Unix Socket Programming Question:
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How can I force a socket to send the data in its buffer?

Answer:

You can't force it. Period. TCP makes up its own mind as to when it can send data. Now, normally when you call write() on a TCP socket, TCP will indeed send a segment, but there's no guarantee and no way to force this. There are lots of reasons why TCP will not send a segment: a closed window and the Nagle algorithm are two things to come immediately to mind.

Setting this only disables one of the many tests, the Nagle algorithm. But if the original poster's problem is this, then setting this socket option will help.

A quick glance at tcp_output() shows around 11 tests TCP has to make as to whether to send a segment or not.

As you've surmised, I've never had any problem with disabling Nagle's algorithm. Its basically a buffering method; there's a fixed overhead for all packets, no matter how small. Hence, Nagle's algorithm collects small packets together (no more than .2sec delay) and thereby reduces the amount of overhead bytes being transferred. This approach works well for rcp, for example: the .2 second delay isn't humanly noticeable, and multiple users have their small packets more efficiently transferred. Helps in university settings where most folks using the network are using standard tools such as rcp and ftp, and programs such as telnet may use it, too.

However, Nagle's algorithm is pure havoc for real-time control and not much better for keystroke interactive applications (control-C, anyone?). It has seemed to me that the types of new programs using sockets that people write usually do have problems with small packet delays. One way to bypass Nagle's algorithm selectively is to use "out-of-band" messaging, but that is limited in its content and has other effects (such as a loss of sequentiality) (by the way, out-of- band is often used for that ctrl-C, too).

So to sum it all up, if you are having trouble and need to flush the socket, setting the TCP_NODELAY option will usually solve the problem. If it doesn't, you will have to use out-of-band messaging, but according to Andrew, "out-of-band data has its own problems, and I don't think it works well as a solution to buffering delays (haven't tried it though). It is not 'expedited data' in the sense that exists in some other protocols; it is transmitted in-stream, but with a pointer to indicate where it is."

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