Read Write lock implementation in C++ - c++

I am trying to use read/write lock in C++ using shared_mutex
typedef boost::shared_mutex Lock;
typedef boost::unique_lock< Lock > WriteLock;
typedef boost::shared_lock< Lock > ReadLock;
class Test {
Lock lock;
WriteLock writeLock;
ReadLock readLock;
Test() : writeLock(lock), readLock(lock) {}
readFn1() {
readLock.lock();
/*
Some Code
*/
readLock.unlock();
}
readFn2() {
readLock.lock();
/*
Some Code
*/
readLock.unlock();
}
writeFn1() {
writeLock.lock();
/*
Some Code
*/
writeLock.unlock();
}
writeFn2() {
writeLock.lock();
/*
Some Code
*/
writeLock.unlock();
}
}
The code seems to be working fine but I have a few conceptual questions.
Q1. I have seen the recommendations to use unique_lock and shared_lock on http://en.cppreference.com/w/cpp/thread/shared_mutex/lock, but I don't understand why because shared_mutex already supports lock and lock_shared methods?
Q2. Does this code have the potential to cause write starvation? If yes then how can I avoid the starvation?
Q3. Is there any other locking class I can try to implement read write lock?

Q1: use of a mutex wrapper
The recommendation to use a wrapper object instead of managing the mutex directly is to avoid unfortunate situation where your code is interrupted and the mutex is not released, leaving it locked forever.
This is the principle of RAII.
But this only works if your ReadLock or WriteLock are local to the function using it.
Example:
readFn1() {
boost::unique_lock< Lock > rl(lock);
/*
Some Code
==> imagine exception is thrown
*/
rl.unlock(); // this is never reached if exception thrown
} // fortunately local object are destroyed automatically in case
// an excpetion makes you leave the function prematurely
In your code this won't work if one of the function is interupted, becaus your ReadLock WriteLock object is a member of Test and not local to the function setting the lock.
Q2: Write starvation
It is not fully clear how you will invoke the readers and the writers, but yes, there is a risk:
as long as readers are active, the writer is blocked by the unique_lock waiting for the mutex to be aquirable in exclusive mode.
however as long as the wrtier is waiting, new readers can obtain access to the shared lock, causing the unique_lock to be further delayed.
If you want to avoid starvation, you have to ensure that waiting writers do get the opportunity to set their unique_lock. For example att in your readers some code to check if a writer is waiting before setting the lock.
Q3 Other locking classes
Not quite sure what you're looking for, but I have the impression that condition_variable could be of interest for you. But the logic is a little bit different.
Maybe, you could also find a solution by thinking out of the box: perhaps there's a suitable lock-free data structure that could facilitate coexistance of readers and writers by changing slightly the approach ?

The types for the locks are ok but instead of having them as member functions create then inside the member functions locktype lock(mymutex). That way they are released on destruction even in the case of an exception.

Q1. I have seen the recommendations to use unique_lock and shared_lock on http://en.cppreference.com/w/cpp/thread/shared_mutex/lock, but I don't understand why because shared_mutex already supports lock and lock_shared methods?
Possibly because unique_lock has been around since c++11 but shared_lock is coming onboard with c++17. Also, [possibly] unique_lock can be more efficient. Here's the original rationale for shared_lock [by the creator] http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2406.html and I defer to that.
Q2. Does this code have the potential to cause write starvation? If yes then how can I avoid the starvation?
Yes, absolutely. If you do:
while (1)
writeFn1();
You can end up with a time line of:
T1: writeLock.lock()
T2: writeLock.unlock()
T3: writeLock.lock()
T4: writeLock.unlock()
T5: writeLock.lock()
T6: writeLock.unlock()
...
The difference T2-T1 is arbitrary, based on amount of work being done. But, T3-T2 is near zero. This is the window for another thread to acquire the lock. Because the window is so small, it probably won't get it.
To solve this, the simplest method is to insert a small sleep (e.g. nanosleep) between T2 and T3. You could do this by adding it to the bottom of writeFn1.
Other methods can involve creating a queue for the lock. If a thread can't get the lock, it adds itself to a queue and the first thread on the queue gets the lock when the lock is released. In the linux kernel, this is implemented for a "queued spinlock"
Q3. Is there any other locking class I can try to implement read write lock?
While not a class, you could use pthread_mutex_lock and pthread_mutex_unlock. These implement recursive locks. You could add your own code to implement the equivalent of boost::scoped_lock. Your class can control the semantics.
Or, boost has its own locks.

Related

C++ atomics: how to allow only a single thread to access a function?

