我正在学习 pthread 并遇到了读写器锁。场景非常简单;一个由所有线程共享的全局变量,读取器不断打印同一个全局变量的当前值,而写入器将更新同一个变量。我可以通过使用两个互斥锁(pthread_mutex_t)来实现这种同步,但我想使用“一个”读写器锁来实现同样的结果。但是,使用一个读写器锁,如此处所示(程序的输出,如下),阅读器只能看到 x 的第一个值,而看不到对全局变量的任何更新。请在这里放一些光。
代码:
#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <pthread.h>
#include <poll.h>
#define ACCESS_ONCE(x) (*(volatile typeof(x) *)&(x))
int x = 0;
pthread_rwlock_t lock_rw = PTHREAD_RWLOCK_INITIALIZER;
void *reader_thread(void *arg)
{
int i;
int newx, oldx;
newx = oldx = -1;
pthread_rwlock_t *p = (pthread_rwlock_t *)arg;
if (pthread_rwlock_rdlock(p) != 0) {
perror("reader_thread: pthread_rwlock_rdlock error");
exit(__LINE__);
}
for (i = 0; i < 100; i++) {
newx = ACCESS_ONCE(x);
if (newx != oldx) {
printf("reader_lock: x: %d\n",x);
}
oldx = newx;
poll(NULL, 0, 1);
}
if (pthread_rwlock_unlock(p) != 0) {
perror("reader thread: pthred_rwlock_unlock error");
exit(__LINE__);
}
return NULL;
}
void *writer_thread(void *arg)
{
int i;
pthread_rwlock_t *p = (pthread_rwlock_t *)arg;
if (pthread_rwlock_wrlock(p) != 0) {
perror("writer thread: pthread_rwlock_wrlock error");
exit(__LINE__);
}
for (i = 0; i < 3; i++) {
ACCESS_ONCE(x)++;
poll(NULL, 0, 5);
}
if (pthread_rwlock_unlock(p) != 0) {
perror("writer thread: pthread_rwlock_unlock error");
exit(__LINE__);
}
return NULL;
}
int main(void)
{
pthread_t tid1, tid2;
void *vp;
if (pthread_create(&tid1, NULL, reader_thread, &lock_rw) != 0) {
perror("pthread_create error");
exit (__LINE__);
}
if (pthread_create(&tid2, NULL, writer_thread, &lock_rw) != 0) {
perror("pthread_create error");
exit (__LINE__);
}
//wait for the thread to complete
if (pthread_join(tid1, &vp) != 0) {
perror("pthread_join error");
exit (__LINE__);
}
if (pthread_join(tid2, &vp) != 0) {
perror("pthread_join error");
exit (__LINE__);
}
printf("Parent process sees x: %d\n",x);
return 0;
}
gcc pthread_rwlock.c -o rwlock -pthread -Wall -Werror
./rwlock
reader_lock: x: 0
父进程看到 x: 3