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[Bug libstdc++/20758] operator-(const T&, const complex<T>&) vs operator-(const complex<T>&, const complex<T>&)
- From: "kreckel at ginac dot de" <gcc-bugzilla at gcc dot gnu dot org>
- To: gcc-bugs at gcc dot gnu dot org
- Date: 5 Apr 2005 14:01:35 -0000
- Subject: [Bug libstdc++/20758] operator-(const T&, const complex<T>&) vs operator-(const complex<T>&, const complex<T>&)
- References: <20050404213802.20758.kreckel@ginac.de>
- Reply-to: gcc-bugzilla at gcc dot gnu dot org
------- Additional Comments From kreckel at ginac dot de 2005-04-05 14:01 -------
(In reply to comment #7)
> (p.s., FWIW, I *think* log(a1) is the same for imag(a1) == -0 vs +0)
Huhh? Not if real(a1) is negative. The branch cut conventionally runs along
the negative real axis.
For instance, C99 specifies so. In 7.3.3.2 it also specifies that there it
should be continuous with the second quadrant: "implementations shall map a cut
so the function is continuous as the cut is approached coming around the finite
endpoint fo the cut in a counter clockwise direction." At least in the absence
of signs on zero.
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=20758