c46ecbd2   
   From: ulrich.eckhardt@dominolaser.com   
      
   Iker Jamardo Zugaza wrote:   
   >>  SPFooInherited pfi;   
   >>  SPFoo& pf = pfi;   
   >>   
   >> This of course doesn't compile, because the two are not related.   
   >   
   > This code compiles perfectly. While creating the new pf reference, the   
   > smart pointer to Foo is able to check that the conversion from   
   > FooInherited is completely valid:   
   >   
   > FooInherited* fi;   
   > Foo* pf = fi;   
   >   
   > This is what is done internally in the shared_ptr class (more or less   
   > with some reference counter handling and such ;)).   
      
   You have missed an ampersand! The equivalent using raw pointers is not what   
   you write here but this:   
      
    FooInherited* fi;   
    Foo*& pf = fi;   
      
      
   The reason this doesn't compile, just in case it isn't clear, is that you   
   could now make 'pf' point to a 'Foo', which would in turn make 'fi' point to   
   a 'Foo', too, even though it promises to point to a 'FooInherited'.   
      
      
   Uli   
      
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