Re: Undefined reference to...

From:
Bart van Ingen Schenau <bart@ingen.ddns.info>
Newsgroups:
comp.lang.c++
Date:
Thu, 11 Nov 2010 02:05:05 -0800 (PST)
Message-ID:
<e3bf1417-3266-4438-afe5-6ecd367e00a8@p1g2000yqm.googlegroups.com>
On Nov 11, 10:22 am, Andrea Crotti <andrea.crott...@gmail.com> wrote:

"Alf P. Steinbach /Usenet" <alf.p.steinbach+use...@gmail.com> writes:

Problems with the above include

  * 'getLower' attempts to return an object of abstract class.

  * A local variable is declared with incomplete type 'Extended'.


What do you incomplete type?
Because I had the forward declaration maybe?


No, the type Extended was incomplete at the source-line
          Extended e;

because only the forward declaration had been seen by the compiler and
not yet the complete definition.
A C++ compiler only does a very limited amount of look-ahead.
Basically, the only things you can refer to, which come later in the
source file, are members of a class while you are in the body of a
member-function of that same class.

  * If those obstacles weren't in the way, 'getLower' would perform
    a *slicing*, returning only the Base part of 'e', and then of t=

ype

    'Base'.

It's not clear what you're attempting, but it may be that you want a
Meyers' singleton:

  class Base
  {
  public:
      virtual void printOut() = 0;
      static Base& getLower();
  };

  class Extended: public Base
  {
  public:
      void printOut() { cout << "hello"; }
  };

  Base& Base::getLower()
  {
      static Extended e;
      // Whatever code you were intending to have here, then:
      return e;
  }

Note that with this scheme you always get the same object -- a si=

ngleton --

from 'getLower'.

Cheers & hth.,

- Alf


Anyway thanks but no, I don't think that's what I need, the simplest
solution is just to avoid the pure function so I don't have an abstract
class anymore and I can actually return something of that type.


You should be careful with the possibility of slicing.
This code

#include <iostream>
using namespace std;

class Base
{
public:
     virtual void printOut() { cout<< "Base" << endl; }
     static Base getLower();
};

class Extended : public Base
{
public:
     void printOut() { cout<< "Extended" << endl; }
};

static Base Base::getLower() {
     Extended e;
     return e;
}

int main()
{
    Base::getLower().printOut();
}

is guaranteed to always print "Base".
The Base::getLower function returns an object of type Base, not of
some derived type.

The other solution I guess is to use a pointer as Paul suggested.
Thanks


Or to use a smart pointer (like auto_ptr) as James suggested.

Bart v Ingen Schenau

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