Date: Fri, 17 Oct 2014 00:51:19 +0200 From: Luigi Rizzo <rizzo@iet.unipi.it> To: Jeremie Le Hen <jlh@freebsd.org> Cc: freebsd-hackers@freebsd.org Subject: Re: struct bintime Message-ID: <20141016225119.GD10204@onelab2.iet.unipi.it> In-Reply-To: <CAGSa5y2voSswSV4XFVR8%2BOqdftsfPWNSHYiptp-BMJ_hSp5u2A@mail.gmail.com> References: <CAGSa5y2voSswSV4XFVR8%2BOqdftsfPWNSHYiptp-BMJ_hSp5u2A@mail.gmail.com>
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On Fri, Oct 17, 2014 at 12:15:57AM +0200, Jeremie Le Hen wrote:
> Hi,
>
> 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.
>
> Can someone explain it to me please?
>
> static __inline void
> bintime2timeval(const struct bintime *_bt, struct timeval *_tv)
> {
>
> _tv->tv_sec = _bt->sec;
> _tv->tv_usec = ((uint64_t)1000000 * (uint32_t)(_bt->frac >> 32)) >> 32;
> }
bt->frac has 64 bits representing the fractional part of the second,
call it f, with 0 <= f < 1
(uint32_t)(_bt->frac >> 32) is equivalent to \floor{f * 2^32}
you then multiply by 10^6 and do an integer division by 2^32 so
the overall expression is
tv_usec = \floor{ ( 10^6 * \floor{f * 2^32} ) / 2^32 } = \floor{ 10^6*f }
i.e. the number of microseconds.
cheers
luigi
> Thanks!
> --
> Jeremie Le Hen
> jlh@FreeBSD.org
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