On Wed, Feb 1, 2017 at 12:19 PM, Arnd Bergmann <arnd@xxxxxxxx> wrote:
> On Saturday, January 28, 2017 9:48:36 PM CET Linus Walleij wrote:
>> + interrupt-map-mask = <0xff00 0 0 7>;
>> + interrupt-map = <0x4800 0 0 1 &pci_intc 0>, /* Slot 9 */
>> + <0x4900 0 0 2 &pci_intc 1>,
>> + <0x4a00 0 0 3 &pci_intc 2>,
>> + <0x4b00 0 0 4 &pci_intc 3>,
>> + <0x5000 0 0 1 &pci_intc 0>, /* Slot 10 */
>> + <0x5100 0 0 2 &pci_intc 1>,
>> + <0x5200 0 0 3 &pci_intc 2>,
>> + <0x5300 0 0 4 &pci_intc 3>,
>> + <0x5800 0 0 1 &pci_intc 0>, /* Slot 11 */
>> + <0x5900 0 0 2 &pci_intc 1>,
>> + <0x5a00 0 0 3 &pci_intc 2>,
>> + <0x5b00 0 0 4 &pci_intc 3>,
>> + <0x6000 0 0 1 &pci_intc 0>, /* Slot 12 */
>> + <0x6100 0 0 2 &pci_intc 1>,
>> + <0x6200 0 0 3 &pci_intc 2>,
>> + <0x6300 0 0 4 &pci_intc 3>;
>>
>
> The mapping looks wrong here, we normally don't list interrupts per function
> so the mask should be 0xf800.
Yup it works that way too, and indeed the USB hub on function
2 was requesting the right pin and everything.
> Note that the interrupt map is board specific, so this should probably
> go in the board.dts file rather than platform.dtsi.
OK moving it down there.
> For this particular board, the interrupt lines appear to have been badly
> configured so all slots use the same interrupt 0 for IntA. IIRC This also
> means you can probably use <0 0 0 7> as the mask and just specify each of
> the four interrupts once. A properly wired board would swizzle the
> interrupts so that each slot has a different IRQ for its IntA line.
They are swizzled, I just have very bad hardware docs. (Only
code.) It turns out when I'm browsing through old board support
code that there is a comment followed by a piece of code like this:
/*
* No swizzle on SL2312
*/
static u8 __init sl2312_pci_swizzle(struct pci_dev *dev, u8 *pinp)
{
return PCI_SLOT(dev->devfn);
}
/*
* map the specified device/slot/pin to an IRQ. This works out such
* that slot 9 pin 1 is INT0, pin 2 is INT1, and slot 10 pin 1 is INT1.
*/
static int __init sl2312_pci_map_irq(struct pci_dev *dev, u8 slot, u8 pin)
{
int intnr = ((slot + (pin - 1)) & 3) + 4; /* the IRQ number
of PCI bridge */
printk("%s : slot = %d pin = %d \n",__func__,slot,pin);
switch (slot)
{
case 12:
if (pin==1)
{
intnr = 3;
}
else
{
intnr = 0;
}
#ifdef CONFIG_DUAL_PCI
return IRQ_PCI_INTD;
#endif
break;
case 11:
intnr = (2 + (pin - 1)) & 3;
#ifdef CONFIG_DUAL_PCI
return IRQ_PCI_INTC;
#endif
break;
case 10:
intnr = (1 + (pin - 1)) & 3;
#ifdef CONFIG_DUAL_PCI
return IRQ_PCI_INTB;
#endif
break;
case 9:
intnr = (pin - 1) & 3;
break;
}
// if (slot == 10)
// intnr = (1 + (pin - 1)) & 3;
// else if (slot == 9)
// intnr = (pin - 1) & 3;
return (IRQ_PCI_INTA + intnr);
}
If I understand correctly they say that the IRQs are not swizzled
on the PCI bridge side but on the IRQ handler side.
So if I put it in the device tree like so:
+ interrupt-map =
+ <0x4800 0 0 1 &pci_intc 0>, /* Slot 9 */
+ <0x4800 0 0 2 &pci_intc 1>,
+ <0x4800 0 0 3 &pci_intc 2>,
+ <0x4800 0 0 4 &pci_intc 3>,
+ <0x5000 0 0 1 &pci_intc 1>, /* Slot 10 */
+ <0x5000 0 0 2 &pci_intc 2>,
+ <0x5000 0 0 3 &pci_intc 3>,
+ <0x5000 0 0 4 &pci_intc 0>,
+ <0x5800 0 0 1 &pci_intc 2>, /* Slot 11 */
+ <0x5800 0 0 2 &pci_intc 3>,
+ <0x5800 0 0 3 &pci_intc 0>,
+ <0x5800 0 0 4 &pci_intc 1>,
+ <0x6000 0 0 1 &pci_intc 3>, /* Slot 12 */
+ <0x6000 0 0 2 &pci_intc 0>,
+ <0x6000 0 0 3 &pci_intc 1>,
+ <0x6000 0 0 4 &pci_intc 2>;
So the IRQs on the right side are swizzled instead
of the pins being swizzled.
...but that looks a bit insane.
Isn't that exactly the same thing just exposed in some
inverse way?
I'll try to get the RALink MiniPCI I have ín the slot going and
see if it works the same with just good old vanilla
swizzling.
Yours,
Linus Walleij
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