Read what the God Of C++ (Stroustrup) had to say:
regards,Code:int main()
{
X ob;
X *prt = &ob;
No. You get the address from "this"; that doesn't involve a call of
Quote:
// implies ==> X* ptr = ob.X::operator&()
// which implies ==> X* prt = X::operator&(&ob);
// My question is since the call to operator& member function must pass the
// address of the calling object, logically shouldn't this result in a recursive call.
operator&(). The operator&() you define affects only explicit uses of
&.
Also, you be careful about making assumptions about what the compiler
does. No global function called operator&() will be generated here. If
it didn't you might get some "interesting" overload problems.
// I would like to understand how does it work ??? or rather why it works
return 0;
}
Shiv