第一个C/S程序，为什么完成后，server端口一直显示Time_wait？

[DryFisHH]

开始学习网络编程，写了一个很简单的C/S程序，当server端运行后监听12345端口，client端发过来消息“client messge”，server端收到后返回“server message”，运行一切正常，可是每运行一次client，server机上用netstat -an命令看，发现就会多出一个12345端口显示Time_wait状态，第一个listening的12345端口不变，这是为什么啊？郁闷！

//server
#include <winsock2.h>

int main()
{
SOCKET sock,sockt;
WSADATA wsa;
int iret=0;
char msg[25];
struct sockaddr_in sin;

if(WSAStartup(MAKEWORD(2,2),&wsa))
{
OutputDebugString("WSAStartup Error!");
return 0;
}

if((sock=socket(AF_INET,SOCK_STREAM,IPPROTO_TCP))==INVALID_SOCKET)
{
OutputDebugString("Socket Error!");
return 0;
}

sin.sin_addr.s_addr=htons(INADDR_ANY);
sin.sin_family=AF_INET;
sin.sin_port=htons(12345);

if(bind(sock,(struct sockaddr *)&sin,sizeof(sin))==SOCKET_ERROR)
{
OutputDebugString("Bind Error!");
return 0;
}

if(listen(sock,5)==SOCKET_ERROR)
{
OutputDebugString("Listen Error!");
return 0;
}

while(1)
{
if((sockt=accept(sock,NULL,NULL))==SOCKET_ERROR)
{
OutputDebugString("Accept Error!");
continue;
}


if((iret=recv(sockt,msg,sizeof(msg),0))==SOCKET_ERROR)
{
OutputDebugString("Recv Error!");
closesocket(sockt);
continue;
}

if(strstr(msg,"client message"))
{
memset(msg,0,sizeof(msg));
memcpy(msg,"server message",sizeof(msg)-1);

if((iret=send(sockt,msg,sizeof(msg),0))==SOCKET_ERROR)
{
OutputDebugString("Send Error!");
closesocket(sockt);
continue;
}
}
OutputDebugString("Transport Successfully");
closesocket(sockt);
}

return 0;
}
/////////////////////////////////////////////////////////////////////
//client
#include <winsock2.h>
#include <stdio.h>

int main(int argc,char *argv[])
{
if(argc!=2)
return 0;

WSADATA wsa;
SOCKET sock;
struct sockaddr_in sin;
char msg[25]="client message";
int iret;

if(WSAStartup(MAKEWORD(2,2),&wsa))
{
printf("WSAStartup Error: %d\n",WSAGetLastError());
return -1;
}

if((sock=socket(AF_INET,SOCK_STREAM,IPPROTO_TCP))==INVALID_SOCKET)
{
printf("Socket Error: %d\n",WSAGetLastError());
return -1;
}

sin.sin_addr.s_addr=inet_addr(argv[1]);
sin.sin_family=AF_INET;
sin.sin_port=htons(12345);

if(connect(sock,(struct sockaddr *)&sin,sizeof(sin))==SOCKET_ERROR)
{
printf("Connect Error: %d\n",WSAGetLastError());
return -1;
}

if((iret=send(sock,msg,sizeof(msg),0))==SOCKET_ERROR)
{
printf("Send Error: %d\n",WSAGetLastError());
return -1;
}

memset(msg,0,sizeof(msg));
if((iret=recv(sock,msg,sizeof(msg),0))==SOCKET_ERROR)
{
printf("Recv Error: %d\n",WSAGetLastError());
return -1;
}
printf("Re: ");
printf(msg);

closesocket(sock);
WSACleanup();

return 0;
}

[louifox]

while(1)的循环没退出

[flyingjm]

好像是closesocket以后并不立即释放端口，还要占用端口2MSL时间才会释放端口。

[DryFisHH]

我也怀疑是while(1)的缘故，不过要是没有while，下一次client怎么连进来啊？请问通常类似的情况下怎么实现呢？

[DryFisHH]

看了一些别人的代码，好像在这种情况下while(1)是很常用的。另外，经过测试发现状态为Time_wait的经过一段时间确实会消失，不知是不是像flyingjm()说的那样？哪位大虾能给出一个肯定的答案吗？

[wqs_qiu_shi]

2.7. Please explain the TIME_WAIT state.

Remember that TCP guarantees all data transmitted will be delivered,
if at all possible. When you close a socket, the server goes into a
TIME_WAIT state, just to be really really sure that all the data has
gone through. When a socket is closed, both sides agree by sending
messages to each other that they will send no more data. This, it
seemed to me was good enough, and after the handshaking is done, the
socket should be closed. The problem is two-fold. First, there is no
way to be sure that the last ack was communicated successfully.
Second, there may be "wandering duplicates" left on the net that must
be dealt with if they are delivered.

Andrew Gierth ([e-mail]) helped to explain the
closing sequence in the following usenet posting:

Assume that a connection is in ESTABLISHED state, and the client is
about to do an orderly release. The client's sequence no. is Sc, and
the server's is Ss. Client Server
====== ======
ESTABLISHED ESTABLISHED
(client closes)
ESTABLISHED ESTABLISHED
------->>
FIN_WAIT_1
<<--------
FIN_WAIT_2 CLOSE_WAIT
<<-------- (server closes)
LAST_ACK
, ------->>
TIME_WAIT CLOSED
(2*msl elapses...)
CLOSED

Note: the +1 on the sequence numbers is because the FIN counts as one
byte of data. (The above diagram is equivalent to fig. 13 from RFC
793).

