[Monetdb-developers] what should happen when appending to a nil bat?

Stefan Manegold Stefan.Manegold at cwi.nl
Wed Dec 21 13:33:00 CET 2005


I also vote for the second alternative.

If/where desired the first result can then achieved by 

	a.append(b.mark(nil));

Stefan

On Wed, Dec 21, 2005 at 01:26:08PM +0100, Ying Zhang wrote:
> To me, the second possibility seems better, since the tail value of 'b'
> is preserved.
> 
> But how about a third possibility: reject such 'append' and require a
> explicit type casting?
> 
> 
> Regards,
> 
> Jennie
> 
> On Wed, Dec 21, 2005 at 01:08:26PM +0100, Sjoerd Mullender wrote:
> > Consider this scenario:
> > mil>var a := new(oid,void).insert(0 at 0,nil);
> > mil>a.print();
> > #-----------------#
> > # h     t         # name
> > # oid   void      # type
> > #-----------------#
> > [ 0 at 0,    nil     ]
> > mil>var b := new(oid,oid).insert(0 at 0,1 at 0);
> > mil>b.print();
> > #-----------------#
> > # h     t         # name
> > # oid   oid       # type
> > #-----------------#
> > [ 0 at 0,    1 at 0     ]
> > mil>a.append(b);
> > mil>a.print();
> > 
> > The question is, what should the result be?
> > 
> > As I see it, there are two possibilities:
> > #-----------------#
> > # h     t         # name
> > # oid   void      # type
> > #-----------------#
> > [ 0 at 0,    nil     ]
> > [ 1 at 0,    nil     ]
> > or
> > #-----------------#
> > # h     t         # name
> > # oid   oid       # type
> > #-----------------#
> > [ 0 at 0,    nil     ]
> > [ 1 at 0,    1 at 0     ]
> > 
> > The former keeps the nil tail (the tail of a was void without seqbase
> > and stays that way), whereas the latter materializes the tail so that
> > the tail value of b gets inserted.
> > 
> > Opinions?
> > 
> > -- 
> > Sjoerd Mullender
> 
> 
> 
> 
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