I'd like to write a function that is accessible only by a single thread at a time. I don't need busy waits, a brutal 'rejection' is enough if another thread is already running it. This is what I have come up with so far:
std::atomic<bool> busy (false);
bool func()
{
if (m_busy.exchange(true) == true)
return false;
// ... do stuff ...
m_busy.exchange(false);
return true;
}
Is the logic for the atomic exchange correct?
Is it correct to mark the two atomic operations as std::memory_order_acq_rel? As far as I understand a relaxed ordering (std::memory_order_relaxed) wouldn't be enough to prevent reordering.
Your atomic swap implementation might work. But trying to do thread safe programming without a lock is most always fraught with issues and is often harder to maintain.
Unless there's a performance improvement that's needed, then std::mutex with the try_lock() method is all you need, eg:
std::mutex mtx;
bool func()
{
// making use of std::unique_lock so if the code throws an
// exception, the std::mutex will still get unlocked correctly...
std::unique_lock<std::mutex> lck(mtx, std::try_to_lock);
bool gotLock = lck.owns_lock();
if (gotLock)
{
// do stuff
}
return gotLock;
}
Your code looks correct to me, as long as you leave the critical section by falling out, not returning or throwing an exception.
You can unlock with a release store; an RMW (like exchange) is unnecessary. The initial exchange only needs acquire. (But does need to be an atomic RMW like exchange or compare_exchange_strong)
Note that ISO C++ says that taking a std::mutex is an "acquire" operation, and releasing is is a "release" operation, because that's the minimum necessary for keeping the critical section contained between the taking and the releasing.
Your algo is exactly like a spinlock, but without retry if the lock's already taken. (i.e. just a try_lock). All the reasoning about necessary memory-order for locking applies here, too. What you've implemented is logically equivalent to the try_lock / unlock in #selbie's answer, and very likely performance-equivalent, too. If you never use mtx.lock() or whatever, you're never actually blocking i.e. waiting for another thread to do something, so your code is still potentially lock-free in the progress-guarantee sense.
Rolling your own with an atomic<bool> is probably good; using std::mutex here gains you nothing; you want it to be doing only this for try-lock and unlock. That's certainly possible (with some extra function-call overhead), but some implementations might do something more. You're not using any of the functionality beyond that. The one nice thing std::mutex gives you is the comfort of knowing that it safely and correctly implements try_lock and unlock. But if you understand locking and acquire / release, it's easy to get that right yourself.
The usual performance reason to not roll your own locking is that mutex will be tuned for the OS and typical hardware, with stuff like exponential backoff, x86 pause instructions while spinning a few times, then fallback to a system call. And efficient wakeup via system calls like Linux futex. All of this is only beneficial to the blocking behaviour. .try_lock leaves that all unused, and if you never have any thread sleeping then unlock never has any other threads to notify.
There is one advantage to using std::mutex: you can use RAII without having to roll your own wrapper class. std::unique_lock with the std::try_to_lock policy will do this. This will make your function exception-safe, making sure to always unlock before exiting, if it got the lock.

C++11: why does std::condition_variable use std::unique_lock?

I am a bit confused about the role of std::unique_lock when working with std::condition_variable. As far as I understood the documentation, std::unique_lock is basically a bloated lock guard, with the possibility to swap the state between two locks.
I've so far used pthread_cond_wait(pthread_cond_t *cond, pthread_mutex_t *mutex) for this purpose (I guess that's what the STL uses on posix). It takes a mutex, not a lock.
What's the difference here? Is the fact that std::condition_variable deals with std::unique_lock an optimization? If so, how exactly is it faster?
so there is no technical reason?
I upvoted cmeerw's answer because I believe he gave a technical reason. Let's walk through it. Let's pretend that the committee had decided to have condition_variable wait on a mutex. Here is code using that design:
void foo()
{
mut.lock();
// mut locked by this thread here
while (not_ready)
cv.wait(mut);
// mut locked by this thread here
mut.unlock();
}
This is exactly how one shouldn't use a condition_variable. In the regions marked with:
// mut locked by this thread here
there is an exception safety problem, and it is a serious one. If an exception is thrown in these areas (or by cv.wait itself), the locked state of the mutex is leaked unless a try/catch is also put in somewhere to catch the exception and unlock it. But that's just more code you're asking the programmer to write.
Let's say that the programmer knows how to write exception safe code, and knows to use unique_lock to achieve it. Now the code looks like this:
void foo()
{
unique_lock<mutex> lk(mut);
// mut locked by this thread here
while (not_ready)
cv.wait(*lk.mutex());
// mut locked by this thread here
}
This is much better, but it is still not a great situation. The condition_variable interface is making the programmer go out of his way to get things to work. There is a possible null pointer dereference if lk accidentally does not reference a mutex. And there is no way for condition_variable::wait to check that this thread does own the lock on mut.
Oh, just remembered, there is also the danger that the programmer may choose the wrong unique_lock member function to expose the mutex. *lk.release() would be disastrous here.
Now let's look at how the code is written with the actual condition_variable API that takes a unique_lock<mutex>:
void foo()
{
unique_lock<mutex> lk(mut);
// mut locked by this thread here
while (not_ready)
cv.wait(lk);
// mut locked by this thread here
}
This code is as simple as it can get.
It is exception safe.
The wait function can check lk.owns_lock() and throw an exception if it is false.
These are technical reasons that drove the API design of condition_variable.
Additionally, condition_variable::wait doesn't take a lock_guard<mutex> because lock_guard<mutex> is how you say: I own the lock on this mutex until lock_guard<mutex> destructs. But when you call condition_variable::wait, you implicitly release the lock on the mutex. So that action is inconsistent with the lock_guard use case / statement.
We needed unique_lock anyway so that one could return locks from functions, put them into containers, and lock/unlock mutexes in non-scoped patterns in an exception safe way, so unique_lock was the natural choice for condition_variable::wait.
Update
bamboon suggested in the comments below that I contrast condition_variable_any, so here goes:
Question: Why isn't condition_variable::wait templated so that I can pass any Lockable type to it?
Answer:
That is really cool functionality to have. For example this paper demonstrates code that waits on a shared_lock (rwlock) in shared mode on a condition variable (something unheard of in the posix world, but very useful nonetheless). However the functionality is more expensive.
So the committee introduced a new type with this functionality:
`condition_variable_any`
With this condition_variable adaptor one can wait on any lockable type. If it has members lock() and unlock(), you are good to go. A proper implementation of condition_variable_any requires a condition_variable data member and a shared_ptr<mutex> data member.
Because this new functionality is more expensive than your basic condition_variable::wait, and because condition_variable is such a low level tool, this very useful but more expensive functionality was put into a separate class so that you only pay for it if you use it.
It's essentially an API design decision to make the API as safe as possible by default (with the additional overhead being seen as negligible). By requiring to pass a unique_lock instead of a raw mutex users of the API are directed towards writing correct code (in the presence of exceptions).
In recent years the focus of the C++ language has shifted towards making it safe by default (but still allowing users to shoot themselves into their feet if they want to and try hard enough).

