指针和多线程中的错误预期为“void * (*)(void *)”，但参数类型为“pthread_t”

Question

此代码使用多线程检查数独。 当我 运行 程序编译后： 分段错误（核心已转储）

int main(){
char t0,t1,t2;

pthread_t row,col,sub1;

t0=pthread_create(&row,NULL,row,NULL); //eror iz here!
t1=pthread_create(&col,NULL,col,NULL);
t2=pthread_create(&sub1,NULL,sub,NULL);

pthread_join(row, NULL);
pthread_join(col, NULL);
pthread_join(sub1, NULL);

exit(EXIT_SUCCESS);
return 0;
}

和错误：

su.c: In function ‘main’:
su.c:87:2: warning: passing argument 3 of ‘pthread_create’ makes pointer from integer without a cast [enabled by default]
In file included from su.c:4:0:
/usr/include/pthread.h:225:12: note: expected ‘void * (*)(void *)’ but argument is of type ‘pthread_t’
su.c:88:2: warning: passing argument 3 of ‘pthread_create’ makes pointer from integer without a cast [enabled by default]
In file included from su.c:4:0:
/usr/include/pthread.h:225:12: note: expected ‘void * (*)(void *)’ but argument is of type ‘pthread_t’

我的函数是给 0 个参数 :void *sub(); 对不起，我的英语不好

Answer 1

pthread_create的原型是这个

int pthread_create(pthread_t *thread, const pthread_attr_t *attr,
                      void *(*start_routine) (void *), void *arg);

在此，第三个参数必须是一个返回空指针的函数，它接受一个参数空指针。

所以像上面那样创建函数，然后将其传递给 pthread_create 的第三个参数。

void * thr_fn2(void *arg)// your function must be like this.

然后将该函数用作参数。

Answer 2

void * my_row_function(void *param){ 
    Row * myrow = (Row*) param;
    //bla bla
}

int main(){
    Row * a_row= & row8outof9;
    pthread_create(&row,(const pthread_t*)NULL,
             my_row_function, a_row);
    return 0;
}

您需要将一个函数作为第三个参数传递（returns void* 并接受 1 个 void* 类型的参数），该函数将接收一个指针，您基本上必须将其转换回指针类型您在调用 pthread_create

时传入