Question

Is there a simple way of rounding a value either down to or to the nearest (1, 2, or 5) x 10^n where n is an integer? As in one of {…, .05 .1, .2, .5, 1, 2, 5, 10, 20, 50, 100…}

Thanks.

Answer 1

You can take the d=floor(log10(n)) of your number n to get the scale, then divide by 10 to the d to normalize the number to the range of [1.0,10.0). From that point it should be easy to round because there are very limited possibilities. Once you’ve done that, multiply by 10 to the d to restore the number to the original range.

The following function is in C, I don’t know enough about Objective-C to know if this is idiomatic.

double RoundTo125(double value)
{
   double magnitude;
   magnitude = floor(log10(value));
   value /= pow(10.0, magnitude);
   if (value < 1.5)
      value = 1.0;
   else if (value < 3.5)
      value = 2.0;
   else if (value < 7.5)
      value = 5.0;
   else
      value = 10.0;
   value *= pow(10.0, magnitude);
   return value;
}