http://gcc.gnu.org/bugzilla/show_bug.cgi?id=56202
Manuel López-Ibáñez <manu at gcc dot gnu.org> changed:
What |Removed |Added
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CC| |manu at gcc dot gnu.org
--- Comment #2 from Manuel López-Ibáñez <manu at gcc dot gnu.org> 2013-02-04
15:46:39 UTC ---
Hum, this seems to be the second waiting method described in page 41 of
http://luc.devroye.org/chapter_ten.pdf
which specifically says that sum can be infinite. So I think it should be
enough to check that __t == __x and return __x in that case.
do
{
const double __e = -std::log(1.0 - __aurng());
if(__t == __x) return __x;
__sum += __e / (__t - __x);
__x += 1;
}
while (__sum <= _M_param._M_q);
No?
