Date: Mon, 19 Aug 2013 19:49:04 -0700 From: Adrian Chadd <adrian@freebsd.org> To: Vitja Makarov <vitja.makarov@gmail.com> Cc: freebsd-current <freebsd-current@freebsd.org> Subject: Re: Question about socket timeouts Message-ID: <CAJ-VmokxJvSK1JjEyZ4s9LeX25Lag0ET9nq3bEebM89xOsK1BA@mail.gmail.com> In-Reply-To: <CAKGHGPS=HCYfXxPXuUz5G83j5sGieejPU-QHmi9TrmMhmweHLw@mail.gmail.com> References: <CAKGHGPS=HCYfXxPXuUz5G83j5sGieejPU-QHmi9TrmMhmweHLw@mail.gmail.com>
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Yes! Please file a PR! -adrian On 19 August 2013 12:33, Vitja Makarov <vitja.makarov@gmail.com> wrote: > Hi! > > Recently I was playing with small socket timeouts. setsockopt(2) > SO_RCVTIMEO and found a problem with it: if timeout is small enough > read(2) may return before timeout is actually expired. > > I was unable to reproduce this on linux box. > > I found that kernel uses a timer with 1/HZ precision so it converts > time in microseconds to ticks that's ok linux does it as well. The > problem is in details: freebsd uses floor() approach while linux uses > ceil(): > > from FreeBSD's sys/kern/uipc_socket.c: > val = (u_long)(tv.tv_sec * hz) + tv.tv_usec / tick; > if (val == 0 && tv.tv_usec != 0) > val = 1; /* at least one tick if tv > 0 */ > > from Linux's net/core/sock.c: > *timeo_p = tv.tv_sec*HZ + (tv.tv_usec+(1000000/HZ-1))/(1000000/HZ); > > So, for instance, we have a freebsd system running with kern.hz set > 100 and set receive timeout to 25ms that is converted to 2 ticks which > is 20ms. In my test program read(2) returns with EAGAIN set in > 0.019ms. > > So the question is: is that a problem or not? > > -- > vitja. > _______________________________________________ > freebsd-current@freebsd.org mailing list > http://lists.freebsd.org/mailman/listinfo/freebsd-current > To unsubscribe, send any mail to "freebsd-current-unsubscribe@freebsd.org" >
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