Path: utzoo!utgpu!jarvis.csri.toronto.edu!clyde.concordia.ca!uunet!microsoft!jimad
From: jimad@microsoft.UUCP (JAMES ADCOCK)
Newsgroups: comp.lang.c++
Subject: virtual assignment operator is possible?
Keywords: assignment operator, virtual function
Message-ID: <10138@microsoft.UUCP>
Date: 3 Jan 90 22:39:16 GMT
References: <1989Dec30.174745.8174@sdr.slb.com> <10137@microsoft.UUCP>
Reply-To: jimad@microsoft.UUCP (JAMES ADCOCK)
Organization: Microsoft Corp., Redmond WA
Lines: 68

//saito:
//>Suppose I have a base class B, and its derived class D.  In each class,
//>I defined assignment operator, i.e, X& operator=(X&).  Then,

//consider:

#include <stdio.h>
#include <stdlib.h>
#include "new.h"

class B            
{ 
	char* bp;
public: 
	virtual B& assignTo(B& b) const { b.bp = bp; return b; }
	B(char* id="B") { bp = id; }
	B(const B& b) { bp = b.bp; }
	virtual B& operator=(const B& b) { return b.B::assignTo(*this); }
	virtual void Print(char sep='\n') { printf(bp); putchar(sep); }
};

class D : public B 
{ 
	char* dp;
protected:
	virtual B& assignTo(B& b) const 
	{ (*(D*)&b).dp = dp; return B::assignTo(b); } 
	//(except not with MI virtual bases)
public: 
	D(char* bid="B", char* did="D") : B(bid) { dp = did; }
	D(const B& b) : B(b) { dp = "D"; }
	D(const D& d) : B(d) { dp = "D"; }
	virtual B& operator=(const B& b) { dp = "?"; return b.assignTo(*this); }
	virtual D& operator=(const D& d) 
	{ return *(D*)&(d.D::assignTo(*this)); } //(ditto)
	virtual void Print(char sep='\n') 
	{ B::Print(' '); printf(dp); putchar(sep); }
};

int main()
{

B* pb0 = new B("b0");
B* pb1 = new B("b1");
D* pd0 = new D("db0","d0");
D* pd1 = new D("db1","d1");

*pb1 = *pb0; // B& B::operator=(B&) is invoked.
pb1->Print();
*pd1 = *pd0; // D& D::operator=(D&) is invoked.
pd1->Print();

//This is perfectly fine.  However:

B* pdb2 = new D("db2","d2");
B* pdb3 = new D("db3","d3");
*pdb2 = *pdb3; //B& D::operator=(B&) is invoked.  No problem.
pdb2->Print();

// But what should we do if someone tries to assign a B to a D ?

B* pdb4 = new D("db4","d4");
B* pb5  = new B("b5");
*pdb4 = *pb5;
pdb4->Print();

return 0;
}