我在下面有以下代码。我希望一次只有一半的线程进入threadedfunction。如何创建一个信号量来阻止其他进程?每当线程完成使用该函数时,我将如何解除先前阻塞的进程的阻塞?
#include <iostream>
#include <unistd.h>
#include <sys/wait.h>
#include <pthread.h>
using namespace std;
#define NUM_THREADS 4
long int sharedcount;
pthread_mutex_t count_mutex;
//Function that will be run by multiple threads
//Needs to return a void pointer and if it takes arguments
//it needs to be a void pointer
void *ThreadedFunction(void *threadid)
{
int success;
long id = (long)threadid;
//Lock mutex preventing the other threads from ru nning
success = pthread_mutex_lock( &count_mutex );
cout << "Thread " << id << " beginning.\n";
for(int i = 0; i < 100000000; i++)
sharedcount++;
cout << "Thread " << id << " exiting.\n";
cout << sharedcount << endl;
//Unlock the mutex after the thread has finished running
pthread_mutex_unlock( &count_mutex );
//Kill the thread
pthread_exit(NULL);
}
int main ()
{
//Initialize mutex
pthread_mutex_init(&count_mutex, NULL);
//Create an array of threads
pthread_t threads[NUM_THREADS];
int rc;
int i;
sharedcount = 0;
for( i=0; i < NUM_THREADS; i++ )
{
cout << "main() : creating thread, " << i << endl;
//Create thread by storing it in a location in the array. Call the
//function for the threads to run inside. And pass the argument (if any).
//If no arguments pass NULL
rc = pthread_create(&threads[i], NULL, ThreadedFunction, (void *)i);
if (rc)
{
cout << "Error:unable to create thread," << rc << endl;
exit(-1);
}
}
//Have main thread wait for all other threads to stop running.
for(i = 0; i < NUM_THREADS; i++)
pthread_join(threads[i], NULL);
//cout << sharedcount << endl;
pthread_exit(NULL);
}
最佳答案
您可以做的是使用计数信号量(而不是二进制信号量)。计数信号量的初始值大于 1,允许多个线程在信号量上调用“等待”,而不是让这些线程实际阻塞并放入信号量队列。
在您的情况下,我会做的是在主函数中初始化一个信号量,其初始值为 NUM_THREADS/2。然后我会在 threadedFunction 的开头插入一行,在那里我做一个等待(信号量),并在函数的末尾插入一行,在那里你做一个信号量(信号量)。这样,当一个线程即将退出该函数时,它会向一个在信号量上调用 wait 后被阻塞的线程发出信号,并让该线程进入。 希望这会有所帮助。
关于c++ - 实现信号量,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43036839/