Is there a facility in boost to allow for write-biased locking?

If I have the following code:
#include <boost/date_time.hpp>
#include <boost/thread.hpp>
boost::shared_mutex g_sharedMutex;
void reader()
{
boost::shared_lock<boost::shared_mutex> lock(g_sharedMutex);
boost::this_thread::sleep(boost::posix_time::seconds(10));
}
void makeReaders()
{
while (1)
{
boost::thread ar(reader);
boost::this_thread::sleep(boost::posix_time::seconds(3));
}
}
boost::thread mr(makeReaders);
boost::this_thread::sleep(boost::posix_time::seconds(5));
boost::unique_lock<boost::shared_mutex> lock(g_sharedMutex);
...
the unique lock will never be acquired, because there are always going to be readers. I want a unique_lock that, when it starts waiting, prevents any new read locks from gaining access to the mutex (called a write-biased or write-preferred lock, based on my wiki searching). Is there a simple way to do this with boost? Or would I need to write my own?
Note that I won't comment on the win32 implementation because it's way more involved and I don't have the time to go through it in detail. That being said, it's interface is the same as the pthread implementation which means that the following answer should be equally valid.
The relevant pieces of the pthread implementation of boost::shared_mutex as of v1.51.0:
void lock_shared()
{
boost::this_thread::disable_interruption do_not_disturb;
boost::mutex::scoped_lock lk(state_change);
while(state.exclusive || state.exclusive_waiting_blocked)
{
shared_cond.wait(lk);
}
++state.shared_count;
}
void lock()
{
boost::this_thread::disable_interruption do_not_disturb;
boost::mutex::scoped_lock lk(state_change);
while(state.shared_count || state.exclusive)
{
state.exclusive_waiting_blocked=true;
exclusive_cond.wait(lk);
}
state.exclusive=true;
}
The while loop conditions are the most relevant part for you. For the lock_shared function (read lock), notice how the while loop will not terminate as long as there's a thread trying to acquire (state.exclusive_waiting_blocked) or already owns (state.exclusive) the lock. This essentially means that write locks have priority over read locks.
For the lock function (write lock), the while loop will not terminate as long as there's at least one thread that currently owns the read lock (state.shared_count) or another thread owns the write lock (state.exclusive). This essentially gives you the usual mutual exclusion guarantees.
As for deadlocks, well the read lock will always return as long as the write locks are guaranteed to be unlocked once they are acquired. As for the write lock, it's guaranteed to return as long as the read locks and the write locks are always guaranteed to be unlocked once acquired.
In case you're wondering, the state_change mutex is used to ensure that there's no concurrent calls to either of these functions. I'm not going to go through the unlock functions because they're a bit more involved. Feel free to look them over yourself, you have the source after all (boost/thread/pthread/shared_mutex.hpp) :)
All in all, this is pretty much a text book implementation and they've been extensively tested in a wide range of scenarios (libs/thread/test/test_shared_mutex.cpp and massive use across the industry). I wouldn't worry too much as long you use them idiomatically (no recursive locking and always lock using the RAII helpers). If you still don't trust the implementation, then you could write a randomized test that simulates whatever test case you're worried about and let it run overnight on hundreds of thread. That's usually a good way to tease out deadlocks.
Now why would you see that a read lock is acquired after a write lock is requested? Difficult to say without seeing the diagnostic code that you're using. Chances are that the read lock is acquired after your print statement (or whatever you're using) is completed and before state_change lock is acquired in the write thread.

Need to mutex-protect (atomic) assignment sought by condition variable?

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.

Modelling boost::Lockable with semaphore rather than mutex (previously titled: Unlocking a mutex from a different thread)

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.