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Added warning note about equivalence inconsistence between Notify and tuples.

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Daniele Varrazzo 2010-11-01 23:10:20 +00:00
parent 77c0ab02d8
commit 2480f587e1
1 changed files with 18 additions and 0 deletions
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18
psycopg/notify_type.c
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@ -147,6 +147,24 @@ notify_astuple(NotifyObject *self, int with_payload)
return tself;
}
/* note on Notify-tuple comparison.
*
* Such a comparison is required otherwise a check n == (pid, channel)
* would fail. We also want to compare two notifies, and the obvious meaning is
* "check that all the attributes are equal". Unfortunately this leads to an
* inconsistent situation:
* Notify(pid, channel, payload1)
* == (pid, channel)
* == Notify(pid, channel, payload2)
* even when payload1 != payload2. We can probably live with that, but hashing
* makes things worse: hashability is a desirable property for a Notify, and
* to maintain compatibility we should put a notify object in the same bucket
* of a 2-item tuples... but we can't put all the payloads with the same
* (pid, channel) in the same bucket: it would be an extremely poor hash.
* So we maintain compatibility in the sense that notify without payload
* behave as 2-item tuples in term of hashability, but if a payload is present
* the (pid, channel) pair is no more equivalent as dict key to the Notify.
*/
static PyObject *
notify_richcompare(NotifyObject *self, PyObject *other, int op)
{