Date: Tue, 28 Nov 2000 19:10:29 -0600 (CST) From: Mike Meyer <mwm@mired.org> To: "Brian F. Feldman" <green@FreeBSD.org> Cc: cvs-all@freebsd.org Subject: Re: cvs commit: src/usr.sbin/inetd builtins.c Message-ID: <14884.22405.22813.41567@guru.mired.org> In-Reply-To: <80189770@toto.iv>
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Brian F. Feldman <green@FreeBSD.org> types:
> Warner Losh <imp@village.org> wrote:
> > In message <20001126182240.A8051@fw.wintelcom.net> Alfred Perlstein writes:
> > : - if ((to.tv_usec += tv.tv_usec) >= 1000000) {
> > : + to.tv_usec += tv.tv_usec;
> > : + if (to.tv_usec >= 1000000) {
> > : to.tv_usec -= 1000000;
> > : to.tv_sec++;
> > : }
> >
> > Shouldn't this be
> > if (to.tv_usec >= 1000000) {
> > to.tv_sec += to.tv_usec / 1000000;
> > to.tv_usec %= 1000000;
> > }
>
> I asked Poul-Henning the same thing with regard to similar checks in the
> kernel code. The reply was that, although this is the "most correct" way of
> doing things, a properly done calculation of to.tv_usec + tv.tv_usec in
> this (and many other) case(s) will result in a maximum of 1999998, so at
> most one subtraction and increment will be necessary.
>
> If it's not possible to have a higher tv_usec, you save the REALLY expensive
> divide-family instructions.
Does it make sense to be paranoid about this, and do:
while (to.tv_usec >= 1000000) {
to.tv_usec -= 1000000;
to.tv_sec++;
}
Which costs one extra compare in the normal case, but insures that the
value is right if the assumption happens to fail (but will do so
slowly)?
<mike
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