On 12/24/13 11:59 AM, H. S. Teoh wrote:
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.)
No, again, I think nothing like that would work. It's hopelessly naive
(in addition to being plain wrong for simple inputs like low=1, high=10,
step=1).
There are combinations of low, high, and step that don't even make
progress, i.e. step is sufficiently small compared to low to effectively
make low + step == low.
Try this to verify approaches:
#!/Users/aalexandre/bin/rdmd
import std.stdio, std.conv, std.math;
void main(string[] args)
{
auto low = to!double(args[1]);
auto up = to!double(args[2]);
auto step = to!double(args[3]);
writeln("Predicted: ", low + fmod(up-low, step)*step);
for (;;)
{
auto next = low + step;
if (next >= up) break;
low = next;
}
writeln("Actual: ", low);
}
Andrei
