Date: Wed, 10 Apr 2013 00:21:14 -0700 From: Adrian Chadd <adrian@freebsd.org> To: lev@freebsd.org Cc: Poul-Henning Kamp <phk@phk.freebsd.dk>, freebsd-current@freebsd.org Subject: Re: Intel D2500CC motherboard and strange RS232/UART behavior Message-ID: <CAJ-Vmon%2BCym=_h%2BRtC-HrUdcAgMQpn_RuFwE867-T4UTQ0jVXg@mail.gmail.com> In-Reply-To: <1659145198.20130410102838@serebryakov.spb.ru> References: <229402991.20130407172016@serebryakov.spb.ru> <201304091608.09257.jhb@freebsd.org> <105818341.20130410004451@serebryakov.spb.ru> <201304091658.22810.jhb@freebsd.org> <1659145198.20130410102838@serebryakov.spb.ru>
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On 9 April 2013 23:28, Lev Serebryakov <lev@freebsd.org> wrote: > Hello, John. > You wrote 10 =D0=B0=D0=BF=D1=80=D0=B5=D0=BB=D1=8F 2013 =D0=B3., 0:58:22: > >>> Problem is, that every uart device now is independent from each >>> other in good "OOP" style, and it looks like interrupt sharing we >>> need one interrupt handler per irq (not per device), which will now >>> about several UARTs. Something like "multiport" device, bot not >>> exactly. > JB> No, the interrupt code itself will handle shared interrupts (it will > JB> call all handlers). I think in practice that uart is setting > And what will happen, if there is two UARTs asserting interrupt in > same time? First one returns "FILTER_HANDLED", will second handler be > called? > > ISA interrupt sharing IS NOT so simple. sio contains a lot of > obscure code to work. .. surely it's solvable with a bit of ugliness? Eg, looping over them until they all return "not handled" or you hit a limit, or something equally ew. .. assuming that it is broken in the first place, that is. Adrian
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