From: ulrich.eckhardt@dominolaser.com   
      
   Am 25.11.2011 08:25, schrieb Yordan Naydenov:   
   > On 2011-11-24 02:40, Lorenzo Caminiti wrote:   
   >>     ry* ryp = yp->f(); // calls y::f   
   >>     rx* rxp = xp->f(); // calls y::f but returns rx* instead of ry*   
   >   
   >         // Nope! calls y::f() [...] and interprets the memory layout   
   >         // the returned pointer pointed to as that of an object of   
   >         // type rx [...].   
      
   While the gist of this is true, there is one minor fault in your   
   statement. The memory does not get interpreted differently, but the   
   pointer to "y" is adjusted so that it points to the "x" base subobject.   
      
   The difference between the two memory addresses is zero in the example   
   Lorenzo gave, but if you imagine multiple nonempty baseclasses it   
   becomes clear that only one of the base subobjects can start at the same   
   address as the overall object.   
      
   The way you wrote it sounds a bit as if bad (implementation-defined or   
   undefined) things could happen, but all this works correctly without you   
   having to care for it.   
      
      
   Greetings!   
      
   Uli   
      
      
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