On Wed, Dec 15, 2004 at 12:38:24AM +0100, Stefan Manegold wrote:
use bat id's instead (stored in a oid bat for example).
niels
Hi MonetDB developers and users,
sorry for this "blunt" posting, but I thought, the problem, respectively the solution/recommendation that will hopefully be initiated by this question, might be interesting for some more of you:
Triggered by a question from Agustin, I noticed that I (believe it or not ;-) apparently lack some knowledge of / experiences with MonetDB/MIL:
Consider you have a variable "a" holding a (possible transient) BAT, and another variable "b" holding a second BAT, which happens to be a view of the first one. Both BATs haven't been baptized ("rename()"), yet; hence they still hold there default names:
var a := new(void,int).seqbase(0@0).insert(nil,1).insert(nil,2).access(BAT_READ); var b := reverse(a); print(a); #-----------------# # h tmp_29 # name # void int # type #-----------------# [ 0@0, 1 ] [ 1@0, 2 ] print(b); #-----------------# # t tmp_29 # name # int void # type #-----------------# [ 1, 0@0 ] [ 2, 1@0 ]
Then, these BATs should be gathered in a "container"-BAT. Since (persistent) nested BATs have disappeared (for good reasons!), we do this by gathering the BAT's name in the container:
var c := new(void,str).seqbase(10@0); c.insert(nil,bbpname(a)).insert(nil,bbpname(b)); print(c); #-------------------------# # h tmp_28 # name # void str # type #-------------------------# [ 10@0, "tmp_29" ] [ 11@0, "tmp_29" ]
Since the second BAT is only a view of the first one, they share not only the same memory/storage, but also the same (bbp-)name; hence, they are not distinguishable by (only) their (bbp-)names, and thus the container c actually contains the first BAT twice iso each BAT once:
print(bat(fetch(c,0))); #-----------------# # h tmp_29 # name # void int # type #-----------------# [ 0@0, 1 ] [ 1@0, 2 ] print(bat(fetch(c,1))); #-----------------# # h tmp_29 # name # void int # type #-----------------# [ 0@0, 1 ] [ 1@0, 2 ] print(bbpname(a)); [ "tmp_29" ] print(bbpname(b)); [ "tmp_29" ]
Obviously, the problem can be solved/avoided by baptizing the two BATs, and hence making them unique "individuals":
rename(a,"BAT_a"); rename(b,"BAT_b"); print(bbpname(a)); [ "BAT_a" ] print(bbpname(b)); [ "BAT_b" ] c.insert(nil,bbpname(a)).insert(nil,bbpname(b)); print(c); #-------------------------# # h tmp_28 # name # void str # type #-------------------------# [ 10@0, "tmp_29" ] [ 11@0, "tmp_29" ] [ 12@0, "BAT_a" ] [ 13@0, "BAT_b" ] print(bat(fetch(c,0))); !ERROR: interpret_params: print(param 1): invalid BAT. print(bat(fetch(c,1))); !ERROR: interpret_params: print(param 1): invalid BAT. print(bat(fetch(c,2))); #-----------------# # h BAT_b # name # void int # type #-----------------# [ 0@0, 1 ] [ 1@0, 2 ] print(bat(fetch(c,3))); #-----------------# # t BAT_b # name # int void # type #-----------------# [ 1, 0@0 ] [ 2, 1@0 ]
My question is now:
- Is this the only solution, or am I just not aware of another command iso bbpname() that would return unique "default" name for non-renamed view?
Of course, given the fact that only MonetDB itself has control over the default "tmp_*" names, it is always recommendable to assign names that you can control yourself, whenever you use a BAT's name as its only reference...
Stefan
-- | Dr. Stefan Manegold | mailto:Stefan.Manegold@cwi.nl | | CWI, P.O.Box 94079 | http://www.cwi.nl/~manegold/ | | 1090 GB Amsterdam | Tel.: +31 (20) 592-4212 | | The Netherlands | Fax : +31 (20) 592-4312 |
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