define double constant as hexadecimal?

Hexadecimal float and double literals do exist. The syntax is 0x1.(mantissa)p(exponent in decimal) In your case the syntax would be

double x = 0x1.fffffffffffffp-1

It's not safe, but something like:

double a;
*(reinterpret_cast<uint64_t *>(&a)) = 0x3FEFFFFFFFFFFFFFL;

However, this relies on a particular endianness of floating-point numbers on your system, so don't do this!

Instead, just put DBL_EPSILON in <cfloat> (or as pointed out in another answer, std::numeric_limits<double>::epsilon()) to good use.

#include <iostream>
#include <iomanip>
#include <limits>
using namespace std;

int main()
{
    double const    x   = 1.0 - numeric_limits< double >::epsilon();

    cout
        << setprecision( numeric_limits< double >::digits10 + 1 ) << fixed << x
        << endl;
}