On Sun, Feb 14, 2016 at 12:54 PM, Charles R Harris < charlesr.har...@gmail.com> wrote:
> > > On Sun, Feb 14, 2016 at 12:35 PM, Nils Becker <nilsc.bec...@gmail.com> > wrote: > >> 2016-02-13 17:42 GMT+01:00 Charles R Harris <charlesr.har...@gmail.com>: >> >>> The Fortran modulo function, which is the same basic function as in my >>>> branch, does not specify any bounds on the result for floating >>>> numbers, but gives only the formula, modulus(a, b) = a - b*floor(a/b), >>>> which has the advantage of being simple and well defined ;) >>>> >>> >>> >> In the light of the libm-discussion I spent some time looking at floating >> point functions and their accuracy. I would vote in favor of keeping an >> implementation that uses the fmod-function of the system library and bends >> it to adhere to the python convention (sign of divisor). There is probably >> a reason why the fmod-implementation is not as simple as "a - b*floor(a/b)" >> [1]. >> >> One obvious problem with the simple expression arises when a/b = 0.0 in >> floating point. E.g. >> >> In [43]: np.__version__ >> Out[43]: '1.10.4' >> In [44]: x = np.float64(1e-320) >> In [45]: y = np.float64(-1e10) >> In [46]: x % y # this uses libm's fmod on my system >> > I'm not too worried about denormals. However, this might be considered a bug in the floor function In [16]: floor(-1e-330) Out[16]: -0.0 Chuck
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