That's just kind of odd....
--
- Nicholas Paldino [.NET/C# MVP]
-
mv*@spam.guard. caspershouse.co m
"Peter Duniho" <Np*********@nn owslpianmk.comw rote in message
news:op******** *******@petes-computer.local. ..
On Mon, 04 Jun 2007 11:54:43 -0700, Nicholas Paldino [.NET/C# MVP]
<mv*@spam.guard .caspershouse.c omwrote:
You have to explain how the first and last cases are supposed to
work. Rounding .5625 to two places should result in .57 and rounding
6.125 to two places should result in 6.13, using mathematical rounding.
What rounding are you doing?
It seems to me that he wants the rounding to round to multiples of 0.05.
To the OP, you can accomplish this in the specific case like this:
double RoundToFiveHund redths(double num)
{
return Math.Round(num * 2.0, 1) / 2.0;
}
Assuming you are always dealing with multiples that are themselves
divisible into 1.0, you can do it generally like this:
double RoundToMultiple (double num, double multiple)
{
double scale = 1.0 / multiple;
return Math.Round(num * scale) / scale;
}
Note that if you pick something that is theoretically divisible into 1.0,
but which is not representable in binary (eg 1/3, or 0.333...), you may
wind up wind rounding error. But this would be unavoidable, since even if
rounded perfectly, you would not be able to represent the result exactly
in binary form (the same thing holds true for decimal types, though with
different specific values of course).
Pete
