquired. If the call to data.push_back(xxx) throws an exception (out of memory?), then execution will never reach mtx.unlock() and the mutex will remain locked forever. All subsequent threads that attempt to acquire that mutex will drop into a permanent wait state. They'll never wake up because the thread that owns the mutex is toast.
For this reason, acquiring and releasing critical resources like mutexes should be done in a manner that will guarantee they will be released regardless of how execution leaves the current scope. In other languages, this would mean putting the mtx.unlock() in the finally section of a try..finally block:
mtx.lock();
try
{
// do stuff
}
finally
{
mtx.unlock();
}
C++ doesn't have try..finally statements. Instead, C++ leverages its language rules for automatic disposal of locally defined variables. You construct an object in a local variable, the object acquires a mutex lock in its constructor. When execution leaves the current function scope, C++ will make sure that the object is disposed, and the object releases the lock when it is disposed. That's the RAII others have mentioned. RAII just makes use of the existing implicit try..finally block that wraps every C++ function body.
Related
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.
What part of memory gets locked by mutex when .lock() or .try_lock(), is it just the function or is it the whole program that gets locked?
Nothing is locked except the mutex. Everything else continues running (until it tries to lock an already locked mutex that is). The mutex is only there so that two threads cannot run the code between a mutex lock and a mutex unlock at the same time.
A mutex doesn't really lock anything, except for itself. You can think of a mutex as being a gate where you can only unlock it from the inside. When the gate is locked, any thread that tries to lock the mutex will sit there at the gate and wait for the current thread that is behind the gate to unlock it and let them in. When they gate is not locked then when you call lock you can just go in, close and lock the gate, and now no threads can get past the gate until you unlock it and let them in.
A mutex doesn't lock anything. You just use a mutex to communicate to other parts of your code that they should consider whatever you decide needs to be protected from access by several threads at the same time to be off-limits for now.
You could think of a mutex as something like a boolean okToModify. Whenever you want to edit something, you check if okToModify is true. If it is, you set it to false (preventing any other threads from modifying it), change it, then set okToModify back to true to tell the other threads you're done and give them a chance to modify:
// WARNING! This code doesn't actually work as a lock!
// it is just an example of the concept.
struct LockedInt {
bool okToModify; // This would be your mutex instead of a bool.
int integer;
};
struct LockedInt myLockedInt = { true, 0 };
...
while (myLockedInt.okToModify == false)
; // wait doing nothing until whoever is modifying the int is done.
myLockedInt.okToModify = false; // Prevent other threads from getting out of while loop above.
myLockedInt.integer += 1;
myLockedInt.okToModify = true; // Now other threads get out of the while loop if they were waiting and can modify.
The while loop and okToModify = false above is basically what locking a mutex does, and okToModify = true is what unlocking a mutex does.
Now, why do we need mutexes and don't use booleans? Because a thread could be running at the same time as those three lines above. The code for locking a mutex actually guarantees that the waiting for okToModify to become true and setting okToModify = false happen in one go, and therefore no other thread can get "in between the lines", for example by using a special machine-code instruction called "compare-and-exchange".
So do not use booleans instead of mutexes, but you can think of a mutex as a special, thread-safe boolean.
m.lock() doesn't really lock anything. What it does is, it waits to take ownership of the mutex. A mutex always either is owned by exactly one thread or else it is available. m.lock() waits until the mutex becomes available, and then it takes ownership of it in the name of the calling thread.
m.unlock releases the mutex (i.e., it relinquishes ownership), causing the mutex to once again become available.
Mutexes also perform another very important function. In modern C++, when some thread T performs a sequence of assignments of various values to various memory locations, the system makes no guarantees about when other threads U, V, and W will see those assignments, whether the other threads will see the assignments happen in the same order in which thread T performed them, or even, whether the other threads will ever see the assignments.
There are some quite complicated rules governing things that a programmer can do to ensure that different threads see a consistent view of shared memory objects (Google "C++ memory model"), but here's one simple rule:
Whatever thread T did before it releases some mutex M is guaranteed to be visible to any other thread U after thread U subsequently locks the same mutex M.
