文件互斥-白红宇

文件互斥

阅读量：6943 次

发布时间：2019-06-27

本文共 2632 字，大约阅读时间需要 8 分钟。

通过linux下文件互斥地打开，实现线程/进程互斥的访问资源，以此实现多线程编程。

值得注意的是，文件互斥的方式不但适用于多线程编程，还能实现多进程之间的交互。

lock.h

#include 
     
      #include 
      
       #include 
       
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         #include 
         
          void initlock (const char* lockfile);void lock (const char* lockfile);void unlock (const char* lockfile);

lock.c

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          void initlock (const char* lockfile){    int i;    unlink(lockfile);}void lock (const char* lockfile){    int fd;    //添加了O_EXCL之后，没有文件则创建文件，有文件则打开失败进入while循环知道这个文件删除    while((fd = open(lockfile, O_RDONLY | O_CREAT | O_EXCL)) < 0)        sleep(1);    close(fd);}void unlock (const char* lockfile){     unlink(lockfile);}

利用这个文件互斥来实现之前的哲学家进餐问题，只是把互斥锁的部分换成这个lock

#include 
     
      #include 
      
       #include 
       
        #include 
        
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          #include 
          
           #include 
           
            #include "lock.h"#define N 5 // five philosopher#define T_EAT 5#define T_THINK 5#define N_ROOM 4 //同一时间只允许４人用餐#define left(phi_id) (phi_id+N-1)%N#define right(phi_id) (phi_id+1)%Nenum { think , hungry , eat }phi_state[N];static char* chopstick[N] = { "chopstick0", "chopstick1", "chopstick2", "chopstick3", "chopstick4"};void thinking(int id){ sleep(T_THINK); printf("philosopher[%d] is thinking...\n", id);}void eating(int id){ sleep(T_EAT); printf("philosopher[%d] is eating...\n", id);}void take_forks(int id){ //获取左右两边的筷子 //printf("Pil[%d], left[%d], right[%d]\n", id, left(id), right(id)); if((id&1) == 1){ lock(chopstick[left(id)]); lock(chopstick[right(id)]); } else{ lock(chopstick[right(id)]); lock(chopstick[left(id)]); } //printf("philosopher[%d] take_forks...\n", id);}void put_down_forks(int id){ printf("philosopher[%d] is put_down_forks...\n", id); unlock(chopstick[left(id)]); unlock(chopstick[right(id)]);}void* philosopher_work(void *arg){ int id = *(int*)arg; printf("philosopher init [%d] \n", id); while(1){ thinking(id); //sem_wait(&room); take_forks(id); //sem_post(&room); eating(id); put_down_forks(id); }}int main(){ pthread_t phiTid[N]; int i; int err; int *id=(int *)malloc(sizeof(int)*N); //initilize semaphore for (i = 0; i < N; i++) { initlock(chopstick[i]); } for(i=0; i < N; ++i){ //printf("i ==%d\n", i); id[i] = i; err = pthread_create(&phiTid[i], NULL, philosopher_work, (void*)(&id[i])); //这种情况生成的thread id是0,1,2,3,4 if (err != 0) printf("can't create process for reader\n"); } while(1); exit(0); return 0;}