if it is possible a pointer to
X& operator*(unsigned b, const X& a){}
will be
X& (*operator*)(unsigned b, const X& a);
right?
No.
The last declaration attempts to declare a variable that is a pointer to
routine, and name that variable 'operator*'.
You can't use that special name for variables.
But as to the question you probably meant to ask, no, the built-in operators do
not exist as routines you can take the address of.
It is however a simple matter to define routines that wrap the use of built-in
operators, and the standard library even offers some such routines.
Cheers & hth.,
- Alf
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