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Home/ Questions/Q 6556417
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Editorial Team
  • 0
Editorial Team
Asked: May 25, 20262026-05-25T12:56:33+00:00 2026-05-25T12:56:33+00:00

In relation to question reference to abstract class I wrote the following example: #include

  • 0

In relation to question “reference to abstract class” I wrote the following example:

#include <iostream>
#include <vector>

class data
{
  public:
    virtual int get_value() = 0;
};

class data_impl : public data
{
  public:
    data_impl( int value_ ) : value( value_ ) {}
    virtual int get_value() { return value; }
  private:
    int value;
};

class database
{
  public:
    data& get( int index ) { return *v[index]; }
    void  add( data* d ) { v.push_back( d ); }
  private:
    std::vector< data* > v;
};

int main()
{
    data_impl d1( 3 );
    data_impl d2( 7 );

    database db;
    db.add( &d1 );
    db.add( &d2 );

    data& d = db.get( 0 );
    std::cout << d.get_value() << std::endl;
    d = db.get( 1 );
    std::cout << d.get_value() << std::endl;

    data& d_ = db.get( 1 );
    std::cout << d_.get_value() << std::endl;
    d_ = db.get( 0 );
    std::cout << d_.get_value() << std::endl;

    return 0;
}

To my surprise, the example prints:

3
3
7
7

and it looks like the reference assignment does the work differently from my expectation. I would expect:

3
7
7
3

Could you please point what my mistake is?

Thank you!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T12:56:33+00:00Added an answer on May 25, 2026 at 12:56 pm

    In

    data& d = db.get( 0 );
    std::cout << d.get_value() << std::endl;
    d = db.get( 1 );
    std::cout << d.get_value() << std::endl;
    

    the first statement is a reference initialization, while the third statement is a slicing assignment.

    You can not re-seat a reference.

    Cheers & hth.,

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