linux多线程程序设计.txt

多线程程序设计

Linux多线程程序设计,（网络上流传的一道题)     
 有四个线程1,2,3,4,线程1的功能就是输出1,线程2的功能就是输出2,以此类推.....
现在有四个文件.ABCD.初始都为空.
现要让四个文件呈如下格式:
A:  1 2 3 4 1 2....
B:  2 3 4 1 2 3....
C:  3 4 1 2 3 4....
D:  4 1 2 3 4 1....

设计程序.

 

#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <pthread.h>
#include <stdio.h>
#include <errno.h>
#include <string.h>
#include <malloc.h>
#include <stdlib.h>
#include <unistd.h>    //usleep ,sleep
#define MY_THREAD_COUNT 4
#define MY_LOOP_COUNT  3
typedef struct
...{
    int loops;  //表示要循环写入文件的次数
    int fd;
    int curData; //当前要写入文件的数字
    pthread_mutex_t mutex;
}FILENODE,*LPFILENODE;

char *filenames[] = ...{"a.txt","b.txt","c.txt","d.txt"};
pthread_t pids[MY_THREAD_COUNT];
FILENODE filenodes[MY_THREAD_COUNT];
bool IsCompleted()
...{
    for(int i = 0; i < MY_THREAD_COUNT; i++)
    ...{
        if(filenodes[i].loops != MY_LOOP_COUNT)
        return false;
    }
    return true;
}
void* ThreadFunc(void* para)
...{
   int num = *(int*)para;
   printf("Loop thread id: %d  ",num);
   char buf[50] = ...{""};   
   bool bExit = false;
   while(!bExit)
   ...{
       if(IsCompleted())
          break;           
       printf("thread id : %d begin loops files ",num);
       for(int i = 0; i < MY_THREAD_COUNT; i++)
       ...{
        int rtn = pthread_mutex_trylock(&filenodes[i].mutex);
        if(rtn == 0)
        ...{
        if(filenodes[i].loops >= MY_LOOP_COUNT)
        ...{
            pthread_mutex_unlock(&filenodes[i].mutex);
                    continue;
        }
        if(filenodes[i].curData == num)
        ...{    
            sprintf(buf,"%d:%d",num,filenodes[i].curData);   
            write(filenodes[i].fd,buf,strlen(buf));
                filenodes[i].curData += 1;
        }
            if(filenodes[i].curData > 4)
        ...{
                     filenodes[i].curData = 1;
                     filenodes[i].loops++;             
        }    
        pthread_mutex_unlock(&filenodes[i].mutex);
        }   
       }
       usleep(500000);
   }
   printf("thread id : %d, succeed exit! ",num);
   free(para);
}
int main()
...{
   int i = 0;    
   for(i = 0;i<MY_THREAD_COUNT;i++)
   ...{
    filenodes[i].fd = open(filenames[i],O_CREAT|O_WRONLY,S_IRUSR|S_IWUSR);
        filenodes[i].loops = 0;
        filenodes[i].curData = i+1;
        int rtn = pthread_mutex_init(&filenodes[i].mutex,NULL);
    int* para = (int*)malloc(sizeof(int));
    *para = i+1;
        rtn = pthread_create(&pids[i],NULL,ThreadFunc,para);           
   }
   for(i = 0;i<MY_THREAD_COUNT;i++)
   ...{
       pthread_join(pids[i],NULL);
   }
   for(int i = 0; i < MY_THREAD_COUNT;i++)
   ...{
       if(!filenodes[i].fd)
       ...{
           close(filenodes[i].fd);
       }           
   }
  return 0;    
}

通过Shell>: g++ -g  -o Threads main.C -lpthread

生成可执行文件Threads,执行可执行文件后就可以查看结果。对于上面的题目，我做了一个扩冲，就是设定了循环的次数。