On Mon, 3 Aug 2009, Dave Tweed wrote:

> sergio masci wrote:
> > On Sun, 2 Aug 2009, Dave Tweed wrote:
> > > My point was, that if you add a simple boolean flag to the FIFO
> > > structure, you CAN completely fill it, and this actually has negligible
> > > impact on the overall code size, and does NOT affect its thread-safe
> > > properties.
> > 
> > Ok, so please show how you would do this.
> 
> Oh, no, that would be telling! :-)
> 
> I've already written this up as part of the EQ quiz for Circuit Cellar #230
> (September 2009), so you'll be able to read my solution there by the end of
> the month. In the meantime, give it some thought. It's actually quite
> straightforward.

Now I'm curious :-)

But I'll respect decission and I wont discuss it here.

If I send you an offlist message (to your from address) will you 
receive it or will your spam filter reject it as not coming from the 
piclist?

> 
> > As a mater of interest what limitted expression-handling capabilities do 
> > you see this code trying to get around (with the use of the temporary 
> > variable).
> 
> Well, you wrote it with only one operation per line. With the "proc"
> keyword, it wasn't immediately obvious that this was C code. I was also
> suspicious of storing the length, head and tail of the FIFO in the array
> with the data. You wouldn't ordinarily do that unless a language limitation
> was forcing you to. I just assumed it was some personal language of yours
> that was even lower-level than C, and I guessed that it was either easily
> translatable to assembly, or that it had come from assembly to begin with.
> 
> In C, of course, you would simply write something like:
> 
>    tail = (tail + 1) % FIFO_SIZE;
> 
> Or, if your compilier isn't smart enough to optimize the % operator:
> 
>    tail = (tail + 1 == FIFO_SIZE) ? 0 : tail + 1;
> 
> Either of these is perfectly thread-safe; no standards-compliant compiler
> would turn either statement into multiple assignments to tail.

Is this guarenteed anywhere?

I can see:

	tail = (tail + 1) % FIFO_SIZE;

being optimised to (if FIFO_SIZE is a power of 2):

	incf	tail
	movlw	FIFO_SIZE-1
	andwf	tail

and

	tail = (tail + 1 == FIFO_SIZE) ? 0 : tail + 1;

being optimised to:

	incf	tail
	movf	tail, w
	xorlw	FIFO_SIZE
	btfss	STATUS, Z
	clrf	tail

Friendly Regards
Sergio Masci
-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist