I am having problems knowing how the code satisfies the conditions for deadlock? And is there any suggestions or strategies to help make the code deadlock free?
Thread1()
{
while (true)
{
lock1.acquire();
lock2.acquire();
CriticalSection();
lock2.release();
lock1.release();
}
}
Thread2()
{
while (true)
{
lock2.acquire();
lock3.acquire();
CriticalSection();
lock3.release();
lock2.release();
}
}
Thread3()
{
while (true)
{
lock3.acquire();
lock2.acquire();
CriticalSection();
lock1.release();
lock3.release();
}
}
A deadlock happens when a process attempts to acquire a resource A while holding resource B, and another process attempts to acquire B while holding A. So, in your program thread 2 and thread3 can crate a deadlock because they both try to acquire lock2 and lock3. If they acquire these resources in the same order, the program will be deadlock safe. For instance if both threads acquire lock2 before lock3, then no process can try to acquire lock2 while holding lock3. However in your program, thread2 acquires 2 before 3, and thread3 acquires 3 before 2, creating a potential deadlock.
Related
void transaction(Account from, Account to, double amount)
{
Semaphore lock1, lock2;
lock1 =getLock(from);
lock2 = getLock(to);
wait(lock1);
wait(lock2);
withdraw(from, amount);
deposit(to, amount);
signal(lock2);
signal (lock1) ;
}
The transaction function transfers funds between two accounts. To prevent a
race condition, each account has an associated semaphore that is obtained from
a get Lock () function such as the following: .I am not able to understand how the semaphores are working here, can someone help explain the functionality of semaphores and getLock functions here.
What is happening in each execution of wait and signal ?
How can a Deadlock occur if 2 threads invoke Transaction(x,y,25) and Transaction(y,x,30) simultaneously?
In the code below, although I avoid deadlock( due to reverse locking hierarchy). However, due to T2 going into an infinite loop, the mutex m1 and m2 never get released, in this case the system can neither achieve progress nor bounded wait( in terms of time). How can I handle this kind of scenario ? Is there a way to interrupt the thread that is holding the resource? At least with pthread mutexes the ownership of the mutex is with the thread that locked it! How could we unlock the mutex held by the other thread ?
Thread T1 :
while(1) {
pthread_mutex_lock( &m1);
//use resource 1
if (pthread_mutex_trylock(&m2) ) {
//Got the lock, do processing
}
else {
//Release m1
pthread_mutex_unlock(&m1);
continue;
}
pthread_mutex_unlock(&m2);
pthread_mutex_unlock(&m1);
}
Thread T2:
pthread_mutex_lock( &m2);
pthread_mutex_lock(&m1)
//Goes in an infinite loop here
pthread_mutex_unlock(&m1);
pthread_mutex_unlock(&m2);
}
I have N threads performing various task and these threads must be regularly synchronized with a thread barrier as illustrated below with 3 thread and 8 tasks. The || indicates the temporal barrier, all threads have to wait until the completion of 8 tasks before starting again.
Thread#1 |----task1--|---task6---|---wait-----||-taskB--| ...
Thread#2 |--task2--|---task5--|-------taskE---||----taskA--| ...
Thread#3 |-task3-|---task4--|-taskG--|--wait--||-taskC-|---taskD ...
I couldn’t find a workable solution, thought the little book of Semaphores http://greenteapress.com/semaphores/index.html was inspiring. I came up with a solution using std::atomic shown below which “seems” to be working using three std::atomic.
I am worried about my code breaking down on corner cases hence the quoted verb. So can you share advise on verification of such code? Do you have a simpler fool proof code available?
std::atomic<int> barrier1(0);
std::atomic<int> barrier2(0);
std::atomic<int> barrier3(0);
void my_thread()
{
while(1) {
// pop task from queue
...
// and execute task
switch(task.id()) {
case TaskID::Barrier:
barrier2.store(0);
barrier1++;
while (barrier1.load() != NUM_THREAD) {
std::this_thread::yield();
}
barrier3.store(0);
barrier2++;
while (barrier2.load() != NUM_THREAD) {
std::this_thread::yield();
}
barrier1.store(0);
barrier3++;
while (barrier3.load() != NUM_THREAD) {
std::this_thread::yield();
}
break;
case TaskID::Task1:
...
}
}
}
Boost offers a barrier implementation as an extension to the C++11 standard thread library. If using Boost is an option, you should look no further than that.
