Date: Wed, 10 Apr 2013 10:20:34 -0400 From: John Baldwin <jhb@freebsd.org> To: lev@freebsd.org Cc: Adrian Chadd <adrian@freebsd.org>, freebsd-current@freebsd.org, Poul-Henning Kamp <phk@phk.freebsd.dk> Subject: Re: Intel D2500CC motherboard and strange RS232/UART behavior Message-ID: <201304101020.34214.jhb@freebsd.org> In-Reply-To: <1659145198.20130410102838@serebryakov.spb.ru> References: <229402991.20130407172016@serebryakov.spb.ru> <201304091658.22810.jhb@freebsd.org> <1659145198.20130410102838@serebryakov.spb.ru>
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On Wednesday, April 10, 2013 2:28:38 am Lev Serebryakov wrote: > Hello, John. > You wrote 10 =E0=EF=F0=E5=EB=FF 2013 =E3., 0:58:22: >=20 > >> 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? They are all called in turn. > ISA interrupt sharing IS NOT so simple. sio contains a lot of > obscure code to work. INTR_FAST handlers in 4.x didn't use to allow sharing. That changed in 6.x or so. > JB> INTR_EXCL or some such and/or uart doesn't set RF_SHAREABLE when > JB> allocating the IRQ. It is probably the latter. You could try just > JB> adding RF_SHAREABLE to the bus_alloc_resource_any() for the IRQ to > JB> uart and see if that fixes it. > sc->sc_ires =3D bus_alloc_resource_any(dev, SYS_RES_IRQ, &sc->sc_= irid, > RF_ACTIVE | RF_SHAREABLE); >=20 > It is here. Ok, then you need to figure out what is actually failing to install an interrupt handler (e.g. does bus_alloc_resource or bus_setup_intr fail?) =2D-=20 John Baldwin
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