c - 试图将函数指针传递给 pthread

Question

我正在尝试创建一个带有函数指针参数的 pthread，这里首先是将在创建 pthread 时调用的函数..

void *passenger(void *arguements){
        struct arg_struct *args = arguements;
        int passenger = args->p;
        int from_floor = args->f;
        int to_floor = args->t;
        void (*enter)(int,int) = args->en;
        void (*exit)(int,int) = args->ex;
        // wait for the elevator to arrive at our origin floor, then get in
        int waiting = 1;
        while(waiting){
                if(current_floor == from_floor && state == ELEVATOR_OPEN && occupancy==0) {
                        pthread_mutex_lock(&lock);
                        enter(passenger, 0);
                        occupancy++;
                        waiting=0;
                        pthread_mutex_unlock(&lock);
                }
        }

        // wait for the elevator at our destination floor, then get out
        int riding=1;
        while(riding) {
                if(current_floor == to_floor && state == ELEVATOR_OPEN){
                        pthread_mutex_lock(&lock);
                        exit(passenger, 0);
                        occupancy--;
                        riding=0;
                        pthread_barrier_wait(&barr);
                        pthread_mutex_unlock(&lock);
                }
        }
}

这里是调用函​​数

void passenger_request(int passenger, int from_floor, int to_floor,void (*enter)(int,int), void(*exit)(int,int))
{
        pthread_mutex_lock(&passlock);
        struct arg_struct args;
        args.p = passenger;
        args.f = from_floor;
        args.t = to_floor;
        args.en = *enter;
        args.ex = *exit;
        pthread_create(&thread, NULL, &passenger, &args);
        //pthread_join(thread, NULL);
        pthread_mutex_unlock(&passlock);
        // wait for the elevator to arrive at our origin floor, then get in
}

程序在初始化时创建多个乘客时出现段错误，如果我注释掉 pthread_create 行，则不会发生崩溃。我假设这是我为函数指针传递参数的问题，但我不清楚到底发生了什么，因为所有这些指针开始让我感到困惑。任何帮助将不胜感激

还有结构声明..

struct arg_struct{
        int p;
        int f;
        int t;
        void *(*ex)(int,int);
        void *(*en)(int,int);
};

Answer 1

args.en = *enter;
args.ex = *exit;

enter 和exit 是函数指针。不要取消引用它们，而是通过 args 直接传递它们。也就是说，您需要:

args.en = enter;
args.ex = exit;

(假设您已正确定义未显示的 struct arg_struct。