[PATCH v3 04/15] watchdog: orion: Handle IRQ

On Sun, Jan 26, 2014 at 10:42:19PM -0800, Guenter Roeck wrote:

On 01/26/2014 10:30 PM, Ezequiel Garcia wrote:

quoted

Hi Guenter,

On Tue, Jan 21, 2014 at 06:31:02AM -0800, Guenter Roeck wrote:
[..]

quoted

quoted

@@ -131,6 +138,19 @@ static int orion_wdt_probe(struct platform_device *pdev)
   	if (!wdt_reg)
   		return -ENOMEM;

+	irq = platform_get_irq(pdev, 0);
+	if (irq > 0) {

0 is a valid interrupt number, and platform_get_irq returns an error code on errors.
Should be >= 0.

I'm revisiting this one. I believe this is not the hardware interrupt
number, but the one mapped into virq space. So, 0 is not a valid
interrupt number.

Right?

If so, the entire interrupt numbering scheme appears broken. Conceptually it should
not make a difference where the interrupt is coming from. If the virq system
returns 0 for invalid (non-configured) interrupts, and non-configured 'real'
interrupts are reported as -ENXIO, all bets are off. How would a driver know
what to expect ? And how would one be expected to review such non-deterministic
code ?

I wouldn't know that much. I'm just pointing out that '0' doesn't seem
to be a valid IRQ number in this context (i.e. virq space).

FWIW, platform_get_irq() does return -ENXIO for invalid interrupts. If there
is an independent notion of "0 is an invalid interrupt", it is well hidden.

Yes, AFAICS platform_get_irq() returns a negative error or a positive
irq number. I fail to see how it's able return '0' (of course, I can be
wrong).

Anyway, if you think the driver should treat 0 as invalid interrupt, go ahead.
Who am I to know. Just please don't use my Reviewed-by in this case.

Quite frankly not sure how to handle this best. A quick grep through the
code doesn't help either: lots of drivers treat 0 as a valid interrupt
and lots treat it as invalid. So I guess it doesn't really matter...

Can someone shed a light on this?
-- 
Ezequiel Garc?a, Free Electrons
Embedded Linux, Kernel and Android Engineering
http://free-electrons.com