Re: [PATCH 5/5] usb_debug: EXPERIMENTAL - poll hcd device to forcewrites

[Alan Stern]

On Wed, 6 May 2009, Oliver Neukum wrote:

> Something a bit a like this:
> 
> commit a02639fe7d3f9788263305cff0669eac91f54002
> Author: Oliver Neukum <oneukum@linux-d698.(none)>
> Date:   Wed May 6 19:14:30 2009 +0200
> 
>     terrible implementation of usb serial write buffering

For algorithm discussions like this, I find reading code rather
difficult.  English or pseudo-code presentations are a lot more
intelligible.

A little thought yielded the following algorithm.  It assumes there is
a fixed set of URBs allocated, unlike what you have done.  Does it make
sense to take this approach?

Let N be the total number of URBs allocated, each capable of holding up 
to B bytes.  Let NIF be the number of URBs in flight at any time, so 
the number of available URBs is N - NIF.  The number of available bytes 
might be < (N - NIF)*B because the next URB might be partially full.

P is an adjustable parameter of the algorithm.  For simplicity you can
take P = 1, but increasing P (any value below N is okay) would yield
reduced latency at the cost of more partially-filled URB submissions
(so possibly reduced throughput).

    Write routine:
	Copy bytes into the available URB buffers, submitting URBs as 
	they get filled.  At the end, if the next URB is partially full 
	then submit it only if NIF < P.

    Completion routine:
	If the next URB to send is partially filled, submit it.

    write_room routine:
	Return the actual number of bytes remaining in the available 
	URBs, but no more than (N-P)*B.

How does that sound?  Converting \n to \r\n will add some complication 
but not too much.  

Allocating URBs on the fly adds a lot of complication.  There has to be 
a minimum number of pre-allocated URBs; otherwise write_room could 
never return a positive value.  If you allocate additional URBs 
later on, when would you free them?

Alan Stern

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