Redefine a single operator of a defined type. C++ - c++11

Question

I want to redefine the bit shift operator on a 64 bit unsigned integer in c++ in such a way that I can do say, x<<d, where x is a 64 bit integer and d is an integer with |d|<64, to make it equivalent to x<<d for d>0 and x>>|d| for d<0.
The only way I know how to do this is to define a whole new class and overload the << operator, but I think that also means I need to overload all the other operators I need (unless there is a trick I don't know), which seems a bit silly considering I want them to behave exactly as they do for the pre-defined type. It's just the bitshift that I want to change. At present, I have just written a function called 'shift' to do this, which doesn't seem very c++ ish, even though it works fine.
What is the stylistically correct way to do what I need?
Thanks

Answer 1

If you were able to do this, it would be very confusing to other C++ programmers who read your code and see:
int64 x = 92134;
int64 y = x >> 3;
And have it behave differently than their expectations, and behave differently from what the C++ standard defines.
The stylistic choice that agrees most with the C++ code I've seen is to continue using your own myshift() function.
int64 y = myshift(x, 3);

Answer 2

I think it's very horrible (and I propose it just for fun) but... if you accept to wrap the number of bit shifted in a struct...
#include <iostream>
struct foo
{ int num; };
long long int operator<< (const long long int & lli, const foo & f)
{
int d { f.num };
if ( d < 0 )
d = -d;
if ( d >= 64 )
d = 0;
return lli << d;
}
int main()
{
long long int lli { 1 };
std::cout << (lli << foo{+3}) << std::endl; // shift +3
std::cout << (lli << foo{-3}) << std::endl; // shift +3 (-3 -> +3)
std::cout << (lli << foo{+90}) << std::endl; // no shift (over 64)
std::cout << (lli << foo{-90}) << std::endl; // no shift (over 64)
return 0;
}