Re: Deriving from concrete types
In article <1151354600.564226.315500@r2g2000cwb.googlegroups.com>,
todma <toddmarshall2002@yahoo.com> wrote:
Read up on the 'using' syntax.
class MyVector : private std::vector< int > {
public:
using operator [];
using resize;
// etc.
};
Now you can control exactly the members that you wish to include and
you cannot use a MyVector as a substitute for any std::vector< int >
either.
Kirit
Was 'using' created mainly to solve this problem? I guess this is a
motivation to upgrade my compiler.
Was the small debate here over deriving from a class without virtual
destructors a classic debate that is now resolved via the 'using'
keyword?
This relates to my desire to derive classes from vector<> and pair<>.
Is this desire in fact misguided?
thanks,
Todd.
The facts are these according to the standard [approved and 06 draft]
but I quote 06 draft since I can copy and paste from it..
Section 11.2 paragraph 1 of N2009=06-0079 dated 2006-04-21 states:
<quote>
If a class is declared to be a base class (clause 10) for another class
using the public access specifier, the public
members of the base class are accessible as public members of the
derived class and protected members of the base
class are accessible as protected members of the derived class. If a
class is declared to be a base class for another
class using the protected access specifier, the public and protected
members of the base class are accessible as
protected members of the derived class. If a class is declared to be a
base class for another class using the private
access specifier, the public and protected members of the base class
are accessible as private members of the
derived class.
</quote>.
related footnote
<quote>
As specified previously in clause 11, private members of a base class
remain inaccessible even to derived classes unless friend declarations
within the base class definition are used to grant access explicitly.
</quote>
Therefore C++ inheritance is a superset of OOD inheritance/derivation.
Any other definition such as its an 'x a' relatiionship is a speial
case if the above descriptions hold. Deriving from a concrete class is
a legal reuse of code. See <stack> note the container used for the
stack is a PROTECTED member. so that special stacks can be written
without rewriting the entire std::stack class as something else.
If you are generally assuming C++ inheritance -> OOD inheritance you
are mistaken. OOD inheritance is included in C++'s public inheritance,
and is often the most often used use of public inheritance, but not the
only one. If you use C++ public inheritance, I do recommend you
document this is not an OOD inheritance, as unfortunately there is a
large number of programmers that assume, in error , that C++ public
inheritances is OOD inheritance.
To answer your question there is nothing wrong with deriving from
std::vector<T,A> as long as it is not used as an OOD inheritance. The
following is a stack, that allows random access read/write access to a
stack [not removal except the top].
struct my_stack:public std::stack<int,std::vector<int> >
{
int & operator [] (int n) { return c.operator [] (n);}
int & operator [] (int n) const {return c.operator[](n);}
};
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