On Tue, Dec 24, 2013 at 11:35:49AM -0800, Andrei Alexandrescu wrote: > On 12/24/13 10:56 AM, Craig Dillabaugh wrote: > >On Tuesday, 24 December 2013 at 17:10:53 UTC, Andrei Alexandrescu > >wrote: [.[..] > >>The integral cases are easy. We need to crack the floating point > >>case: given numbers low, up, and step, what's the closest number > >>smaller than up that's reached by repeated adds of step to low? > >> > >>Andrei > > > >Doesn't think work, or am I missing something? > > > >low + floor( (up-low)/step ) * step > > I doubt it's anything as simple as that. The magnitudes of up, low, > and step must be taken into account. [...]
What about low + fmod(up-low, step)*step? Is that better? (Assuming that fmod would take relative magnitudes into account, of course.) T -- Only boring people get bored. -- JM
