#585 was: Quick update

[Dirk Hohndel]

On Thu, Jul 03, 2014 at 04:36:15PM +0200, Robert Helling wrote:
> 
> On 03.07.2014, at 15:58, Dirk Hohndel <dirk at hohndel.org> wrote:
> 
> >> 1) A minor one: the “going back in time” is done as a loop with 
> >> 
> >> 	while (i && —i)
> >> 
> >> where i is the index in the dive list. Here, it seems to me i==0 is a perfectly fine value, it is the first entry in the dive list. So the condition should rather be
> >> 
> >> 	while (i >= 0 && --i >= 0)
> > 
> > Yes, but that's equivalent to what we have.
> > while (i && --i) allows i = 1 which gets decremented to 0 before the loop
> > starts.
> > 
> >> Could somebody please confirm/refute this?
> > 
> > refute :-)
> 
> Ahm. I thought the difference is whether the loop runs with the value
> i==0 (which is a perfectly fine number for an index of the dive list).
> Do you still hold to your refutation?

But of course. Here's the code:

        divenr = get_divenr(dive);
	when = dive->when;
	i = divenr;
	while (i && --i) {
		struct dive *pdive = get_dive(i);

if we have a dive in the divelist, then get_divenr will give us a
non-negative divenr.
If divenr is 0 (first dive), then the while condition will FAIL because
there is no previous dive pdive that we could look at (--i will be -1)
If divenr is >0, then the while condition will be true and pdive will
indeed be the previous dive.
And if the divelist is empty (or dive is NULL) get_divenr will give us -1
and the while condition will also be false.

So to condens my answer: no, i=0 is NOT a valid number on which to enter
the loop as then --i turns it into -1 and pdive would become NULL. i=1 is
perfectly fine (second dive in the list), pdive will be get_dive(0) and
therefore the first dive.

The difference is that it's perfectly fine to have i=0 AFTER the --i, but
not before it...

Any more questions?

/D