Question

Does the “delete” statement below “doubly free” an object?

(...object_list is a global vector<object*>...)

vector< object * >::iterator     it, eit, iter;
object *p_object;
vector< object * >   dead_objects;

it  = object_list_.begin();
eit = object_list_.end();

//---collect pointers of all dead objects to dead_objects vector
for ( ; it != eit; it++ )
{
    p_object = *it;
    if ( p_object->is_dead() == false )
        continue;

    dead_objects.push_back( p_object );
}

//---free every dead object from the global object_list
for ( iter = dead_objects.begin(); iter != dead_objects.end(); iter++ )
{
    p_object = *iter;

    it  = object_list_.begin();
    eit = object_list_.end();

    for ( ; it != eit; it++ )
    {
        if ( *it != p_object )
            continue;

        object_list_.erase( it );
        delete p_object;
        break;
    }
}

I ask the question because the erase() statement above should have called the destructor of an object and freed it already, shouldn’t it?

Answer 1

erase() does call the destructor on the object, but the destructor of a pointer type (such as object * here) does nothing — it does NOT call delete on the pointer. If you want it to call delete, you need to use some object (such as auto_ptr<object *>) which does call delete.