Linux下pthread的读写锁的优先级问题
有这么一个情况：有一个C实现的HashMap，需要在多个线程之间共享。对它的读操作远远大于写操作。所以采用了pthread的读写锁来保障并发读写时的一致性。现在测试发现的问题是：因为读操作太多，导致写操作一直拿不到锁。按理说不应该啊，假如有三个线程，线程1 先申请读锁并成功拿到，然后线程2申请写锁那么必然会陷入等待，之后线程3去申请读锁，那么应该是陷入等待才对，因为pthread_rwlo
有这么一个情况：有一个C实现的HashMap，需要在多个线程之间共享。对它的读操作远远大于写操作。所以采用了pthread的读写锁来保障并发读写时的一致性。
现在测试发现的问题是：因为读操作太多，导致写操作一直拿不到锁。按理说不应该啊，假如有三个线程，线程1 先申请读锁并成功拿到，然后线程2申请写锁那么必然会陷入等待，之后线程3去申请读锁，那么应该是陷入等待才对，因为pthread_rwlock_rdlock的man pages上说"The calling thread acquires the read lock if a writer does not hold the lock and there are no writers blocked on the lock"，可是实际我的测试结果是线程3拿到读锁了。
为了模拟这个场景，我着实想了好一阵子。因为我怎么确定线程2已经陷入等待状态了呢？后来我的测试是这么做的
线程1 线程2 线程3
rdlock
barrier barrier barrier
wrlock while(true) {rdlock;unlock;}
代码如下：
#include <iostream>
#include <pthread.h>
#include <stdio.h>
#include <assert.h>
static pthread_barrier_t barr;
static pthread_barrier_t barr2;
static pthread_rwlock_t rwlock;
void * thr1_entry(void *arg){
int threadCount=*(int*)arg;
std::cout<<"this is thread "<<threadCount<<std::endl;
if(pthread_rwlock_rdlock(&rwlock)!=0)
return NULL;
std::cout<<"thread1 got the read lock "<<std::endl;
{int rc=pthread_barrier_wait(&barr);
if(rc != 0 && rc != PTHREAD_BARRIER_SERIAL_THREAD) {
printf("Could not wait on barrier\n");
}}
std::cout<<"thread "<<threadCount<<" work done"<<std::endl;
int rc=pthread_barrier_wait(&barr2);
if(rc != 0 && rc != PTHREAD_BARRIER_SERIAL_THREAD) {
printf("Could not wait on barrier\n");
}
std::cout<<"thread "<<threadCount<<" return"<<std::endl;
}
void * thr2_entry(void *arg){
int threadCount=*(int*)arg;
std::cout<<"this is thread "<<threadCount<<std::endl;
if(threadCount!=1){int rc=pthread_barrier_wait(&barr);
if(rc != 0 && rc != PTHREAD_BARRIER_SERIAL_THREAD) {
printf("Could not wait on barrier\n");
}}
if(threadCount==1){
if(pthread_rwlock_rdlock(&rwlock)!=0)
return NULL;
std::cout<<"thread1 got the read lock "<<std::endl;
{int rc=pthread_barrier_wait(&barr);
if(rc != 0 && rc != PTHREAD_BARRIER_SERIAL_THREAD) {
printf("Could not wait on barrier\n");
}}
} else if(threadCount==2){
pthread_rwlock_wrlock(&rwlock);
} else if(threadCount==3){
while(true){
sleep(5);
pthread_rwlock_rdlock(&rwlock);
std::cout<<"thread3 got lock"<<std::endl;
pthread_rwlock_unlock(&rwlock);
std::cout<<"thread3 released lock"<<std::endl;
}
}
std::cout<<"thread "<<threadCount<<" work done"<<std::endl;
int rc=pthread_barrier_wait(&barr2);
if(rc != 0 && rc != PTHREAD_BARRIER_SERIAL_THREAD) {
printf("Could not wait on barrier\n");
}
std::cout<<"thread "<<threadCount<<" return"<<std::endl;
}
int main(int argc,char* argv[]){
pthread_t thr1,thr2,thr3;
if(pthread_barrier_init(&barr, NULL, 3)) {
printf("Could not create a barrier\n");
return -1;
}
if(pthread_barrier_init(&barr2, NULL, 3)) {
printf("Could not create a barrier\n");
return -1;
}
pthread_rwlockattr_t attr;
if(pthread_rwlockattr_init(&attr)){
printf("Could not create a rwlock attr\n");
return -1;
}
// int perf=-1;
//pthread_rwlockattr_getkind_np(&attr,&perf);
// std::cout<<perf<<std::endl;
pthread_rwlockattr_setkind_np(&attr,PTHREAD_RWLOCK_PREFER_WRITER_NP);
if(pthread_rwlock_init(&rwlock,&attr)){
printf("Could not create a rwlock\n");
return -1;
}
int threadCount[]={1,2,3};
if(pthread_create(&thr1, NULL, &thr1_entry, (void*)&threadCount[0])) {
printf("Could not create thread %d\n", threadCount);
return -1;
}
if(pthread_create(&thr2, NULL, &thr2_entry, (void*)&threadCount[1])) {
printf("Could not create thread %d\n", threadCount);
return -1;
}
if(pthread_create(&thr3, NULL, &thr2_entry, (void*)&threadCount[2])) {
printf("Could not create thread %d\n", threadCount);
return -1;
}
pthread_join(thr1,NULL);
pthread_join(thr2,NULL);
pthread_join(thr3,NULL);
pthread_barrier_destroy(&barr);
pthread_rwlock_destroy(&rwlock);
return 0;
}
系统是Fedora 14，线程库就是默认的NPTL。pthread_rwlockattr_setkind_np是专门用来设置读写优先级的，但是我用完之后无效。怪哉，我哪用错了？
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