Date: Fri, 17 Oct 2014 00:48:14 +0200 From: Dimitry Andric <dim@FreeBSD.org> To: Jeremie Le Hen <jlh@FreeBSD.org> Cc: freebsd-hackers@FreeBSD.org Subject: Re: struct bintime Message-ID: <353C79A7-1D1D-48CF-8B42-7BA369D22321@FreeBSD.org> In-Reply-To: <CAGSa5y2voSswSV4XFVR8%2BOqdftsfPWNSHYiptp-BMJ_hSp5u2A@mail.gmail.com> References: <CAGSa5y2voSswSV4XFVR8%2BOqdftsfPWNSHYiptp-BMJ_hSp5u2A@mail.gmail.com>
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On 17 Oct 2014, at 00:15, Jeremie Le Hen <jlh@FreeBSD.org> wrote:
> I need to get microseconds from a struct bintime. I found
> bintime2timeval() in sys/time.h which more or less does this, but I
> don't understand how the computation works.
>=20
> Can someone explain it to me please?
>=20
> static __inline void
> bintime2timeval(const struct bintime *_bt, struct timeval *_tv)
> {
>=20
> _tv->tv_sec =3D _bt->sec;
> _tv->tv_usec =3D ((uint64_t)1000000 * (uint32_t)(_bt->frac >> =
32)) >> 32;
> }
Maybe it's easier to grok if you read it as:
usec =3D (10^6 * (frac / 2^32)) / 2^32
or even:
usec =3D 10^6 * (frac / 2^64)
and the other way around:
frac =3D (usec / 10^6) * 2^64
E.g., bintime divides a second into 2^64 slices, instead of 10^6 or
10^9. And the division by 2^64 in bintime2timeval() is split up in two
shifts, to avoid losing precision.
-Dimitry
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