Hi, Can anyone posting a sample code for C++
overloading "->" operator.

I wrote the following code, but it doesn't work.
------------------------------------------------------------------
Class A : public base
{
 int id;
 B *p;
}

Class B : public base
{
 int id;
}

class base
{
public :
 operator->(base *p);
}

A *a;
B *b;

a->p = b; // this doesn't call operator->()
a->b; // but this line woundn't compile

_________________________________________________________________

Please forward posting to xwu@ucrengr.ucr.edu also if possible.

Thanks.

Xingwu

In article <31501@galaxy.ucr.edu> xwu@ucrengr.ucr.edu (xing wu) writes:
>Hi, Can anyone posting a sample code for C++
>overloading "->" operator.


#include <stdio.h>

class A
{
public:
 int id;
 A(int i) : id(i) {}
};

class APtr
{
 A* aptr;
public:
 APtr(A* pa) : aptr(pa) {}
 A* operator->() const { printf("(%lX) ->", (long)aptr); return aptr;}
 A& operator*() const  { printf("*(%lX)", (long)aptr); return *aptr;}
 A& operator[](int i) c
 const  { printf("(%lX)[%d]", (long)aptr, i); return aptr[i];}
};

main()
{
 A a(123);
 APtr p(&a);

 printf(" id = %d\n", p->id);
 printf(" . id = %d\n", (*p).id);
 printf(" . id = %d\n", p[0].id);

 return 0;
}

In article 31501@galaxy.ucr.edu, xwu@ucrengr.ucr.edu (xing wu) writes:
>class base
>{
>public :
> operator->(base *p);
>}
>
>A *a;
>B *b;
>
>a->p = b; // this doesn't call operator->()
>a->b; // but this line woundn't compile
>

Try this:

(*a)->p;

An operator binds to an object, not a pointer to an object (I guess).


---
Roxsan Payette      lmcropa@lmc.ericsson.se
Ericsson Communication Inc (LMC/U/SF)   Tel. (514) 738-8300 #2508
All opinions are my own.    Fax. (514) 345-7982