请用普通的互斥锁编程实现一个读写锁

[问答题]

请用普通的互斥锁编程实现一个读写锁

陈木木

下面是可参考的伪代码:

count_mutex = mutex_init();
write_mutex = mutex_init();
read_count = 0;

void read_lock {
 lock(count_mutex);
 read_count++;
 if (read_count == 1) {
  lock(write_mutex);
 }
 unlock(count_mutex);
}

void read_unlock {
 lock(count_mutex);
 read_count--;
 if (read_count == 0) {
  unlock(write_mutex);
 }
 unlock(count_mutex);
}

void write_lock {
 lock(write_mutex);
}

void write_unlock {
 unlock(write_mutex);
}

发表于 2015-05-05 14:44:40 回复(1)

pcd

以下为实现与演示

#include <stdio.h>

#include <pthread.h>
#include <time.h>

pthread_mutex_t read_mutex;
pthread_mutex_t write_mutex;
int read_count = 0;

void read_lock(){
pthread_mutex_lock(&read_mutex);
read_count++;
if(read_count==1){
pthread_mutex_lock(&write_mutex);
}
pthread_mutex_unlock(&read_mutex);
}

void read_unlock(){
read_count--;
if(read_count==0){
pthread_mutex_unlock(&write_mutex);
}
}

void write_lock(){
pthread_mutex_lock(&write_mutex);
}

void write_unlock(){
pthread_mutex_unlock(&write_mutex);
}

void* fun1(void* v){
read_lock();
int i;
for(i=0; i<10 ;i++)
{
printf("reading A\n");
sleep(1);

}
read_unlock();

return NULL;
}

void* fun2(void* v){
read_lock();
int i;
for(i=0; i<10 ;i++)
{
printf("reading B\n");
sleep(1);
}
read_unlock();
return NULL;
}

void* fun3(void* v){
sleep(1);
write_lock();
printf("writing C\n");
write_unlock();

return NULL;
}

int main(int argc, char* argv[]){

pthread_mutex_init(&read_mutex, NULL);
pthread_mutex_init(&write_mutex, NULL);

pthread_t p1,p2,p3;

pthread_create(&p1, NULL, fun1, NULL);
pthread_create(&p2, NULL, fun2, NULL);
pthread_create(&p3, NULL, fun3, NULL);

pthread_join(p1, NULL);
pthread_join(p2, NULL);
pthread_join(p3, NULL);

pthread_mutex_destroy(&read_mutex);
pthread_mutex_destroy(&write_mutex);

return 0;
}

发表于 2018-05-07 11:06:25 回复(0)