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Subject: Re: struct bintime
From: Dimitry Andric <dim@FreeBSD.org>
In-Reply-To: <CAGSa5y2voSswSV4XFVR8+OqdftsfPWNSHYiptp-BMJ_hSp5u2A@mail.gmail.com>
Date: Fri, 17 Oct 2014 00:48:14 +0200
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To: Jeremie Le Hen <jlh@FreeBSD.org>
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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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