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Question

Look:
int main()
if(fork())
exit(0);
setsid();
//the second fork
if(fork())
exit(0);
//=========
int fd=open("/dev/tty",O_RDWR);
ioctl(fd, TIOCSCTTY, 1);
while(1)
sleep(20);
write(fd,"hello",10);
//daemon code
I am heard if I don't write the second fork,the process can connect
a control terminal,but I find I write the second fork or don't write
the codes,the screen couldn't display the string "hello",Why?
How to correct my codes?

Answer 1

leetow2006, have you read the manual page daemon(7)?  There a number of steps are given that system daemons should perform.  It says in step 7 that the second fork serves "to ensure the daemon can never re-acquire a terminal again."  So this seems to be a precautionary step, just in case.  Detaching from the terminal is done before that with setsid(), so ditching a controlling terminal seems unrelated to stdin/out and more related to the process hierarchy.  I am no expert, but my guess is that the daemon might be killed when the terminal exits if it does not detach, and might even be brought back  into the foreground and killed with Ctrl-C for example.  You don't want that for a daemon.