TITLE: Inheritance or templates?


RC = Raghavan Chakravarti (rchak@saucer.cc.umr.edu)
RB = Russell Blackadar <russell@news.mdli.com>
JS = John Max Skaller (maxtal@Physics.usyd.edu.au), 26 Mar 95


RC:

I have a type casting doubt. Say i have 

  class A {
     public:
       int a;
       int b;
       A(int i,int j) { a=i; b=j; }
  };

  class B {
     public:
       int a;
       int b;
       B(int i,int j) { a=i; b=j; }
  };

  B Bobj;

  cout << ((A) bObj).a;    		// gives error
  cout << ( (A*) (void*) &bObj)->a;   	// prints 1 correctly


JS:

	[Note its] better style to write
	A(int i, int j) : a(i), b(j) {}


	[Prints 1 correctly] ... on some machines.

RC:

My question why is the first one not allowed ? 
What is the implication of this ?


JS:

	The kind of cast you are doing is called a "reinterpretation"
in which you say to yourself "I know the layout of B is the same
as A -- at least for the first few members" and then you tell
the compiler that.

	This is considered to warrant a specific keyword and a new
kind of cast:

	cout << reinterpret_cast<A&>(bObj).a;  // note use of reference!
	cout << reinterpret_cast<A*>(&bObj).a; 

Using reinterpret casts is not generally a good idea.


RC:

Secondly if I have 2 classes which have similar data members is it wise to
typecast as done previously ?  


RB:

No.  That's what inheritance is for.


JS:

	Not really. Its what templates are for:

	template<class T> void print(T const& object)
	{
		cout << object.a << ", " << object.b << endl;
	}

	print(aObj);
	print(bObj);


[ I added caps for emphasis below. -adc ]

INHERITANCE ISN'T FOR WHEN TWO CLASSES ARE SIMILAR, ITS FOR WHEN
THE TYPE OF ONE CLASS _IS_ THE TYPE OF THE OTHER.
INHERITANCE IS AN INVASIVE TECHNIQUE TO BE USED WITH CAUTION.