Why is it impossible, without attempting I/O, to detect that TCP socket was gracefully closed by peer?

Solution 1:

I have been using Sockets often, mostly with Selectors, and though not a Network OSI expert, from my understanding, calling shutdownOutput() on a Socket actually sends something on the network (FIN) that wakes up my Selector on the other side (same behaviour in C language). Here you have detection: actually detecting a read operation that will fail when you try it.

In the code you give, closing the socket will shutdown both input and output streams, without possibilities of reading the data that might be available, therefore loosing them. The Java Socket.close() method performs a "graceful" disconnection (opposite as what I initially thought) in that the data left in the output stream will be sent followed by a FIN to signal its close. The FIN will be ACK'd by the other side, as any regular packet would¹.

If you need to wait for the other side to close its socket, you need to wait for its FIN. And to achieve that, you have to detect Socket.getInputStream().read() < 0, which means you should not close your socket, as it would close its InputStream.

From what I did in C, and now in Java, achieving such a synchronized close should be done like this:

1. Shutdown socket output (sends FIN on the other end, this is the last thing that will ever be sent by this socket). Input is still open so you can read() and detect the remote close()
2. Read the socket InputStream until we receive the reply-FIN from the other end (as it will detect the FIN, it will go through the same graceful diconnection process). This is important on some OS as they don't actually close the socket as long as one of its buffer still contains data. They're called "ghost" socket and use up descriptor numbers in the OS (that might not be an issue anymore with modern OS)
3. Close the socket (by either calling Socket.close() or closing its InputStream or OutputStream)

As shown in the following Java snippet:

public void synchronizedClose(Socket sok) {
    InputStream is = sok.getInputStream();
    sok.shutdownOutput(); // Sends the 'FIN' on the network
    while (is.read() > 0) ; // "read()" returns '-1' when the 'FIN' is reached
    sok.close(); // or is.close(); Now we can close the Socket
}

Of course both sides have to use the same way of closing, or the sending part might always be sending enough data to keep the while loop busy (e.g. if the sending part is only sending data and never reading to detect connection termination. Which is clumsy, but you might not have control on that).

As @WarrenDew pointed out in his comment, discarding the data in the program (application layer) induces a non-graceful disconnection at application layer: though all data were received at TCP layer (the while loop), they are discarded.

¹: From "Fundamental Networking in Java": see fig. 3.3 p.45, and the whole §3.7, pp 43-48

Solution 2:

I think this is more of a socket programming question. Java is just following the socket programming tradition.

From Wikipedia:

TCP provides reliable, ordered delivery of a stream of bytes from one program on one computer to another program on another computer.

Once the handshake is done, TCP does not make any distinction between two end points (client and server). The term "client" and "server" is mostly for convenience. So, the "server" could be sending data and "client" could be sending some other data simultaneously to each other.

The term "Close" is also misleading. There's only FIN declaration, which means "I am not going to send you any more stuff." But this does not mean that there are no packets in flight, or the other has no more to say. If you implement snail mail as the data link layer, or if your packet traveled different routes, it's possible that the receiver receives packets in wrong order. TCP knows how to fix this for you.

Also you, as a program, may not have time to keep checking what's in the buffer. So, at your convenience you can check what's in the buffer. All in all, current socket implementation is not so bad. If there actually were isPeerClosed(), that's extra call you have to make every time you want to call read.

Solution 3:

The underlying sockets API doesn't have such a notification.

The sending TCP stack won't send the FIN bit until the last packet anyway, so there could be a lot of data buffered from when the sending application logically closed its socket before that data is even sent. Likewise, data that's buffered because the network is quicker than the receiving application (I don't know, maybe you're relaying it over a slower connection) could be significant to the receiver and you wouldn't want the receiving application to discard it just because the FIN bit has been received by the stack.

Solution 4:

Since none of the answers so far fully answer the question, I'm summarizing my current understanding of the issue.

When a TCP connection is established and one peer calls close() or shutdownOutput() on its socket, the socket on the other side of the connection transitions into CLOSE_WAIT state. In principle, it's possible to find out from the TCP stack whether a socket is in CLOSE_WAIT state without calling read/recv (e.g., getsockopt() on Linux: http://www.developerweb.net/forum/showthread.php?t=4395), but that's not portable.

Java's Socket class seems to be designed to provide an abstraction comparable to a BSD TCP socket, probably because this is the level of abstraction to which people are used to when programming TCP/IP applications. BSD sockets are a generalization supporting sockets other than just INET (e.g., TCP) ones, so they don't provide a portable way of finding out the TCP state of a socket.

There's no method like isCloseWait() because people used to programming TCP applications at the level of abstraction offered by BSD sockets don't expect Java to provide any extra methods.