【发布时间】:2017-08-11 05:22:00
【问题描述】:
在找话题的时候,我从here找到了代码。如您所见,两个函数线程都使用相同的mutex。那么,即使先前的线程拥有互斥锁,如何向其他线程发出信号或捕获信号并继续其功能?如何/为什么没有死锁?有点混乱。
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
pthread_mutex_t count_mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t condition_var = PTHREAD_COND_INITIALIZER;
void *functionCount1();
void *functionCount2();
int count = 0;
#define COUNT_DONE 10
#define COUNT_HALT1 3
#define COUNT_HALT2 6
main()
{
pthread_t thread1, thread2;
pthread_create( &thread1, NULL, &functionCount1, NULL);
pthread_create( &thread2, NULL, &functionCount2, NULL);
pthread_join( thread1, NULL);
pthread_join( thread2, NULL);
printf("Final count: %d\n",count);
exit(EXIT_SUCCESS);
}
// Write numbers 1-3 and 8-10 as permitted by functionCount2()
void *functionCount1()
{
for(;;)
{
// Lock mutex and then wait for signal to relase mutex
pthread_mutex_lock( &count_mutex ); // <---- Same mutex
// Wait while functionCount2() operates on count
// mutex unlocked if condition varialbe in functionCount2() signaled.
pthread_cond_wait( &condition_var, &count_mutex );
count++;
printf("Counter value functionCount1: %d\n",count);
pthread_mutex_unlock( &count_mutex );
if(count >= COUNT_DONE) return(NULL);
}
}
// Write numbers 4-7
void *functionCount2()
{
for(;;)
{
pthread_mutex_lock( &count_mutex ); // <---- Same mutex
if( count < COUNT_HALT1 || count > COUNT_HALT2 )
{
// Condition of if statement has been met.
// Signal to free waiting thread by freeing the mutex.
// Note: functionCount1() is now permitted to modify "count".
pthread_cond_signal( &condition_var );
}
else
{
count++;
printf("Counter value functionCount2: %d\n",count);
}
pthread_mutex_unlock( &count_mutex );
if(count >= COUNT_DONE) return(NULL);
}
}
【问题讨论】:
标签: c multithreading pthreads