有没有其他同步方式
Is there an alternative way to sync
是否有其他方法可以确保线程已准备好接收广播信号。我想替换 main 中的 Sleep(1) 函数。
#include <iostream>
#include <pthread.h>
#define NUM 4
using namespace std;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
pthread_t tid[NUM];
void *threads(void *arg){
int tid = (int)arg;
while(true){
pthread_mutex_lock(&mutex);
pthread_cond_wait(&cond,&mutex);
//do some work
cout<<"Thread: "<<tid<<endl;;
pthread_mutex_unlock(&mutex);
}
}
int main(){
for(int i=0;i<NUM;i++){
pthread_create(&tid[i],NULL,threads,(void*)i);
}
Sleep(1);
pthread_cond_broadcast(&cond);
Sleep(1);
pthread_cond_broadcast(&cond);
Sleep(1);
pthread_cond_broadcast(&cond);
return 0;
}
我在 pthread_cond_wait 之前尝试过内存屏障,我想过使用计数器,但还没有任何效果。
条件变量通常连接到谓词。在其他线程中,检查谓词是否已经满足(检查同时持有保护谓词的互斥锁),如果是,则不要等待条件变量。在 main 中,获取互斥锁,持有互斥锁的同时改变谓词。然后释放互斥量并在 condvar 上发出信号或广播。这是一个类似的问题:
Synchronisation before pthread_cond_broadcast
下面是一些示例代码:
#include <iostream>
#include <pthread.h>
#include <unistd.h>
#include <cassert>
#define NUM 4
#define SIZE 256
using std::cout;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
pthread_t tid[NUM];
int work_available;
void *threads(void *arg)
{
int tid = *((int*)arg);
while (1) {
pthread_mutex_lock(&mutex);
while (work_available == 0) {
// While loop since cond_wait can have spurious wakeups.
pthread_cond_wait(&cond, &mutex);
cout << "Worker " << tid << " woke up...\n";
cout << "Work available: " << work_available << '\n';
}
if (work_available == -1) {
cout << "Worker " << tid << " quitting\n";
pthread_mutex_unlock(&mutex); // Easy to forget, better to use C++11 RAII mutexes.
break;
}
assert(work_available > 0);
work_available--;
cout << "Worker " << tid << " took one item of work\n";
pthread_mutex_unlock(&mutex);
//do some work
sleep(2); // simulated work
pthread_mutex_lock(&mutex);
cout << "Worker " << tid << " done with one item of work.\n";
pthread_mutex_unlock(&mutex);
}
}
int main()
{
work_available = 0;
int args[NUM];
for (int i=0; i<NUM; i++) {
args[i] = i;
pthread_create(&tid[i], NULL, threads, (void*)&args[i]);
}
const int MAX_TIME = 10;
for (int i = 0; i < MAX_TIME; i++)
{
pthread_mutex_lock(&mutex);
work_available++;
cout << "Main thread, work available: " << work_available << '\n';
pthread_mutex_unlock(&mutex);
pthread_cond_broadcast(&cond);
sleep(1);
}
pthread_mutex_lock(&mutex);
cout << "Main signalling threads to quit\n";
work_available = -1;
pthread_mutex_unlock(&mutex);
pthread_cond_broadcast(&cond);
for (int i = 0; i < NUM; i++)
{
pthread_join(tid[i], NULL);
}
return 0;
}
是否有其他方法可以确保线程已准备好接收广播信号。我想替换 main 中的 Sleep(1) 函数。
#include <iostream>
#include <pthread.h>
#define NUM 4
using namespace std;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
pthread_t tid[NUM];
void *threads(void *arg){
int tid = (int)arg;
while(true){
pthread_mutex_lock(&mutex);
pthread_cond_wait(&cond,&mutex);
//do some work
cout<<"Thread: "<<tid<<endl;;
pthread_mutex_unlock(&mutex);
}
}
int main(){
for(int i=0;i<NUM;i++){
pthread_create(&tid[i],NULL,threads,(void*)i);
}
Sleep(1);
pthread_cond_broadcast(&cond);
Sleep(1);
pthread_cond_broadcast(&cond);
Sleep(1);
pthread_cond_broadcast(&cond);
return 0;
}
我在 pthread_cond_wait 之前尝试过内存屏障,我想过使用计数器,但还没有任何效果。
条件变量通常连接到谓词。在其他线程中,检查谓词是否已经满足(检查同时持有保护谓词的互斥锁),如果是,则不要等待条件变量。在 main 中,获取互斥锁,持有互斥锁的同时改变谓词。然后释放互斥量并在 condvar 上发出信号或广播。这是一个类似的问题: Synchronisation before pthread_cond_broadcast
下面是一些示例代码:
#include <iostream>
#include <pthread.h>
#include <unistd.h>
#include <cassert>
#define NUM 4
#define SIZE 256
using std::cout;
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
pthread_t tid[NUM];
int work_available;
void *threads(void *arg)
{
int tid = *((int*)arg);
while (1) {
pthread_mutex_lock(&mutex);
while (work_available == 0) {
// While loop since cond_wait can have spurious wakeups.
pthread_cond_wait(&cond, &mutex);
cout << "Worker " << tid << " woke up...\n";
cout << "Work available: " << work_available << '\n';
}
if (work_available == -1) {
cout << "Worker " << tid << " quitting\n";
pthread_mutex_unlock(&mutex); // Easy to forget, better to use C++11 RAII mutexes.
break;
}
assert(work_available > 0);
work_available--;
cout << "Worker " << tid << " took one item of work\n";
pthread_mutex_unlock(&mutex);
//do some work
sleep(2); // simulated work
pthread_mutex_lock(&mutex);
cout << "Worker " << tid << " done with one item of work.\n";
pthread_mutex_unlock(&mutex);
}
}
int main()
{
work_available = 0;
int args[NUM];
for (int i=0; i<NUM; i++) {
args[i] = i;
pthread_create(&tid[i], NULL, threads, (void*)&args[i]);
}
const int MAX_TIME = 10;
for (int i = 0; i < MAX_TIME; i++)
{
pthread_mutex_lock(&mutex);
work_available++;
cout << "Main thread, work available: " << work_available << '\n';
pthread_mutex_unlock(&mutex);
pthread_cond_broadcast(&cond);
sleep(1);
}
pthread_mutex_lock(&mutex);
cout << "Main signalling threads to quit\n";
work_available = -1;
pthread_mutex_unlock(&mutex);
pthread_cond_broadcast(&cond);
for (int i = 0; i < NUM; i++)
{
pthread_join(tid[i], NULL);
}
return 0;
}