《TCP/IP详解卷1 协议》第2版中写道：
Linux% sock -s -v -q1 -O30000 6666
The -q1 option sets the backlog of the listening endpoint to 1.The -O30000
option causes the program to sleep for 30,000s(basically a long time,about 8
hours) before accepting any client connections.If we now try to connect to this
server continually,the first four connections are completed immediately.After
that,two connections are completed every 9s.Other operating systems vary
considerably in how this is handled.In Solaris 8 and FreeBSD 4.7,for example,
two connections are handled immediately and the third times out;subsequent
connections time out as well.
我不懂的是命令设置了backlog=1，那么后面的
"If we now try to connect to this server continually,the first four connections
are completed immediately.After that,two connections are completed every 9s."
和
"In Solaris 8 and FreeBSD 4.7,for example,
two connections are handled immediately and the third times out"
是怎么回事？
求详解，谢谢大神！
因为backlog =1 那么，你的第一个连接可以马上处理，而第二个放到backlog中，但是第三个来的时候就会超时丢掉了
参见18.11节，Incoming Connection Request Queue
The Solaris values in this figure are what we expect. The traditional BSD values are (for some unknown reason) the backlog value times 3, divided by 2, plus 1.
不好意思，才注意到你是在说第二版（linux实现）