Re: [RFC, 2.6] a simple FIFO implementation

[Buddy Lucas]

On Thu, 16 Sep 2004 20:30:30 +0200, Stelian Pop <stelian@xxxxxxxxxx> wrote:
> On Thu, Sep 16, 2004 at 06:07:07PM +0200, Buddy Lucas wrote:
> 
> > > > Indeed, that would exactly be the reason *why* this would fail. ;-)
> > > >
> > > > The expression fifo->size - fifo->tail + fifo->head might be negative
> > > > at some point, right? (fifo->head has wrapped to some small value and
> > > > fifo->tail > fifo->size)
> > >
> > > And what is the value of an unsigned int holding that 'negative' value ? :)
> > >
> >
> > Which unsigned int?! ;-) The expression a - b is negative for unsigned
> > ints a and b where a < b. So, your unsigned ints "total" and
> > "remaining" won't be negative of
> > course, but they won't reflect what is actually left in the buffer;
> > they will equal the
> > value of len (in some cases) after fifo->head has wrapped (because of the
> > unsignedness) and fifo->tail has not. Which would not be correct.

Argh! Headless chicken routine.
 
> It is, thanks to modular arithmetic.
> 
> Let's imagine we use unsigned char instead of unsigned int (simpler
> to explain), and we have a 200 bytes buffer.
> 
> At the beginning:
>         head = tail = 0
> We add 200 bytes:
>         head = 0, tail = 200
> We extract 200 bytes:
>         head = 200, tail = 200
> We add 200 bytes more:
>         head = 200, tail = (200 + 200) % 256 = 144
> Now the buffer length is:
>         144 - 200 = (-56) % 256 = 200
> 
> Thanks to modular arithmetic magic, we get the correct answer: 200.

Indeed. I didn't notice you were getting at my head. ;-) 
Sorry for the noise, Stelian.

> 
> 
> Stelian.
> --
> Stelian Pop <stelian@xxxxxxxxxx>
>
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