Come fare loop fili Win32/MFC in lockstep?

Question

Sono un neofita del multithreading in Windows, quindi questa potrebbe essere una domanda banale: qual è il modo più semplice per assicurarsi che i thread eseguano un ciclo in sequenza?

Ho provato a passare un array condiviso di Event s a tutti i thread e utilizzando WaitForMultipleObjects alla fine del ciclo per sincronizzarli, ma questo mi dà un deadlock dopo uno, a volte due cicli. Ecco una versione semplificata del mio codice corrente (con solo due fili, ma vorrei renderlo scalabile):

typedef struct 
{ 
    int rank; 
    HANDLE* step_events; 
} IterationParams; 

int main(int argc, char **argv) 
{ 
    // ... 

    IterationParams p[2]; 
    HANDLE step_events[2]; 
    for (int j=0; j<2; ++j) 
    { 
     step_events[j] = CreateEvent(NULL, FALSE, FALSE, NULL); 
    } 

    for (int j=0; j<2; ++j) 
    { 
     p[j].rank = j; 
     p[j].step_events = step_events; 
     AfxBeginThread(Iteration, p+j); 
    } 

    // ... 
} 

UINT Iteration(LPVOID pParam) 
{ 
    IterationParams* p = (IterationParams*)pParam; 
    int rank = p->rank; 

    for (int i=0; i<100; i++) 
    { 
     if (rank == 0) 
     { 
      printf("%dth iteration\n",i); 
      // do something 
      SetEvent(p->step_events[0]); 
      WaitForMultipleObjects(2, p->step_events, TRUE, INFINITE); 
     } 
     else if (rank == 1) 
     { 
      // do something else 
      SetEvent(p->step_events[1]); 
      WaitForMultipleObjects(2, p->step_events, TRUE, INFINITE); 
     } 
    } 
    return 0; 
} 

(so che sto mescolando C e C++, in realtà è l'eredità di codice C che ho Sto cercando di parallelizzare.)

Leggere i documenti su MSDN, penso che questo dovrebbe funzionare. Tuttavia, il thread 0 stampa solo una volta, occasionalmente due volte, quindi il programma si blocca. È un modo corretto per sincronizzare i thread? In caso contrario, cosa consiglieresti (non esiste davvero un supporto integrato per una barriera in MFC?).

---

EDIT: questa soluzione è SBAGLIATO, tra cui anche Alessandro's fix. Ad esempio, si consideri questo scenario:

1. Discussione 0 imposta il suo evento e chiede Attendere, blocchi
2. filettatura 1 definisce i suoi eventi e le chiamate Attendere, blocchi
3. discussione 0 rendimenti Attendere, ripristina la sua manifestazione, e completa un ciclo senza Thread 1 ottenendo il controllo
4. Il thread 0 imposta il proprio evento e chiama Wait. Poiché Thread 1 non ha ancora avuto la possibilità di reimpostare il suo evento, Thread Wait di Wait ritorna immediatamente ei thread non sono sincronizzati.

Quindi la domanda rimane: come si fa a tranquillamente assicurarsi che i fili rimanere in sincronia?

Answer 1

Introduzione

ho implementato un semplice programma C++ per la vostra considerazione (testato in Visual Studio 2010). Sta usando solo le API Win32 (e la libreria standard per l'output della console e un po 'di randomizzazione). Dovresti essere in grado di rilasciarlo in un nuovo progetto di console Win32 (senza intestazioni precompilate), compilare ed eseguire.

---

Soluzione

#include <tchar.h> 
#include <windows.h> 


//--------------------------------------------------------- 
// Defines synchronization info structure. All threads will 
// use the same instance of this struct to implement randezvous/ 
// barrier synchronization pattern. 
struct SyncInfo 
{ 
    SyncInfo(int threadsCount) : Awaiting(threadsCount), ThreadsCount(threadsCount), Semaphore(::CreateSemaphore(0, 0, 1024, 0)) {}; 
    ~SyncInfo() { ::CloseHandle(this->Semaphore); } 
    volatile unsigned int Awaiting; // how many threads still have to complete their iteration 
    const int ThreadsCount; 
    const HANDLE Semaphore; 
}; 


