linux：如何防止程序开启两个实例（进程）？

有时候不小心开启同一个程序两次会造成意想不到的错误，下面介绍一种方法，可以防止出现这种情况，测试可用。

在程序中添加如下代码

 1 #include <unistd.h>
 2 #include <signal.h>
 3 #include <sys/types.h>
 4 #include <sys/stat.h>
 5 #include <errno.h>
 6 #include <fcntl.h>
 7 #include <fstream>
 8
 9
10 #define PIDFILE     "/daemon_shouhu.pid"         //定义你自己的file名
11 #define    write_lock(fd, offset, whence, len) 12     lock_reg(fd, F_SETLK, F_WRLCK, offset, whence, len)
13
14 int lock_reg(int fd, int cmd, int type, off_t offset, int whence, off_t len)
15 {
16     struct flock    lock;
17
18     lock.l_type = type;        /* F_RDLCK, F_WRLCK, F_UNLCK */
19     lock.l_start = offset;        /* byte offset, relative to l_whence */
20     lock.l_whence = whence;        /* SEEK_SET, SEEK_CUR, SEEK_END */
21     lock.l_len = len;        /* #bytes (0 means to EOF) */
22
23     return( fcntl(fd, cmd, &lock) );    /* -1 upon error */
24 }
25
26
27
28 int Check_Running()
29 {
30     int fd, val;
31
32     if ((fd = open(PIDFILE, O_WRONLY|O_CREAT, "at")) < 0)
33     {
34         printf(" Open file error \n");
35     }
36
37     if (write_lock(fd, 0, SEEK_SET, 0) < 0)
38     {
39         if (errno == EACCES || errno == EAGAIN)
40         {
41             printf(" Server Already running! \n");
42             _exit(-1);
43             return -1;
44         }
45         else
46         {
47             printf(" Write_lock error! \n");
48         }
49     }
50
51     if (ftruncate(fd, 0) < 0)
52     {
53         printf("  ftruncate error  \n");
54     }
55
56     char buf[10] = {0};
57     sprintf(buf, "%d\n", getpid());
58     if (write(fd, buf, strlen(buf)) != (int)strlen(buf))
59         printf(" Write error \n");
60
61     if ((val = fcntl(fd, F_GETFD, 0)) < 0)
62         printf(" fcntl F_GETFD error \n");
63
64     val |= FD_CLOEXEC;
65     if (fcntl(fd, F_SETFD, val) < 0)
66         printf(" fcntl F_SETFD error \n");
67
68     return 0;
69 }

然后在程序入口添加Check_Running()就可以了，重新编译后执行，当第二次打开程序时会输出" Server Already running! "，第二个进程无法打开。

linux：如何防止程序开启两个实例（进程）？