我从APUE中选择了以下示例:
void daemonize(const char* cmd)
{
int i,fd0,fd1,fd2;
pid_t pid;
struct rlimit r1;
struct sigaction sa;
//clear all file masks
umask(0);
//get max number of file descriptors
if(getrlimit(RLIMIT_NOFILE,&r1) < 0)
{
perror("error getting file descriptor size");
return;
}
//become a session leader
if((pid = fork()) < 0)
{
perror("error forking");
return;
}
else if(pid == 0)
{
setsid();
}
else
{
exit(0); //parent exits
}
sa.sa_handler = SIG_IGN;
sigemptyset(&sa.sa_mask);
sa.sa_flags = 0;
if(sigaction(SIGHUP,&sa,NULL) < 0)
{
return;
}
if((pid = fork()) < 0)
{
return;
}
else if(pid != 0)
{
exit(0); //parent
}
//child continues
syslog(LOG_ERR,"chile continuing with pid : %d",getpid());
//change the working directory
if(chdir("/") < 0)
{
return;
}
if(r1.rlim_max == RLIM_INFINITY)
r1.rlim_max = 1024;
for(i=0;i<r1.rlim_max;i++)
close(i);
//attach the file descriptors to /dev/null
fd0 = open("/dev/null",O_RDWR);
fd1 = dup(0);
fd2 = dup(0);
//initialize the log file
openlog(cmd, LOG_CONS,LOG_DAEMON);
if(fd0!=0 || fd1!=1 || fd2!=2)
{
syslog(LOG_ERR,"unexpected file descriptors %d %d %d\n",fd0,fd1,fd2);
exit(1);
}
}
int main()
{
daemonize("date");
pause(); //how is this working???
}
我不明白主函数pause()
是如何工作的?我期望的是,由于我们已经为exit(0)
中的父进程完成了daemonize()
,所以它应该已经退出并导致main()进程的正常终止。它应该永远不会返回到main()
,并且甚至不会发生对pause()
的调用。为什么它没有终止以及为什么叫pause()
?
答案 0 :(得分:3)
该代码分叉两次,产生一个父级,一个子级和一个孙子级。前两个exit(0)
;最后返回的是daemonize
。