Now consider what happens if the last of those packets is dropped in
the network. The client has done with the connection; it has no more
data or control info to send, and never will have. But the server does
not know whether the client received all the data correctly; that's
what the last ACK segment is for. Now the server may or may not care
whether the client got the data, but that is not an issue for TCP; TCP
is a reliable rotocol, and must distinguish between an orderly
connection close where all data is transferred, and a connection abort
where data may or may not have been lost.

So, if that last packet is dropped, the server will retransmit it (it
is, after all, an unacknowledged segment) and will expect to see a
suitable ACK segment in reply. If the client went straight to CLOSED,
the only possible response to that retransmit would be a RST, which
would indicate to the server that data had been lost, when in fact it
had not been.

(Bear in mind that the server's FIN segment may, additionally, contain
data.)

DISCLAIMER: This is my interpretation of the RFCs (I have read all the
TCP-related ones I could find), but I have not attempted to examine
implementation source code or trace actual connections in order to
verify it. I am satisfied that the logic is correct, though.

More commentarty from Vic:

The second issue was addressed by Richard Stevens ([e-mail],
author of "Unix Network Programming", see ``1.5 Where can I get source
code for the book [book title]?''). I have put together quotes from
some of his postings and email which explain this. I have brought
together paragraphs from different postings, and have made as few
changes as possible.

From Richard Stevens ([e-mail]):

If the duration of the TIME_WAIT state were just to handle TCP's full-
duplex close, then the time would be much smaller, and it would be
some function of the current RTO (retransmission timeout), not the MSL
(the packet lifetime).

A couple of points about the TIME_WAIT state.

o The end that sends the first FIN goes into the TIME_WAIT state,
because that is the end that sends the final ACK. If the other
end's FIN is lost, or if the final ACK is lost, having the end that
sends the first FIN maintain state about the connection guarantees
that it has enough information to retransmit the final ACK.

o Realize that TCP sequence numbers wrap around after 2**32 bytes
have been transferred. Assume a connection between A.1500 (host A,
port 1500) and B.2000. During the connection one segment is lost
and retransmitted. But the segment is not really lost, it is held
by some intermediate router and then re-injected into the network.
(This is called a "wandering duplicate".) But in the time between
the packet being lost & retransmitted, and then reappearing, the
connection is closed (without any problems) and then another
connection is established between the same host, same port (that
is, A.1500 and B.2000; this is called another "incarnation" of the
connection). But the sequence numbers chosen for the new
incarnation just happen to overlap with the sequence number of the
wandering duplicate that is about to reappear. (This is indeed
possible, given the way sequence numbers are chosen for TCP
connections.) Bingo, you are about to deliver the data from the
wandering duplicate (the previous incarnation of the connection) to
the new incarnation of the connection. To avoid this, you do not
allow the same incarnation of the connection to be reestablished
until the TIME_WAIT state terminates.

Even the TIME_WAIT state doesn't complete solve the second problem,
given what is called TIME_WAIT assassination. RFC 1337 has more
details.

o The reason that the duration of the TIME_WAIT state is 2*MSL is
that the maximum amount of time a packet can wander around a
network is assumed to be MSL seconds. The factor of 2 is for the
round-trip. The recommended value for MSL is 120 seconds, but
Berkeley-derived implementations normally use 30 seconds instead.
This means a TIME_WAIT delay between 1 and 4 minutes. Solaris 2.x
does indeed use the recommended MSL of 120 seconds.

A wandering duplicate is a packet that appeared to be lost and was
retransmitted. But it wasn't really lost ... some router had
problems, held on to the packet for a while (order of seconds, could
be a minute if the TTL is large enough) and then re-injects the packet
back into the network. But by the time it reappears, the application
that sent it originally has already retransmitted the data contained
in that packet.

Because of these potential problems with TIME_WAIT assassinations, one
should not avoid the TIME_WAIT state by setting the SO_LINGER option
to send an RST instead of the normal TCP connection termination
(FIN/ACK/FIN/ACK). The TIME_WAIT state is there for a reason; it's
your friend and it's there to help you :-)

I have a long discussion of just this topic in my just-released
"TCP/IP Illustrated, Volume 3". The TIME_WAIT state is indeed, one of
the most misunderstood features of TCP.

I'm currently rewriting "Unix Network Programming" (see ``1.5 Where
can I get source code for the book [book title]?''). and will include
lots more on this topic, as it is often confusing and misunderstood.

An additional note from Andrew:

Closing a socket: if SO_LINGER has not been called on a socket, then
close() is not supposed to discard data. This is true on SVR4.2 (and,
apparently, on all non-SVR4 systems) but apparently not on SVR4; the
use of either shutdown() or SO_LINGER seems to be required to
guarantee delivery of all data.

[DryFisHH]

懂了，谢谢。事实证明Time_wait的时间确实是1到4分钟，多谢指教。