If you have to rely on standard library facilities, you can roll your own implementation based on std::mutex and std::condition_variable without too much of a hassle.
class Barrier {
int wait_count;
int const target_wait_count;
std::mutex mtx;
std::condition_variable cond_var;
Barrier(int threads_to_wait_for)
: wait_count(0), target_wait_count(threads_to_wait_for) {}
void wait() {
std::unique_lock<std::mutex> lk(mtx);
++wait_count;
if(wait_count != target_wait_count) {
// not all threads have arrived yet; go to sleep until they do
cond_var.wait(lk,
[this]() { return wait_count == target_wait_count; });
} else {
// we are the last thread to arrive; wake the others and go on
cond_var.notify_all();
}
// note that if you want to reuse the barrier, you will have to
// reset wait_count to 0 now before calling wait again
// if you do this, be aware that the reset must be synchronized with
// threads that are still stuck in the wait
}
};
This implementation has the advantage over your atomics-based solution that threads waiting in condition_variable::wait should get send to sleep by your operating system's scheduler, so you don't block CPU cores by having waiting threads spin on the barrier.
A few words on resetting the barrier: The simplest solution is to just have a separate reset() method and have the user ensure that reset and wait are never invoked concurrently. But in many use cases, this is not easy to achieve for the user.
For a self-resetting barrier, you have to consider races on the wait count: If the wait count is reset before the last thread returned from wait, some threads might get stuck in the barrier. A clever solution here is to not have the terminating condition depend on the wait count variable itself. Instead you introduce a second counter, that is only increased by the thread calling the notify. The other threads then observe that counter for changes to determine whether to exit the wait:
void wait() {
std::unique_lock<std::mutex> lk(mtx);
unsigned int const current_wait_cycle = m_inter_wait_count;
++wait_count;
if(wait_count != target_wait_count) {
// wait condition must not depend on wait_count
cond_var.wait(lk,
[this, current_wait_cycle]() {
return m_inter_wait_count != current_wait_cycle;
});
} else {
// increasing the second counter allows waiting threads to exit
++m_inter_wait_count;
cond_var.notify_all();
}
}
This solution is correct under the (very reasonable) assumption that all threads leave the wait before the inter_wait_count overflows.
With atomic variables, using three of them for a barrier is simply overkill that only serves to complicate the issue. You know the number of threads, so you can simply atomically increment a single counter every time a thread enters the barrier, and then spin until the counter becomes greater or equal to N. Something like this:
void barrier(int N) {
static std::atomic<unsigned int> gCounter = 0;
gCounter++;
while((int)(gCounter - N) < 0) std::this_thread::yield();
}
If you don't have more threads than CPU cores and a short expected waiting time, you might want to remove the call to std::this_thread::yield(). This call is likely to be really expensive (more than a microsecond, I'd wager, but I haven't measured it). Depending on the size of your tasks, this may be significant.
If you want to do repeated barriers, just increment the N as you go:
unsigned int lastBarrier = 0;
while(1) {
switch(task.id()) {
case TaskID::Barrier:
barrier(lastBarrier += processCount);
break;
}
}
I would like to point out that in the solution given by #ComicSansMS ,
wait_count should be reset to 0 before executing cond_var.notify_all();
This is because when the barrier is called a second time the if condition will always fail, if wait_count is not reset to 0.
UINT __stdcall CExternal::WorkThread( void * pParam)
{
HRESULT hr;
CTaskBase* pTask;
CComPtr<IHTMLDocument3> spDoc3;
CExternal* pThis = reinterpret_cast<CExternal*>(pParam);
if (pThis == NULL)
return 0;
// Init the com
::CoInitializeEx(0,COINIT_APARTMENTTHREADED);
hr = ::CoGetInterfaceAndReleaseStream(
pThis->m_pStream_,
IID_IHTMLDocument3,
(void**)&spDoc3);
if(FAILED(hr))
return 0;
while (pThis->m_bShutdown_ == 0)
{
if(pThis->m_TaskList_.size())
{
pTask = pThis->m_TaskList_.front();
pThis->m_TaskList_.pop_front();
if(pTask)
{
pTask->doTask(spDoc3); //do my custom task
delete pTask;
}
}
else
{
Sleep(10);
}
}
OutputDebugString(L"start CoUninitialize\n");
::CoUninitialize(); //release com
OutputDebugString(L"end CoUninitialize\n");
return 0;
}
The above the code that let my thread hang, the only output is "start CoUninitialize".
m_hWorker_ = (HANDLE)_beginthreadex(NULL, 0, WorkThread, this, 0, 0);
This code starts my thread, but the thread can't exit safely, so it waits. What the problem with this code?
