[FIXED] What happens if one doesn't call [POSIX's] `recv` "fast enough"?

Issue

I want to account for a possible scenario where clients of my TCP/IP stream socket service send data to my service faster than it manages to move the data to its buffers (I am talking about application buffers, naturally) with recv and work with it.

So basically, what happens in such scenarios?

The way I see it, some facility beneath my service has to receive pieces of the incoming stream and store these somewhere until I issue 'recv', right? Most certainly the operating system. What happens if it runs out of memory to store the pieces while my service is not receiving them fast enough?

I don't want to re-open old questions, but I can't seem to find an answer to this seemingly obvious one?

Solution

TCP provides flow control . The TCP stack (both on the sender and receiver side) will be able to buffer some data for you, and this is usually done in the OS kernel.

When the receiver buffers fill up, the sender will know about it, and stop sending more data, eventually leading to the sending application blocking(or otherwise not being able to send more data) until space becomes available again.

Shortly described, every TCP packet(segment) sent includes the size of data that can be buffered - the window size. This means the other end at all times know how much data it can send without the receiver throwing it away because the buffers are full. If the window size becomes 0, buffers are full and no more data will be sent (and in case of the sender being blocking, a send() call will block), Theres procedures for probing whether the tcp window is still 0, so sending can resume again when the data has been consumed.

There's some more details here

Answered By - nos

Answer Checked By - Mary Flores (PHPFixing Volunteer)