//--------------------------------------------------------- 
// Thread-specific parameters. Note that Sync is a reference 
// (i.e. all threads share the same SyncInfo instance). 
struct ThreadParams 
{ 
    ThreadParams(SyncInfo &sync, int ordinal, int delay) : Sync(sync), Ordinal(ordinal), Delay(delay) {}; 
    SyncInfo &Sync; 
    const int Ordinal; 
    const int Delay; 
}; 


//--------------------------------------------------------- 
// Called at the end of each itaration, it will "randezvous" 
// (meet) all the threads before returning (so that next 
// iteration can begin). In practical terms this function 
// will block until all the other threads finish their iteration. 
static void RandezvousOthers(SyncInfo &sync, int ordinal) 
{ 
    if (0 == ::InterlockedDecrement(&(sync.Awaiting))) { // are we the last ones to arrive? 
     // at this point, all the other threads are blocking on the semaphore 
     // so we can manipulate shared structures without having to worry 
     // about conflicts 
     sync.Awaiting = sync.ThreadsCount; 
     wprintf(L"Thread %d is the last to arrive, releasing synchronization barrier\n", ordinal); 
     wprintf(L"---~~~---\n"); 

     // let's release the other threads from their slumber 
     // by using the semaphore 
     ::ReleaseSemaphore(sync.Semaphore, sync.ThreadsCount - 1, 0); // "ThreadsCount - 1" because this last thread will not block on semaphore 
    } 
    else { // nope, there are other threads still working on the iteration so let's wait 
     wprintf(L"Thread %d is waiting on synchronization barrier\n", ordinal); 
     ::WaitForSingleObject(sync.Semaphore, INFINITE); // note that return value should be validated at this point ;) 
    } 
} 


//--------------------------------------------------------- 
// Define worker thread lifetime. It starts with retrieving 
// thread-specific parameters, then loops through 5 iterations 
// (randezvous-ing with other threads at the end of each), 
// and then finishes (the thread can then be joined). 
static DWORD WINAPI ThreadProc(void *p) 
{ 
    ThreadParams *params = static_cast<ThreadParams *>(p); 
    wprintf(L"Starting thread %d\n", params->Ordinal); 

    for (int i = 1; i <= 5; ++i) { 
     wprintf(L"Thread %d is executing iteration #%d (%d delay)\n", params->Ordinal, i, params->Delay); 
     ::Sleep(params->Delay); 
     wprintf(L"Thread %d is synchronizing end of iteration #%d\n", params->Ordinal, i); 
     RandezvousOthers(params->Sync, params->Ordinal); 
    } 

    wprintf(L"Finishing thread %d\n", params->Ordinal); 
    return 0; 
} 


//--------------------------------------------------------- 
// Program to illustrate iteration-lockstep C++ solution. 
int _tmain(int argc, _TCHAR* argv[]) 
{ 
    // prepare to run 
    ::srand(::GetTickCount()); // pseudo-randomize random values :-) 
    SyncInfo sync(4); 
    ThreadParams p[] = { 
     ThreadParams(sync, 1, ::rand() * 900/RAND_MAX + 100), // a delay between 200 and 1000 milliseconds will simulate work that an iteration would do 
     ThreadParams(sync, 2, ::rand() * 900/RAND_MAX + 100), 
     ThreadParams(sync, 3, ::rand() * 900/RAND_MAX + 100), 
     ThreadParams(sync, 4, ::rand() * 900/RAND_MAX + 100), 
    }; 

    // let the threads rip 
    HANDLE t[] = { 
     ::CreateThread(0, 0, ThreadProc, p + 0, 0, 0), 
     ::CreateThread(0, 0, ThreadProc, p + 1, 0, 0), 
     ::CreateThread(0, 0, ThreadProc, p + 2, 0, 0), 
     ::CreateThread(0, 0, ThreadProc, p + 3, 0, 0), 
    }; 

    // wait for the threads to finish (join) 
    ::WaitForMultipleObjects(4, t, true, INFINITE); 

    return 0; 
} 

---

Esempio di output

L'esecuzione di questo programma sulla mia macchina (dual-core) produce il seguente output:

