Re: How to extend a class without redeclaring all its methods?

From:
"Alf P. Steinbach /Usenet" <alf.p.steinbach+usenet@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 04 Aug 2010 23:10:27 +0200
Message-ID:
<i3cl05$li9$1@news.eternal-september.org>
* Francois Grieu, on 04.08.2010 23:00:

In article<i3chnv$ke5$1@news.eternal-september.org>, "Alf P. Steinbach /Usenet"<alf.p.steinbach+usenet@gmail.com> wrote:

* Francois Grieu, on 04.08.2010 21:46:

Hello, I know C well, but not C++, and have a naive question.

I have a class defined by a library that I do not want to modify;
in my case the mpz_class of gmpxx.h (see<http://gmplib.org> ).

I want to add a member function to this class, without the hassle
of writing something for each existing function of that class.
I can make a new class with a new name, and do not need to add
any field. If have tried the following and it nearly works:

#include<iostream>
using namespace std;

#include<gmpxx.h>

// attempt at a straight extension of gmpxx's mpz_class
class myz_class : public mpz_class {
public:
      // smallest n such that 2^^n is greater than absolute value
      mp_bitcnt_t bitlen()


This should be 'const'.

{
          mp_bitcnt_t n = mpz_sgn(get_mpz_t());
          if (n!=0)
              n = mpz_sizeinbase(get_mpz_t(), 2);
          return n;
      };
};

int main (void)
{
      mpz_class p;
      myz_class y;
      p = 0;
// y = 0; // does not compile, error: no match for 'operator='
      for (int i=0; i<5; ++i)
          {
          cout
              << "# p="<< p
              << " abs(p)="<< abs(p)
              << " y="<< y
              << " abs(y)="<< abs(y)
              << " bitlen(y)="<< y.bitlen()
              << "\n";
          ++p;
          ++y; // works!
          }
      return 0;
}

I have two problems:

1) it appears that the overloaded operator = did not make it from
mpz_class to myz_class even though operator ++ did.


The problem is that the compiler automatically generates a copy assignment
operator for you, and it hides the operator= implementation(s) from the base class.


Ah that "copy assignment" gizmo makes = behave differently than ++

See the FAQ about such hiding.


I read<http://www.parashift.com/c++-faq-lite/assignment-operators.html#faq-12.4>
and only vaguely get what happens in the case of my code. Will learn.


That one is relevant yes.

What you're up against, though, is this:

<url: http://www.parashift.com/c++-faq-lite/strange-inheritance.html#faq-23.9>

A simple solution:

    using mpz_class::operator=;


I guess that you mean
      y mpz_class::operator= 0;


No, I meant what I wrote. Place that in your class definition.

2) short of using a macro, how do I change things in order to
write bitlen(y) much like I write abs(y)?


typedef ptrdiff_t Size;

inline Size bitlen( myz_class const& v )
{
      return v.bitlen();
}


At least I fully get that one. Many thanks.


You're welcome.

Cheers & hth.,

- Alf

PS: Kai-Uwe Bux, else-thread, managed to see the Big Picture. I'd follow that
advice! ;-)

--
blog at <url: http://alfps.wordpress.com>

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