Re: sockets affected by IPsec always block (2.6.23)

[Stefan Rompf]

Am Donnerstag, 6. Dezember 2007 09:53 schrieb David Miller:

> > I think the words "shall fail" and "immediately" are quite clear.
>
> They are, but the context in which they apply is vague.

"socket is connection-mode" => SOCK_STREAM

> I can equally generate examples where the non-blocking behavior you
> are a proponent of would break non-blocking UDP apps during a
> sendmsg() call when we hit IPSEC resolution.  Yet similar language on
> blocking semantics exists for sendmsg() in the standards.

I am not a good enough kernel hacker to exactly understand the code flow in 
udp_sendmsg(). However, it seems that it first checks destination validity 
via ip_route_output_flow() and queues the message then. The sendmsg() 
documentation only talks about buffer space. I can see your dilemma.

The reason why I'm pushing this issue another time is that I know quite a 
bit about system level application development. A very typical design pattern 
for non-naive single or multi threaded programs is that they set all 
communication sockets to be nonblocking and use a select()/epoll() based loop 
to dispatch IO. This often includes initiating a TCP connect() and 
asynchronously waiting for it to finish or fail from the main loop.

The dangerous situation here is that in 99% of all cases things will just work 
because the phase 2 SA exists. In 0.8%, the SA will be established in <1 sec. 
However, in the rest of time the server application that you have considered 
to be stable will end up sleeping with all threads in a connect() call that 
is supposed to return immediatly.

> The world is shades of gray, implying anything else is foolhardy and
> that's how I'm handling this.

Even though I consider programmers that ignore the result code on a 
nonblocking UDP sendmsg() fools, I agree. May be the best compromise is what 
Herbert Xu suggested in <20071205001230.GA11391@xxxxxxxxxxxxxxxxxxx> in this 
thread: At least, for connect() O_NONBLOCK ist ALWAYS respected. Because this 
is where the chance for breakage is highest.

Stefan
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