modf returns 1 as the fractional:

Question

I have this static method, it receives a double and \"cuts\" its fractional tail leaving two digits after the dot. works almost all the time. I have noticed that when i

Answer 1

Here is a way without rounding:

double double_cut(double d)
{
    long long x = d * 100;
    return x/100.0;
}

Even if you want rounding according to 3rd digit after decimal point, here is a solution:

double double_cut_round(double d)
{
    long long x = d * 1000;

    if (x > 0)
        x += 5;
    else
        x -= 5;

    return x / 1000.0;
}

Answer 2

Remember floating point number cannot represent decimal numbers exactly. 2.3 * 100 actually gives 229.99999999999997. Thus modf returns 229 and 0.9999999999999716.

However, cout's format will only display floating point numbers to 6 decimal places by default. So the 0.9999999999999716 is shown as 1.

---

You could use (roughly) the upper error limit that a value represents in floating point to avoid the 2.3 error:

#include <cmath>
#include <limits>
static double dRound(double d) {
   double inf = copysign(std::numeric_limits<double>::infinity(), d);
   double theNumberAfter = nextafter(d, inf);
   double epsilon = theNumberAfter - d;

   int factor = 100;
   d *= factor;
   epsilon *= factor/2;
   d += epsilon;

   double returnVal;
   modf(number, &returnVal);
   return returnVal / factor;
}

Result: http://www.ideone.com/ywmua