Question

From what I understand, when a socket buffer (skb) is allocated by the kernel stack (on Tx path), the ‘head’ and ‘tail’ pointers point to the start & end of reserved memory space.

My qs is: what exactly determines this amount: (tail – head) octets?
Is it max MTU? Doesn’t seem like it as a little experimentation shows that the # of bytes (tail-head) can be quite small (320 bytes, etc). So is it the socket app that plays a role here?

Also, AFAIK, the idea is to reserve enough memory such that no further allocs are required further down the protocol stack, i.e., reserved space can be more than the payload. If so, what about cases where further encapsulation is done (like VoIP, SSL, etc?)?

Update:
Does this help answer the qs?

1. Found in net/ipv4/tcp_output.c:

   …
   tcp_make_synack()
   { …
   skb = sock_wmalloc(sk, MAX_TCP_HEADER + 15 + s_data_desired, 1, GFP_ATOMIC);
   …
   }

This seems to be the SYN-ACK part of the TCP state m/c.
So, by this, ‘size’ seems to work out to around 320.

1. TCP segments data packets..
   default ‘mss’ is 512 ? <– based on tcp_base_mss ?

TIA!

Answer 1

Small correction first – the buffer is between head and end, not tail (which is paired with data).

Your basic idea is correct. TCP allocates according to the data it needs, plus room for lower layer headers. Since these headers are not really known in advance, TCP allocates according to the maximum possible (which is very likely to be much more than actually needed).

If this ends up not enough, the packet can be reallocated (pskb_expand_head can be used).
In most cases, code that adds data to the packet first checks if there’s enough headspace (which will hopefully be the case), and reallocates if not.

The “default” MSS isn’t very important – it’s only used when no better value can be found. Normally, the MSS will be MTU-40, which is usually 1460.