[9fans] qread() with a coalescing queue

[erik quanstrom <quanstro@quanstro.net>]

in qbread, why does qbread only take full blocks for coalescing queues?
(these are only used for the console, and for tcp)  for example:

	if(q->state & Qcoalesce){
		/* when coalescing, 0 length blocks just go away */
		b = q->bfirst;
		if(BLEN(b) <= 0){
			freeb(qremove(q));
			goto again;
		}

		/*  grab the first block plus as many
		 *  following blocks as will completely
		 *  fit in the read.
		 */
		n = 0;
		l = &first;
		m = BLEN(b);
		for(;;) {
			*l = qremove(q);
			l = &b->next;
			n += m;

			b = q->bfirst;
			if(b == nil)
				break;
			m = BLEN(b);
			if(n+m > len)
				break;
		}
	}
	/* code to put extra bytes back on the queue below */

in the typical use with tcp, we're going to have a protocol like 9p or tls running
on top of tcp, and if it does a read of n bytes, it really does need n bytes.  so
splitting reads rather than blocks would seem less efficient because
copying a partial block <= mss should be much faster than a syscall.  (nsec()
with the fd open takes ~7-900ns, but a 1460-byte copy takes ~130ns @ a modest
10gb/s, much less if cache pressure doesn't dominate.)

i noticed this because even with the tcp rx queue full (slow receiver) i was seeing
twice as many reads  as i would have expected.  the following code works fine.

	if(q->state & Qcoalesce){
		/* when coalescing, 0 length blocks just go away */
		b = q->bfirst;
		if(BLEN(b) <= 0){
			freeb(qremove(q));
			goto again;
		}

		/*
		 * grab the first block and as many following
		 * blocks as will partially fit in the read
		 */
		n = 0;
		l = &first;
		m = BLEN(b);
		for(;;) {
			*l = qremove(q);
			l = &b->next;
			n += BLEN(b);
			if(n >= len || (b = q->bfirst) == nil)
				break;
		}
	}
	/* code to put extra bytes back on the queue below */

is there something or some consideration here that i'm missing?

- erik