[英]Barrier Synchronization between 2 process using mutex
我需要使用互斥锁(仅)在2个线程之间实现屏障同步。 屏障同步是2个线程将在预定义的步骤互相等待,然后再继续。
我能够使用seamaphore做到这一点,但是我如何仅使用互斥锁来实现这一点 。 提示我需要2个互斥锁而不是1个。
使用Seamaphore:
#include <pthread.h>
#include <semaphore.h>
using namespace std;
sem_t s1;
sem_t s2;
void* fun1(void* i)
{
cout << "fun1 stage 1" << endl;
cout << "fun1 stage 2" << endl;
cout << "fun1 stage 3" << endl;
sem_post (&s1);
sem_wait (&s2);
cout << "fun1 stage 4" << endl;
}
void* fun2(void* i)
{
cout << "fun2 stage 1" << endl;
cout << "fun2 stage 2" << endl;
// sleep(5);
sem_post (&s2);
sem_wait (&s1);
cout << "fun2 stage 3" << endl;
}
main()
{
sem_init(&s1, 0, 0);
sem_init(&s2, 0, 0);
int value;
sem_getvalue(&s2, &value);
cout << "s2 = " << value << endl;
pthread_t iThreadId;
cout << pthread_create(&iThreadId, NULL, &fun2, NULL) << endl;
// cout << pthread_create(&iThreadId, NULL, &fun2, NULL) << endl;
pthread_create(&iThreadId, NULL, &fun1, NULL);
sleep(10);
}
将以上代码编译为“ g ++ barrier.cc -lpthread”
没有杂物和锁怎么办? 仅使用原子操作 :
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <signal.h>
static sigset_t _fSigSet;
static volatile int _cMax=20, _cWait = 0;
static pthread_t _aThread[1000];
void * thread(void *idIn)
{
int nSig, iThread, cWait, id = (int)idIn;
printf("Start %d\n", id, cWait, _cMax);
// do some fake weork
nanosleep(&(struct timespec){0, 500000000}, NULL);
// barrier
cWait = __sync_add_and_fetch(&_cWait, 1);
printf("Middle %d, %d/%d Waiting\n", id, cWait, _cMax);
if (cWait < _cMax)
{
// if we are not the last thread, sleep on signal
sigwait(&_fSigSet, &nSig); // sleepytime
}
else
{
// if we are the last thread, don't sleep and wake everyone else up
for (iThread = 0; iThread < _cMax; ++iThread)
if (iThread != id)
pthread_kill(_aThread[iThread], SIGUSR1);
}
// watch em wake up
cWait = __sync_add_and_fetch(&_cWait, -1);
printf("End %d, %d/%d Active\n", id, cWait, _cMax);
return 0;
}
int main(int argc, char** argv)
{
pthread_attr_t attr;
int i, err;
sigemptyset(&_fSigSet);
sigaddset(&_fSigSet, SIGUSR1);
sigaddset(&_fSigSet, SIGSEGV);
printf("Start\n");
pthread_attr_init(&attr);
if ((err = pthread_attr_setstacksize(&attr, 16384)) != 0)
{
printf("pthread_attr_setstacksize failed: err: %d %s\n", err, strerror(err));
exit(0);
}
for (i = 0; i < _cMax; i++)
{
if ((err = pthread_create(&_aThread[i], &attr, thread, (void*)i)) != 0)
{
printf("pthread_create failed on thread %d, error code: %d %s\n", i, err, strerror(err));
exit(0);
}
}
for (i = 0; i < _cMax; ++i)
pthread_join(_aThread[i], NULL);
printf("\nDone.\n");
return 0;
}
我不确定您是否需要两个互斥锁,其中一个互斥锁和一个条件变量以及一个额外的标志可能就足够了。 这个想法是您通过获取互斥锁进入关键部分,然后检查是否是第一个线程,如果是,则等待条件。 如果您是第二个线程,那么您将唤醒等待的线程,并且两个线程都将离开。
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