Semaphore与multithreading中的条件变量？

问题 ：我必须增加x1和x2变量，这应该由不同的线程完成，并且在两个变量的上一个增量都没有完成之前，不应该调用两个变量的下一个增量。

问题：

x1 = 0; x2 = 0; x1++; and x2++; should run parallel on different threads and wait for printing. should print: x1 = 1 and x2 = 1 x1++; and x2++; should run parallel on different threads and wait for printing. should print: x1 = 2 and x2 = 2 x1++; and x2++; should run parallel on different threads and wait for printing. should print: x1 = 3 and x2 = 3 x1++; and x2++; should run parallel on different threads and wait for printing. should print: x1 = 4 and x2 = 4 … … … x1++; and x2++; should run parallel on different threads and wait for printing. should print: x1 = 10 and x2 = 10 close the threads 

建议的解决方案使用pthread条件 ：初始化4个互斥锁和4个条件变量，并使用2个互斥锁来锁定主函数并为每个线程rest。 两个线程都将等待main函数传递condtion信号以调用它们，并且在计算之后，它们将信号传递回主线程以进一步移动。 提供1秒的睡眠，要求线程正确调用并准备接收并在计算后返回信号。

Pthread条件代码：

  #include  #include  pthread_t pth1,pth2; //Values to calculate int x1 = 0, x2 = 0; pthread_mutex_t m1, m2, m3, m4 = PTHREAD_MUTEX_INITIALIZER; pthread_cond_t c1, c2, c3, c4 = PTHREAD_COND_INITIALIZER; void *threadfunc1(void *parm) { pthread_mutex_lock(&m1); for (;;) { pthread_cond_wait(&c1, &m1); x1++; pthread_mutex_lock(&m3); pthread_cond_signal(&c3); pthread_mutex_unlock(&m3); } pthread_mutex_unlock(&m1); return NULL ; } void *threadfunc2(void *parm) { pthread_mutex_lock(&m2); for (;;) { pthread_cond_wait(&c2, &m2); x2++; pthread_mutex_lock(&m4); pthread_cond_signal(&c4); pthread_mutex_unlock(&m4); } pthread_mutex_unlock(&m2); return NULL ; } int main () { pthread_create(&pth1, NULL, threadfunc1, "foo"); pthread_create(&pth2, NULL, threadfunc2, "foo"); sleep(1); int loop = 0; pthread_mutex_lock(&m3); pthread_mutex_lock(&m4); while (loop < 10) { // iterated as a step loop++; printf("Initial : x1 = %d, x2 = %d\n", x1, x2); pthread_mutex_lock(&m1); pthread_cond_signal(&c1); pthread_mutex_unlock(&m1); pthread_mutex_lock(&m2); pthread_cond_signal(&c2); pthread_mutex_unlock(&m2); pthread_cond_wait(&c3, &m3); pthread_cond_wait(&c4, &m4); printf("Final : x1 = %d, x2 = %d\n", x1, x2); } printf("Result : x1 = %d, x2 = %d\n", x1, x2); pthread_mutex_unlock(&m3); pthread_mutex_unlock(&m4); pthread_cancel(pth1); pthread_cancel(pth2); return 1; } 

使用信号量的建议解决方案：初始化4个信号量并调用单独的线程以单独增加变量。 2个信号量，用于将消息传递给线程以开始递增; 2个信号量，用于将消息传递到主线程，以完成递增。 主线程将等待来自两个子线程的信号量发布，显示两个变量的增量完成，然后主线程将消息传递给两个子线程，允许进一步递增。

信号量代码：

  #include  #include  #include  //Threads pthread_t pth1,pth2; //Values to calculate int x1 = 0, x2 = 0; sem_t c1,c2,c3,c4; void *threadfunc1(void *parm) { for (;;) { x1++; sem_post(&c1); sem_wait(&c3); } return NULL ; } void *threadfunc2(void *parm) { for (;;) { x2++; sem_post(&c2); sem_wait(&c4); } return NULL ; } int main () { sem_init(&c1, 0, 0); sem_init(&c2, 0, 0); sem_init(&c3, 0, 0); sem_init(&c4, 0, 0); pthread_create(&pth1, NULL, threadfunc1, "foo"); pthread_create(&pth2, NULL, threadfunc2, "foo"); sem_wait(&c1); sem_wait(&c2); sem_post(&c3); sem_post(&c4); int loop = 0; while (loop < 8) { // iterated as a step loop++; printf("Initial : x1 = %d, x2 = %d\n", x1, x2); sem_wait(&c1); sem_wait(&c2); printf("Final : x1 = %d, x2 = %d\n", x1, x2); sem_post(&c3); sem_post(&c4); } sem_wait(&c1); sem_wait(&c2); sem_destroy(&c1); sem_destroy(&c2); sem_destroy(&c3); sem_destroy(&c4); printf("Result : x1 = %d, x2 = %d\n", x1, x2); pthread_cancel(pth1); pthread_cancel(pth2); return 1; } 

请建议我，哪一种更好的实施方式或哪些方面可以改进？ 任何建议都会有所帮助。 提前致谢。 对不起，如果我重复自己，因为，这个问题已成为我的噩梦..请帮助。