Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!tut.cis.ohio-state.edu!ucbvax!hplabs!hp-pcd!hplsla!jima
From: jima@hplsla.HP.COM (Jim Adcock)
Newsgroups: comp.lang.c++
Subject: Re: 'this' question
Message-ID: <6590244@hplsla.HP.COM>
Date: 11 Sep 89 17:45:35 GMT
References: <14232@polyslo.CalPoly.EDU>
Organization: HP Lake Stevens, WA
Lines: 42

//> Wang:
//> When I write:
//> 
//> type_student* me = new type_student();
//> 
//> Is the address of 'me' guaranteed to be the same as the value of 'this'
//> for the 'me' object?

//Depends on what you consider the "address" of me, and what you consider the 
//"me" object.  I presume by the "me object" you mean the structure referred to
//by *me.  Note that in C++ the object "me" is typically considered to be "me",
//not "*me."  This is different from other object oriented languages which
//use pointers or handles for all objects.  Similarly, I presume that by the
//"address of me" you mean the address of the underlying object pointed to
// by "me."  In which case, consider:

#include <stdio.h>

class bogus_student
{
  const char* name;
public:
  bogus_student(const char* Name) : name(Name) {}
  void print_name(){printf("I'm a student named %s, ",name);}
  void print_addresses()
  {printf("my this=%X but my address=%X\n", this, &(*this));}
  bogus_student* operator&(){return (bogus_student*)0x1234;}
};

void main()
{
  bogus_student& me = *(new bogus_student)("me");
  me.print_name();  me.print_addresses();

  bogus_student* me2 = new bogus_student("me2");
  me2->print_name();  me2->print_addresses();
}


//Hopefully, people would not overload the address-of operator "&" in ways
//that will be incompatible with normal usage.  Unfortunately, in reality
//they will.