任务描述：
0 推荐在openEuer上实现 1 描述操作系统中“读者-写者”问题，理解问题的本质，提交你理解或查找到的文本资料 2 利用多线程完成reader 和writer 3 在main中测试若干个reader 和writer的测试，提交截图说明代码的正确性
- 如果一个进程正在某个文件上写入内容，而另一个进程也同时开始在同一文件上写入内容，则系统将进入不一致状态。在特定的时间仅应允许一个进程更改文件中存在的数据的值。
- 另一个问题是，如果一个进程正在读取文件，而另一个进程同时在同一文件上写入，则可能会导致读取不干净，因为在该文件上写入的进程会更改文件的值，但是读取该文件的过程将读取文件中存在的旧值。因此，应避免这种情况。
1.问题描述
有两组并发进程：读者和写者，共享一个文件F，要求:
- 允许多个读者同时执行读操作
- 任一写者在完成写操作之前不允许其它读者或写者工作
- 写者执行写操作前，应让已有的写者和读者全部退出
2.问题分析
- 共享数据（文件、记录）由Reader和Writer是两组并发进程共享
- Reader和Writer两组并发进程
- 同组的Reader进程可同时对文件进行读，不用互斥
- 写-写互斥
- 读-写互斥
对于线程，有以下流程：定义--创建初始化-线程运行函数--线程退出
对于信号量，有以下流程：定义--创建初始化--信号量wait和signal--信号量销毁
代码实现
实现多读者（多线程）操作
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>
sem_t rmutex,wmutex;
static void *readerThread(void *arg);
static void *reader3Thread(void *arg);
static void *reader2Thread(void *arg);
static void *writerThread(void *arg);
int readcount = 0;
int n = 0;
int nowLen = 1;
char contentArticle[10][100];
int main(){
pthread_t readerTidp,writerTidp,reader3Tidp,reader2Tidp;
void *retval;
if(sem_init(&rmutex,0,1)==-1||sem_init(&wmutex,0,1)==-1){
printf("sem_init error\n");
return -1;
}//init semaphore
if(pthread_create(&readerTidp,NULL,readerThread,NULL) !=0||pthread_create(&writerTidp,NULL,writerThread,NULL) !=0||pthread_create(&reader3Tidp,NULL,reader3Thread,NULL) !=0||pthread_create(&reader2Tidp,NULL,reader2Thread,NULL) !=0){
printf("pthread_create error\n");
return -2;
}//init pthread
pthread_join(readerTidp,&retval);
pthread_join(reader3Tidp,&retval);
pthread_join(reader2Tidp,&retval);
pthread_join(writerTidp,&retval);
sem_destroy(&rmutex);
sem_destroy(&wmutex);
return 0;
}
static void *readerThread(void *arg){
for(int i = 0;i < 10;i++)
{
sem_wait(&rmutex);
if(readcount == 0)sem_wait(&wmutex);
readcount = readcount+1;
sem_post(&rmutex);
//read operatiom
printf("\n\nI'm reader first Reader thread :...the global variable n equals to %d\n",n);
for(int j = 0;j < nowLen-1;j++)
{
for(int k = 0;k < 26;k++)
printf("%c",contentArticle[j][k]);
printf("\n");
}
printf("now the count 0f reader is %d\n",readcount);
printf("now the length 0f content is %d\n",nowLen-1);
sleep(5);
sem_wait(&rmutex);
readcount = readcount-1;
if(readcount == 0)sem_post(&wmutex);
sem_post(&rmutex);
sleep(1);
}
}
static void *reader3Thread(void *arg){
for(int i = 0;i < 10;i++)
{
sem_wait(&rmutex);
if(readcount == 0)sem_wait(&wmutex);
readcount = readcount+1;
sem_post(&rmutex);
//read operatiom
printf("\n\nI'm reader third Reader thread :...the global variable n equals to %d\n",n);
for(int j = 0;j < nowLen-1;j++)
{
for(int k = 0;k < 26;k++)
printf("%c",contentArticle[j][k]);
printf("\n");
}
printf("now the count 0f reader is %d\n",readcount);
printf("now the length 0f content is %d\n",nowLen-1);
sleep(5);
sem_wait(&rmutex);
readcount = readcount-1;
if(readcount == 0)sem_post(&wmutex);
sem_post(&rmutex);
sleep(8);
}
}
static void *reader2Thread(void *arg){
for(int i = 0;i < 10;i++)
{
sem_wait(&rmutex);
if(readcount == 0)sem_wait(&wmutex);
readcount = readcount+1;
sem_post(&rmutex);
//read operatiom
printf("\n\nI'm reader second Reader thread :...the global variable n equals to %d\n",n);
for(int j = 0;j < nowLen-1;j++)
{
for(int k = 0;k < 26;k++)
printf("%c",contentArticle[j][k]);
printf("\n");
}
printf("now the count 0f reader is %d\n",readcount);
printf("now the length 0f content is %d\n",nowLen-1);
sem_wait(&rmutex);
readcount = readcount-1;
if(readcount == 0)sem_post(&wmutex);
sem_post(&rmutex);
sleep(4);
}
}
static void *writerThread(void *arg){
for(int i = 0;i < 10;i++)
{
sem_wait(&wmutex);
//writer operation
n = n+1;
for(int k = 0;k < 26;k++)
contentArticle[nowLen-1][k] = 'z'-k;
nowLen++;
printf("\n\nWriter thread :writing opration the global variable n equals to %d \n",n);
sleep(2);
sem_post(&wmutex);
sleep(3);
}
}
代码运行截图如下：
标签：readcount,int,void,写者,线程,printf,rmutex,sem,多线程 From： https://www.cnblogs.com/syf0105/p/16872908.html