[LAU] OT(ish): Strange coding problem (audio related)

[Peter Nelson]

On Sat, 2011-01-29 at 12:07 +0100, Philipp Überbacher wrote:
> Excerpts from fons's message of 2011-01-28 16:11:52 +0100:
> > On Fri, Jan 28, 2011 at 02:02:36PM +0100, Philipp Überbacher wrote:
> > 
> > > rant_begin
> > >     Why can't log mean the same thing everywhere? Why does it need to be
> > >     base e here and base 10 there? Why is there no consistency?
> > >     And why is there no proper logarithmus dualis function? Because you
> > >     can simply do log(n)/log(2)? We've just seen how well this works.
> > >     How about:
> > >         log() - base 10
> > >         ln() - base e - logarithmus naturalis
> > >         ld() - base 2 - logarithmus dualis
> > > rant_end
> > 
> > Libm has log(), log10, and log2().
> 
> Took me a while to figure out that libm is part of glibc :)
> Good to know that those functions are available on probably pretty much
> all linux systems.
> 
> > > The next obvious question is: Does the inaccuracy reliably result in
> > > values bigger than 11?
> > 
> > No.
> > 
> > If the input is a power of two, and you expect an integer as
> > a result, just do
> > 
> >   k = (int)(log2(x) + 1e-6)
> 
> log2() suffers from the same problem? I somewhat dislike the idea of
> adding a constant.
> 
> > or
> > 
> >   k = (int)(log(x)/log(2) + 1e-6)
> > 
> > or
> > 
> >   int m, k;
> >   for (k = 0, m = 1; m < x; k++, m <<= 1);
> > 
> >   which will round up if x is not a power of 2.
> 
> Neat. I thought about it myself yesterday but my ideas weren't exactly
> brilliant. One idea was to divide by 2, the other to use a small 
> lookup table for powers of 2. I don't really know about efficiency, but
> I guess bit shifting is as efficient as it gets?
> Anyway, it's a neat way to avoid the problem and the rounding properties
> of mult/div in case of not power of 2 could be useful as well.

Well, now I'm just being pedantic :-), but as a quick test using rdtsc
(i.e. profiling to be taken with a grain of salt):

1: log(x)/log(2)
2: (1 << k) < x
3: m < x & m <<= 1

This is 1000 iterations; cycling through x of 512, 1024, 2048, 4096 (to
prevent the compiler optimizing log(x)/log(2) to a single call
throughout the whole test). The fourth line is the sum of the iterator
in each loop.

./a.out (unoptimized)
1 3132000 cycles
2 261468 cycles
3 285273 cycles
1 50500, 2 50500, 3 50500

./a.out (-O3)
1 2598345 cycles
2 170055 cycles
3 173673 cycles
1 50500, 2 50500, 3 50500

I must admit, I'm surprised that fons way shows slightly slower in my
test!

As an aside, here's the result with (-O3 -ffast-math)

./a.out
1 150894 cycles,
2 169983 cycles,
3 158850 cycles,
1 47750, 2 50500, 3 50500

Yeah, faster, but wrong :-)

Peter.