The problem is not in this code, although it violates core COM requirements. Which says that you should release interface pointers when you no longer use them, calling IUnknown::Release(), and that an apartment-threaded thread must pump a message loop. Especially the message loop is important, you'll get deadlock when the owner thread of a single-threaded object (like a browser) is not pumping.
CoUninitialize() is forced to clean up the interface pointer wrapped by spDoc3 since you didn't do this yourself. It is clear from the code that the owner of the interface pointer actually runs on another thread, something to generally keep in mind since that pretty much defeats the point of starting your own worker thread. Creating your own STA thread doesn't fix this, it is still the wrong thread.
So the proxy needs to context switch to the apartment that owns the browser object. With the hard requirement that this apartment pumps a message loop so that the call can be dispatched on the right thread in order to safely call the Release() function. With very high odds that this thread isn't pumping messages anymore when your program is shutting down. Something you should be able to see in the debugger, locate the owner thread in the Debug + Windows + Threads window and see what it is doing.
Deadlock is the common outcome. The only good way to fix it is to shut down threads in the right order, this one has to shut down before the thread that owns the browser object. Shutting down a multi-threaded program cleanly can be quite difficult when threads have an interdependency like this. The inspiration behind the C++11 std::quick_exit() addition.
I'm trying to implement a sort of thread pool whereby I keep threads in a FIFO and process a bunch of images. Unfortunately, for some reason my cond_wait doesn't always wake even though it's been signaled.
// Initialize the thread pool
for(i=0;i<numThreads;i++)
{
pthread_t *tmpthread = (pthread_t *) malloc(sizeof(pthread_t));
struct Node* newNode;
newNode=(struct Node *) malloc(sizeof(struct Node));
newNode->Thread = tmpthread;
newNode->Id = i;
newNode->threadParams = 0;
pthread_cond_init(&(newNode->cond),NULL);
pthread_mutex_init(&(newNode->mutx),NULL);
pthread_create( tmpthread, NULL, someprocess, (void*) newNode);
push_back(newNode, &threadPool);
}
for() //stuff here
{
//...stuff
pthread_mutex_lock(&queueMutex);
struct Node *tmpNode = pop_front(&threadPool);
pthread_mutex_unlock(&queueMutex);
if(tmpNode != 0)
{
pthread_mutex_lock(&(tmpNode->mutx));
pthread_cond_signal(&(tmpNode->cond)); // Not starting mutex sometimes?
pthread_mutex_unlock(&(tmpNode->mutx));
}
//...stuff
}
destroy_threads=1;
//loop through and signal all the threads again so they can exit.
//pthread_join here
}
void *someprocess(void* threadarg)
{
do
{
//...stuff
pthread_mutex_lock(&(threadNode->mutx));
pthread_cond_wait(&(threadNode->cond), &(threadNode->mutx));
// Doesn't always seem to resume here after signalled.
pthread_mutex_unlock(&(threadNode->mutx));
} while(!destroy_threads);
pthread_exit(NULL);
}
Am I missing something? It works about half of the time, so I would assume that I have a race somewhere, but the only thing I can think of is that I'm screwing up the mutexes? I read something about not signalling before locking or something, but I don't really understand what's going on.
Any suggestions?
Thanks!
Firstly, your example shows you locking the queueMutex around the call to pop_front, but not round push_back. Typically you would need to lock round both, unless you can guarantee that all the pushes happen-before all the pops.
Secondly, your call to pthread_cond_wait doesn't seem to have an associated predicate. Typical usage of condition variables is:
pthread_mutex_lock(&mtx);
while(!ready)
{
pthread_cond_wait(&cond,&mtx);
}
do_stuff();
pthread_mutex_unlock(&mtx);
In this example, ready is some variable that is set by another thread whilst that thread holds a lock on mtx.
If the waiting thread is not blocked in the pthread_cond_wait when pthread_cond_signal is called then the signal will be ignored. The associated ready variable allows you to handle this scenario, and also allows you to handle so-called spurious wake-ups where the call to pthread_cond_wait returns without a corresponding call to pthread_cond_signal from another thread.
I'm not sure, but I think you don't have to (you must not) lock the mutex in the thread pool before calling pthread_cond_signal(&(tmpNode->cond)); , otherwise, the thread which is woken up won't be able to lock the mutex as part of pthread_cond_wait(&(threadNode->cond), &(threadNode->mutx)); operation.