I am passing few variables by pointer (cv::Mat's and bool's) to several threads and am trying to understand when it is necessary to use mutex. I have found that without using it on the cv::Mat's, my program will crash (likely because one thread is writing to the same area of memory that the other is reading from), so I have implemented mutex for these variables and it has fixed the problem.
But now the mutex is one more variable I am passing by pointer to each thread. So in this case, the use of the handling of the mutex variable is the same as the other variables I need to mutex, so what's so special about the mutex that I wouldn't need a mutex as well (and of course that goes on forever and the conept doesn't work).
To be clear, the code I have works fine, this is more for educational purposes.
Example:
//Common frames
cv::Mat captureimage, displayimage;
std::mutex capturemutex, displaymutex;
//Start image capture thread
std::thread t_imagecapture( CaptureImageThread, &captureimage, &capturemutex, &exitsignal );
//Start image processor thread
std::thread t_imageprocessor( ProcessImageThread, &captureimage, &capturemutex, &exitsignal );
//Start display thread
std::thread t_displayupdate( DisplayUpdateThread, &displayimage, &displaymutex, &exitsignal );
A mutex is an atomic lock. It uses lowlevel methods (CPU), for example, it could test-and-set the lock without being interrupted, so it doesn't need an external lock to perform this. And once the lock is set, no other thread can do that, so the mutex can protect the access of other resources.
I understand how to use condition variables (crummy name for this construct, IMO, as the cv object neither is a variable nor indicates a condition). So I have a pair of threads, canonically set up with Boost.Thread as:
bool awake = false;
boost::mutex sync;
boost::condition_variable cv;
void thread1()
{
boost::unique_lock<boost::mutex> lock1(sync);
while (!awake)
cv.wait(lock1);
lock1.unlock(); // this line actually not canonical, but why not?
// proceed...
}
void thread2()
{
//...
boost::unique_lock<boost::mutex> lock2;
awake = true;
lock2.unlock();
cv.notify_all();
}
My question is: does thread2 really need to be protecting the assignment to awake? It seems to me the notify_all() call should be sufficient. If the data being manipulated and checked against were more than a simple "ok to proceed" flag, I see the value in the mutex, but here it seems like overkill.
A secondary question is that asked in the code fragment: Why doesn't the Boost documentation show the lock in thread1 being unlocked before the "process data" step?
EDIT: Maybe my question is really: Is there a cleaner construct than a CV to implement this kind of wait?
does thread2 really need to be protecting the assignment to awake?
Yes. Modifying an object from one thread and accessing it from another without synchronisation gives undefined behaviour. Even if it's just a bool.
For example, on some multiprocessor systems the write might only affect local memory; without an explicit synchronisation operation, other threads might never see the change.
Why doesn't the Boost documentation show the lock in thread1 being unlocked before the "process data" step?
If you unlocked the mutex before clearing the flag, then you might miss another signal.
Is there a cleaner construct than a CV to implement this kind of wait?
In Boost and the standard C++ library, no; a condition variable is flexible enough to handle arbitrary shared state and not particularly over-complicated for this simple case, so there's no particular need for anything simpler.
More generally, you could use a semaphore or a pipe to send a simple signal between threads.
Formally, you definitely need the lock in both threads: if any thread
modifies an object, and more than one thread accesses it, then all
accesses must be synchronized.
In practice, you'll probably get away with it without the lock; it's
almost certain that notify_all will issue the necessary fence or
membar instructions to ensure that the memory is properly synchronized.
But why take the risk?
As to the absense of the unlock, that's the whole point of the scoped
locking pattern: the unlock is in the destructor of the object, so
that the mutex will be unlocked even if an exception passes through.
I'm using the C++ boost::thread library, which in my case means I'm using pthreads. Officially, a mutex must be unlocked from the same thread which locks it, and I want the effect of being able to lock in one thread and then unlock in another. There are many ways to accomplish this. One possibility would be to write a new mutex class which allows this behavior.
