共享内存 进程通信

参考地址：http://www.ibm.com/developerworks/cn/linux/l-ipc/part5/index1.html?ca=drs-

main.c

/*-------------map_normalfile1.c-----------*/
#include <sys/mman.h>
#include <sys/types.h>
#include <fcntl.h>
#include <unistd.h>
typedef struct{
  char name[4];
  int  age;
}people;
main(int argc, char** argv) // map a normal file as shared mem:
{
  int fd,i;
  people *p_map;
  char temp;

  fd=open(argv[1],O_CREAT|O_RDWR|O_TRUNC,00777);
  lseek(fd,sizeof(people)*5-1,SEEK_SET);
  write(fd,"",1);

  p_map = (people*) mmap( NULL,sizeof(people)*10,PROT_READ|PROT_WRITE,
        MAP_SHARED,fd,0 );
  close( fd );
  temp = 'a';
  for(i=0; i<10; i++)
  {
    temp += 1;
    memcpy( ( *(p_map+i) ).name, &temp,2 );
    ( *(p_map+i) ).age = 20+i;
  }
  printf(" initialize over \n ");
  sleep(10);
  munmap( p_map, sizeof(people)*10 );
  printf( "umap ok \n" );
}

 

client.c

/*-------------map_normalfile2.c-----------*/
#include <sys/mman.h>
#include <sys/types.h>
#include <fcntl.h>
#include <unistd.h>
typedef struct{
  char name[4];
  int  age;
}people;
main(int argc, char** argv)  // map a normal file as shared mem:
{
  int fd,i;
  people *p_map;
  fd=open( argv[1],O_CREAT|O_RDWR,00777 );
  p_map = (people*)mmap(NULL,sizeof(people)*10,PROT_READ|PROT_WRITE,
       MAP_SHARED,fd,0);
  for(i = 0;i<10;i++)
  {
  printf( "name: %s age %d;\n",(*(p_map+i)).name, (*(p_map+i)).age );
  }
  munmap( p_map,sizeof(people)*10 );
}

 

Makefile

all:main
main:
    gcc -g -Wall -O0 main.c -o main
    
client:
    gcc -g -Wall -O0 client.c -o client
    
clean:
    rm *.o main client
        

 

分别把两个程序编译成可执行文件map_normalfile1和map_normalfile2后，在一个终端上先运行./map_normalfile2 /tmp/test_shm，程序输出结果如下：

initialize over
umap ok

 

在map_normalfile1输出initialize over 之后，输出umap ok之前，在另一个终端上运行map_normalfile2 /tmp/test_shm，将会产生如下输出(为了节省空间，输出结果为稍作整理后的结果)：

name: b	age 20;	name: c	age 21;	name: d	age 22;	name: e	age 23;	name: f	age 24;
name: g	age 25;	name: h	age 26;	name: I	age 27;	name: j	age 28;	name: k	age 29;

 

在map_normalfile1 输出umap ok后，运行map_normalfile2则输出如下结果：

name: b	age 20;	name: c	age 21;	name: d	age 22;	name: e	age 23;	name: f	age 24;
name:	age 0;	name:	age 0;	name:	age 0;	name:	age 0;	name:	age 0;