Starting thread 1 
Starting thread 2 
Starting thread 4 
Thread 1 is executing iteration #1 (712 delay) 
Starting thread 3 
Thread 2 is executing iteration #1 (798 delay) 
Thread 4 is executing iteration #1 (477 delay) 
Thread 3 is executing iteration #1 (104 delay) 
Thread 3 is synchronizing end of iteration #1 
Thread 3 is waiting on synchronization barrier 
Thread 4 is synchronizing end of iteration #1 
Thread 4 is waiting on synchronization barrier 
Thread 1 is synchronizing end of iteration #1 
Thread 1 is waiting on synchronization barrier 
Thread 2 is synchronizing end of iteration #1 
Thread 2 is the last to arrive, releasing synchronization barrier 
---~~~--- 
Thread 2 is executing iteration #2 (798 delay) 
Thread 3 is executing iteration #2 (104 delay) 
Thread 1 is executing iteration #2 (712 delay) 
Thread 4 is executing iteration #2 (477 delay) 
Thread 3 is synchronizing end of iteration #2 
Thread 3 is waiting on synchronization barrier 
Thread 4 is synchronizing end of iteration #2 
Thread 4 is waiting on synchronization barrier 
Thread 1 is synchronizing end of iteration #2 
Thread 1 is waiting on synchronization barrier 
Thread 2 is synchronizing end of iteration #2 
Thread 2 is the last to arrive, releasing synchronization barrier 
---~~~--- 
Thread 4 is executing iteration #3 (477 delay) 
Thread 3 is executing iteration #3 (104 delay) 
Thread 1 is executing iteration #3 (712 delay) 
Thread 2 is executing iteration #3 (798 delay) 
Thread 3 is synchronizing end of iteration #3 
Thread 3 is waiting on synchronization barrier 
Thread 4 is synchronizing end of iteration #3 
Thread 4 is waiting on synchronization barrier 
Thread 1 is synchronizing end of iteration #3 
Thread 1 is waiting on synchronization barrier 
Thread 2 is synchronizing end of iteration #3 
Thread 2 is the last to arrive, releasing synchronization barrier 
---~~~--- 
Thread 2 is executing iteration #4 (798 delay) 
Thread 3 is executing iteration #4 (104 delay) 
Thread 1 is executing iteration #4 (712 delay) 
Thread 4 is executing iteration #4 (477 delay) 
Thread 3 is synchronizing end of iteration #4 
Thread 3 is waiting on synchronization barrier 
Thread 4 is synchronizing end of iteration #4 
Thread 4 is waiting on synchronization barrier 
Thread 1 is synchronizing end of iteration #4 
Thread 1 is waiting on synchronization barrier 
Thread 2 is synchronizing end of iteration #4 
Thread 2 is the last to arrive, releasing synchronization barrier 
---~~~--- 
Thread 3 is executing iteration #5 (104 delay) 
Thread 4 is executing iteration #5 (477 delay) 
Thread 1 is executing iteration #5 (712 delay) 
Thread 2 is executing iteration #5 (798 delay) 
Thread 3 is synchronizing end of iteration #5 
Thread 3 is waiting on synchronization barrier 
Thread 4 is synchronizing end of iteration #5 
Thread 4 is waiting on synchronization barrier 
Thread 1 is synchronizing end of iteration #5 
Thread 1 is waiting on synchronization barrier 
Thread 2 is synchronizing end of iteration #5 
Thread 2 is the last to arrive, releasing synchronization barrier 
---~~~--- 
Finishing thread 4 
Finishing thread 3 
Finishing thread 2 
Finishing thread 1 

Si noti che per semplicità che ogni filo ha durata casuale dell'iterazione, ma tutte le iterazioni di quel thread useranno la stessa durata casuale (es. non cambia tra le iterazioni).

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Come funziona?

Il "nucleo" della soluzione è nella funzione "RandezvousOthers". Questa funzione bloccherà su un semaforo condiviso (se il thread su cui questa funzione è stata chiamata non è l'ultimo a chiamare la funzione) o ripristina la struttura Sync e sblocca tutti i thread bloccati su un semaforo condiviso (se il thread su cui è presente la funzione è stata chiamata è stata l'ultima a chiamare la funzione).

Answer 2

Per farlo funzionare, impostare il secondo parametro di CreateEvent su TRUE. Ciò renderà gli eventi "reset manuale" e impedirà allo Waitxxx di ripristinarlo. Quindi posizionare uno ResetEvent all'inizio del ciclo.

Answer 3

Ho trovato questo SyncTools (scarica SyncTools.zip) googling "finestre di sincronizzazione barriera". Usa una CriticalSection e un Event per implementare una barriera per N thread.