连续写入套接字后的Perl套接字编程问题

Question

我IO::Socket::INET用来创建这样的套接字:

$lsn1 = IO::Socket::INET->new(
                            PeerAddr => '192.168.0.2', 
                            PeerPort => 1850, 
                            Proto    => 'tcp', 
                            Type     => SOCK_STREAM
   ) || die "Can't connect to 192.168.0.2:1850 : $!\n"; 

$lsn2 = IO::Socket::INET->new(
                            PeerAddr => '192.168.0.2', 
                            PeerPort = >1852, 
                            Proto    => 'tcp', 
                            Type     => SOCK_STREAM
   ) || die "Can't connect to 192.168.0.2:1852 : $!\n";

那么,我想读取和写入两个套接字的数据,所以序列是:

1. $lsn1->print(msg1);  send message 1 to server from $lsn1.
2. $line = <$lsn2>;     receive message 2 from server from $lsn2.
3. $lsn2->print(msg3);  send message 3 to server from $lsn2.
4. $lsn2->print(msg4);  send message 4 to server from $lsn2.
5. $line = <$lsn2>;     receive message 5 from server. But it is all zeros! However I can 
                        see the data on wireshark.

一切都很好,直到第5步:在服务器端后收到我的消息4并发回这应该是通过捕获MSG5 $line = <$lsn2>,而不是捕获有意义的价值,它捕获所有0.我用wireshark看看发生了什么,在服务器发给我msg5之后发送了我身边的RST ACK.

服务器中的函数收到msg4并立即发送回msg5.如果我摆脱了该函数中msg4的发送,那么发送msg5之后的是FIN ACK.

谁能告诉我为什么RST ACK会这样？perl脚本可以打开两个套接字并像我一样读写.

Answer 1

您可能没有完全阅读消息 2，而在套接字上留下了一些数据（换行符、空字符等）。然后，当您尝试读取消息 5 时，您会读取此数据，并且您的程序会在服务器实际发送消息 5 之前终止。