For example:
class inter_thread_mutex{
bool locked;
boost::mutex mx;
boost::condition_variable cv;
public:
void lock(){
boost::unique_lock<boost::mutex> lck(mx);
while(locked) cv.wait(lck);
locked=true;
}
void unlock(){
{
boost::lock_guard<boost::mutex> lck(mx);
if(!locked) error();
locked=false;
}
cv.notify_one();
}
// bool try_lock(); void error(); etc.
}
I should point out that the above code doesn't guarantee FIFO access, since if one thread calls lock() while another calls unlock(), this first thread may acquire the lock ahead of other threads which are waiting. (Come to think of it, the boost::thread documentation doesn't appear to make any explicit scheduling guarantees for either mutexes or condition variables). But let's just ignore that (and any other bugs) for now.
My question is, if I decide to go this route, would I be able to use such a mutex as a model for the boost Lockable concept. For example, would anything go wrong if I use a boost::unique_lock< inter_thread_mutex > for RAII-style access, and then pass this lock to boost::condition_variable_any.wait(), etc.
On one hand I don't see why not. On the other hand, "I don't see why not" is usually a very bad way of determining whether something will work.
The reason I ask is that if it turns out that I have to write wrapper classes for RAII locks and condition variables and whatever else, then I'd rather just find some other way to achieve the same effect.
EDIT:
The kind of behavior I want is basically as follows. I have an object, and it needs to be locked whenever it is modified. I want to lock the object from one thread, and do some work on it. Then I want to keep the object locked while I tell another worker thread to complete the work. So the first thread can go on and do something else while the worker thread finishes up. When the worker thread gets done, it unlocks the mutex.
And I want the transition to be seemless so nobody else can get the mutex lock in between when thread 1 starts the work and thread 2 completes it.
Something like inter_thread_mutex seems like it would work, and it would also allow the program to interact with it as if it were an ordinary mutex. So it seems like a clean solution. If there's a better solution, I'd be happy to hear that also.
EDIT AGAIN:
The reason I need locks to begin with is that there are multiple master threads, and the locks are there to prevent them from accessing shared objects concurrently in invalid ways.
So the code already uses loop-level lock-free sequencing of operations at the master thread level. Also, in the original implementation, there were no worker threads, and the mutexes were ordinary kosher mutexes.
The inter_thread_thingy came up as an optimization, primarily to improve response time. In many cases, it was sufficient to guarantee that the "first part" of operation A, occurs before the "first part" of operation B. As a dumb example, say I punch object 1 and give it a black eye. Then I tell object 1 to change it's internal structure to reflect all the tissue damage. I don't want to wait around for the tissue damage before I move on to punch object 2. However, I do want the tissue damage to occur as part of the same operation; for example, in the interim, I don't want any other thread to reconfigure the object in such a way that would make tissue damage an invalid operation. (yes, this example is imperfect in many ways, and no I'm not working on a game)
So we made the change to a model where ownership of an object can be passed to a worker thread to complete an operation, and it actually works quite nicely; each master thread is able to get a lot more operations done because it doesn't need to wait for them all to complete. And, since the event sequencing at the master thread level is still loop-based, it is easy to write high-level master-thread operations, as they can be based on the assumption that an operation is complete (more precisely, the critical "first part" upon which the sequencing logic depends is complete) when the corresponding function call returns.
Finally, I thought it would be nice to use inter_thread mutex/semaphore thingies using RAII with boost locks to encapsulate the necessary synchronization that is required to make the whole thing work.
man pthread_unlock (this is on OS X, similar wording on Linux) has the answer:
NAME
pthread_mutex_unlock -- unlock a mutex
SYNOPSIS
#include <pthread.h>
int
pthread_mutex_unlock(pthread_mutex_t *mutex);
DESCRIPTION
If the current thread holds the lock on mutex, then the
pthread_mutex_unlock() function unlocks mutex.
Calling pthread_mutex_unlock() with a mutex that the
calling thread does not hold will result in
undefined behavior.
