c - 在 C 中对 pthreads 使用互斥锁/解锁和广播

Question

这是我的第一篇文章，我很兴奋。

我的问题是我正在用 C 创建一个石头剪刀布程序，其中父进程创建 2 个线程。然后，2 个线程随机扔一 block 石头、布或剪刀，并将值返回给父线程，父线程将对其进行计数，并返回 3 轮的结果，然后进行最终计数。

我的问题是我无法让线程正确启动，我让它们在我的 thread_function1 中等待，但随后它们只完成一轮，即使这样我也没有在结果中得到两个线程。如果有人可以阐明一些观点，我将非常感激!谢谢

#include <stdlib.h>
#include <stdio.h>
#include <errno.h>
#include <pthread.h>
#include <sys/time.h>

#define NTHREADS 2
struct timeval tv;
void *thread_function1();
void *thread_function2();
char *guess_string(int g);
int wins[3];
int cmd_ready = 0;
int x1=0, x2=1, count=0;
int  guess, object, turns, i, j, k,l, winner, cmd, go, y;
pthread_mutex_t cv_m;
pthread_mutex_t count_mutex;
pthread_cond_t cv;
int myindex;
int flag; 
int throws[3];
int main(int argc, char *argv[])
{


  wins[0] = 0; wins[1] = 0;
  if ((argc != 2) || ((turns = atoi(argv[1])) <= 0))
    {
      fprintf(stderr,"Usage: %s turns\n", argv[0]);
      return 0;
    }

    pthread_t thread_id1, thread_id2;


    if (pthread_create(&thread_id1, NULL, thread_function1,&x1) != 0)
        perror("pthread_create"),
        exit(1); 

    if (pthread_create(&thread_id2, NULL, thread_function1,&x2) != 0)
        perror("pthread_create"),
        exit(1); 

    printf("Beginning %d Rounds...\nFight!\n", turns);
    printf("Child 1 TID: %d\n", (unsigned int) thread_id1);
    printf("Child 2 TID: %d\n", (unsigned int) thread_id2 );

    for(k=0; k<turns; k++)
    {

            pthread_mutex_lock (&cv_m);
            cmd = go; 
            cmd_ready = 2;
            pthread_cond_broadcast(&cv);    
            pthread_mutex_unlock(&cv_m);
            printf("------------------------\n");
            printf("Round: %d\n", k+1);

            printf("Child %d throws %s!\n",myindex+1, guess_string(myindex));

            pthread_mutex_lock (&count_mutex);

            winner = find_winner(throws[0], throws[1]);
            while(count == 2){
            if(winner >= 0)
            {
                printf("Child %d Wins!\n", winner+1);
                wins[winner]++;
                printf("6\n");
            }else
            {
                printf("Game is a Tie!\n");
            }
            go--;
            count = 0; 
            pthread_mutex_unlock(&count_mutex);
            }
    }

    pthread_join(thread_id1,NULL); 
    pthread_join(thread_id2,NULL);

    printf("------------------------\n");
    printf("------------------------\n");
    printf("Result:\n");
    printf("Child 1: %d\n", wins[0]);
    printf("Child 2: %d\n", wins[1]);
    printf("Ties: %d\n", turns - (wins[0] + wins[1]));
    printf("Child %d Wins!\n", (wins[0] > wins[1]) ? 1 : 2);

    pthread_mutex_destroy(&cv_m);
    pthread_cond_destroy(&cv);
    pthread_exit(NULL);
    return 0; 

}

void *thread_function1(void *p)
{
    struct timeval tv;
    myindex = *(int *)p;
    gettimeofday(&tv, NULL);
    srand(tv.tv_sec + tv.tv_usec + getpid());
    printf("1\n");
    pthread_mutex_lock (&cv_m);
    while(cmd_ready == 0)
    {
        printf("2\n");
        pthread_cond_wait(&cv, &cv_m);
    }
    printf("3\n");

    throws[myindex] = rand() % 3;
    cmd_ready--;
    printf("Ready: %d\n",cmd_ready);
    pthread_mutex_unlock (&cv_m);
    printf("4\n");

    pthread_mutex_lock (&count_mutex);
    count++;
    printf("Count %d\n", count);
    pthread_mutex_unlock(&count_mutex);

    while(count == 2){
        printf("5\n");
        return NULL;
    }

}

char *guess_string(int g){
    switch(g){
    case 0:
        return "Rock";
        break;
    case 1:
        return "Paper";
        break;
    case 2:
        return "Scissors";
        break;
    }
}

int find_winner(int g1, int g2){
    if(g1 == g2)
        return -1; 
    else if ((g1 == 2) && (g2 == 0))
        return 1; 
    else if ((g1 == 0) && (g2 == 2))
        return 0;
    else
        return (g1 > g2) ? 0 : 1;
}

Answer 1

您似乎没有使用 pthread_mutex_init 或 pthread_cond_init 初始化互斥锁或条件。

两个线程都在不 protected 情况下修改变量 myindex，更新该变量的第二个线程将是似乎返返回告的线程。

您还指望您的线程在 main 获取锁并发出广播之前开始等待条件，您可能会遇到 main 首先到达那里而您的线程尚未准备好的情况。

这应该是一个开始。