Overloading the assignment operator, right now!

[Sean Kelly]

Bruno Medeiros wrote:
>
> One must be wary of what Oskar mentioned: that the assignment operator 
> is not the only assignment, there is also the implicit assignment of a 
> function call's argument passing. I think this can be easily solved by 
> having the function call use the copy operator, if it is overloaded, for 
> argument creation. Hum.., now that this is mentioned, how does it work 
> in C++? Does C++ do it that way?

Pretty much.  For example:

     class C
     {
     public:
         C(int i) : x(i) {}
         C(C const& o) : x(0) {}
         int x;
     };

     void f1(C val)  { printf("%d\n", val.x); }
     void f2(C* val) { printf("%d\n", val->x); }
     void f3(C& val) { printf("%d\n", val.x); }

     C c(1); f1(c); f2(&c); f3(c);

All of the above functions pass their parameters by value (just as in D) 
but only the f1 has C as a parameter.  f2 and f3 accept a pointer and a 
reference to C, respectively.  So calling f1 will create a copy of C via 
C's copy ctor, while f2 and f3 will copy the address of C and end up 
referencing the original object.  So executing the above should print:

     0
     1
     1


Sean