...
My counter-question would be - what kind of synchronization problem are you trying to solve with this? Most probably there is an easier solution.
Neither pthreads nor boost::thread (built on top of it) guarantee any order in which a contended mutex is acquired by competing threads.
Sorry, but I don't understand. what will be the state of your mutex in line [1] in the following code if another thread can unlock it?
inter_thread_mutex m;
{
m.lock();
// [1]
m.unlock();
}
This has no sens.
There's a few ways to approach this. Both of the ones I'm going to suggest are going to involve adding an additional piece of information to the object, rather adding a mechanism to unlock a thread from a thread other than the one that owns it.
1) you can add some information to indicate the object's state:
enum modification_state { consistent, // ready to be examined or to start being modified
phase1_complete, // ready for the second thread to finish the work
};
// first worker thread
lock();
do_init_work(object);
object.mod_state = phase1_complete;
unlock();
signal();
do_other_stuff();
// second worker thread
lock()
while( object.mod_state != phase1_complete )
wait()
do_final_work(obj)
object.mod_state = consistent;
unlock()
signal()
// some other thread that needs to read the data
lock()
while( object.mod_state != consistent )
wait();
read_data(obj)
unlock()
Works just fine with condition variables, because obviously you're not writing your own lock.
2) If you have a specific thread in mind, you can give the object an owner.
// first worker
lock();
while( obj.owner != this_thread() ) wait();
do_initial_work(obj);
obj.owner = second_thread_id;
unlock()
signal()
...
This is pretty much the same solution as my first solution, but more flexible in the adding/removing of phases, and less flexible in the adding/removing of threads.
To be honest, I'm not sure how inter thread mutex would help you here. You'd still need a semaphore or condition variable to signal the passing of the work to the second thread.
Small modification to what you already have: how about storing the id of the thread which you want to take the lock, in your inter_thread_whatever? Then unlock it, and send a message to that thread, saying "I want you execute whatever routine it is that tries to take this lock".
Then the condition in lock becomes while(locked || (desired_locker != thisthread && desired_locker != 0)). Technically you've "released the lock" in the first thread, and "taken it again" in the second thread, but there's no way that any other thread can grab it in between, so it's as if you've transferred it directly from one to the other.
There's a potential problem, that if a thread exits or is killed, while it's the desired locker of your lock, then that thread deadlocks. But you were already talking about the first thread waiting for a message from the second thread to say that it has successfully acquired the lock, so presumably you already have a plan in mind for what happens if that message is never received. To that plan, add "reset the desired_locker field on the inter_thread_whatever".
This is all very hairy, though, I'm not convinced that what I've proposed is correct. Is there a way that the "master" thread (the one that's directing all these helpers) can just make sure that it doesn't order any more operations to be performed on whatever is protected by this lock, until the first op is completed (or fails and some RAII thing notifies you)? You don't need locks as such, if you can deal with it at the level of the message loop.
I don't think it is a good idea to say that your inter_thread_mutex (binary_semaphore) can be seen as a model of Lockable. The main issue is that the main feature of your inter_thread_mutex defeats the Locakble concept. If inter_thread_mutex was a model of lockable you will expect in In [1] that the inter_thread_mutex m is locked.
// thread T1
inter_thread_mutex m;
{
unique_lock<inter_thread_mutex> lk(m);
// [1]
}
But as an other thread T2 can do m.unlock() while T1 is in [1], the guaranty is broken.
Binary semaphores can be used as Lockables as far as each thread tries to lock before unlocking. But the main goal of your class is exactly the contrary.
This is one of the reason semaphores in Boost.Interprocess don't use lock/unlock to name the functions, but wait/notify. Curiously these are the same names used by conditions :)
A mutex is a mechanism for describing mutually exclusive blocks of code. It does not make sense for these blocks of code to cross thread boundaries. Trying to use such a concept in such an counter intuitive way can only lead to problems down the line.
It sounds very much like you're looking for a different multi-threading concept, but without more detail it's hard to know what.