Check double variable if it contains an integer, and not floating point

#define _EPSILON_ 0.000001

bool close_to_int(double &d)
{
    double integer,
           fraction = modf(d, &integer);
    if(fraction < _EPSILON_)
    {
        d = integer;
        return true;
    }
    if((1.0 - fraction) < _EPSILON_)
    {
        d = integer + 1;
        return true;
    }
    return false;
}

This looks at both side of the integer value and sets the value of d if it is within the limits of an integer value.

Just compare ceil and floor value of d

return floor(d)==ceil(d);

So, for d1=555, the above statement will return 555==555, i.e, true, so it's an integer.

And for d2=555.6, the above statement will return 555==556, i.e, false, so it's a double.

modf uses std::nearbyint(num) that why you should use nearbyint which return a double without decimal and may be faster.

#include <iostream>
#include <cmath>

int main() {
    double      number = 55.12;

    if (!(number - std::nearbyint(number))) {
      std::cout << "Is integer!";
    } else {
      std::cout << "Has decimal!";
    }
    
    return 0;
}