UNIX for Dummies Questions & Answers

I have a UNIX daemon process that's been started by a parent process, an application server.
The behavior of this daemon process is to inherit and use the app
server's file descriptors (ports/sockets).
When I shutdown the app server, the daemon continues to run, because there may be other applications that want to use it.
However, the app server won't restart, because the daemon process is holding its parents' ports.
There is a debate as to whether its the app server's or the daemon's responsibility to close the file descriptors (say, using fcntl(...)).
Which process has the responsibility of closing the open file
descriptors?
Richard Steven, in his Advanced Unix programming book says that the daemon should do it, but the people who created the daemon say the Posix way is better.
Any takers?
Thanks in advance,
Kunal

I'm not sure why you think fcntl() fits into this. To close an fd you use close().

It is absolutely required that a daemon close fd's 0, 1, and 2. After that, as the lawyers say, "reasonable minds may disagree". In the days when we were limited to 64 fd's, it was reasonable to simply loop invoking close() on them all. If I was to write a daemon, yes I guess that I would use get getconf(_SC_OPEN_MAX) to get the max possible fd and loop invoking close() on every last one. This could be thousands of close() calls that aren't needed, but close() fails very quickly when invoked against a non-open fd.

If they are closing stdin, stdout, and stderr, I would have to say that their stance is reasonable. But I do believe they would have a difficult time producing any language in posix that supports them. The posix standard is on-line and we have a link to it on our home page. Exactly which section do they cite?

It should be very easy to solve your problem though. Can't you change the app server to not leave extra fd's open?

Or if this daemon is called, say, daemonx, just write a program that closes all fd's and then exec()'s daemonx. Call your program pre_daemonx. Have your app server call pre_daemonx.

Is the name "daemonx" hard-coded into an unchangable app server? No problem. Rename "daemonx" to "real.daemonx" and call your program "daemonx".

Whichever path you take here, fixing this should be a 10 minute problem.