Why does trunc(1) output as 0?

Question

Can someone explain me why in c++ happens such a thing:

double tmp;
...                    // I do some operations with tmp 
                       // after which it has to be equal to one
cout << tmp;           // prints 1
cout << trunc(tmp);    // prints 0
cout << trunc(tmp*10); // prints 9

I am using this for separation part right of decimal part from the number for example if i have: 5.010 ... i want to have 0.010 .. so I am using:

double remainder = tmp - trunc(tmp);

---

I am posting the whole code....the suggestion with floor does not worked

short getPrecision(double num, short maxPrecision) {

  // Retrieve only part right of decimal point
  double tmp  = fabs(num - trunc(num));
  double remainder = tmp;

  // Count number of decimal places
  int c = 0;
  while (remainder > 0 && c < maxPrecision) {
    tmp *= 10;
    remainder = tmp - trunc(tmp);
    c++;
  }
  return c;
}

When I run this function for example with 5.1 the remanider is 0 instead of 1

Answer 1

After some calculations it has to be one? Well, it could as well be 0.99999999999999999. Floating point operations are not precise, you should always take that into account.

Answer 2

Please see picture at http://en.cppreference.com/w/cpp/numeric/math/trunc. The chart there explains the inconsistency with truncing 1. Probably the same applies to 10 as well

This should help you achieving what you need:

double remainder = tmp - floor(tmp);