Re: [musl] Superfluous shift in qsort()?

[Markus Wichmann <nullplan@gmx.net>]

On Wed, Jul 01, 2020 at 11:23:09PM +0200, Valentin Ochs wrote:
> On Wed, Jul 01, 2020 at 04:44:56PM -0400, Rich Felker wrote:
> > On Wed, Jul 01, 2020 at 08:50:26PM +0200, Markus Wichmann wrote:
> > > Hi all,
> > >
> > > I noticed something while reading code today: Near the end of qsort(),
> > > we have this gem:
> > >
> > > shl(p, 2);
> > > pshift -= 2;
> > > p[0] ^= 7;
> > > shr(p, 1);
> > >
> > > Now, I don't know if I am missing something, but don't the shl and the
> > > shr partially cancel out? Isn't this the same as
> > >
> > > shl(p, 1);
> > > p[0] ^= 3;
> > >
> > > As it is, it isn't wrong, just weird.
> >
> > Assuming non-overflow, I think they're equivalent (also assuming you
> > keep the pshift-=2).
>
> Yes, it looks that way. I'm afraid I don't have any further insight -
> it's been quite a while since I thought about the qsort code, and I've
> not been doing much work on algorithms over the last couple of years.
> The only thing I can think of is that one could figure out which
> behaviour with regard to overflow in shl() should be the valid one. I
> suspect that replacing it would be valid and this is some transformation
> I did while implementing smoothsort without realizing that this can be
> simplified.
>
> Cheers,
> Valentin
>

Overflow on shl() is completely impossible. To overflow a shl(p, 2), we
would need the penultimate bit in p to be set. Every bit in p stands in for
a tree of that order, so if bit n is set, the heap contains a tree with
a number of elements equal to the n'th Leonardo number.

I don't know how big the Leonardo number corresponding to the
penultimate bit is, but I do know that halfway through the upper word
(wasn't the factor something like 1.44 or so?), the Leonardo numbers get
too big to be contained in a machine word. Meaning that tree would be
way too large to be addressed.

I concur that this looks like a missed optimization opportunity, and not
a deliberate design decision.

Ciao,
Markus