Aren't lseg_eq and lseg_ne broken? - Mailing list pgsql-hackers

From Tom Lane
Subject Aren't lseg_eq and lseg_ne broken?
Date
Msg-id 27545.1038592778@sss.pgh.pa.us
Whole thread Raw
Responses Re: Aren't lseg_eq and lseg_ne broken?  (Bruce Momjian <pgman@candle.pha.pa.us>)
List pgsql-hackers
By chance I just noticed that lseg equality is coded as

Datum
lseg_eq(PG_FUNCTION_ARGS)
{   LSEG       *l1 = PG_GETARG_LSEG_P(0);   LSEG       *l2 = PG_GETARG_LSEG_P(1);
   PG_RETURN_BOOL(FPeq(l1->p[0].x, l2->p[0].x) &&                  FPeq(l1->p[1].y, l2->p[1].y) &&
FPeq(l1->p[0].x,l2->p[0].x) &&                  FPeq(l1->p[1].y, l2->p[1].y));
 
}

Surely this should be
   PG_RETURN_BOOL(FPeq(l1->p[0].x, l2->p[0].x) &&                  FPeq(l1->p[0].y, l2->p[0].y) &&
FPeq(l1->p[1].x,l2->p[1].x) &&                  FPeq(l1->p[1].y, l2->p[1].y));
 

since I don't think I like this result:

regression=# select '[(0, 0), (1, 1)]'::lseg = '[(0, 42), (2, 1)]'::lseg;?column?
----------t
(1 row)

lseg_ne has the identical bug.

Checking the CVS archives, I see that this error dates back to the
original Berkeley code, so I'm a bit hesitant to just change it.
Is there any possibility that it really should work this way?
        regards, tom lane


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