Re: Exception while handling MEM Hole on OMAP3 / ARM Cortex A8

[Rabin Vincent <rabin@rab.in>]

On Mon, Aug 17, 2009 at 06:47:18PM +0100, Russell King - ARM Linux wrote:
> On Mon, Aug 17, 2009 at 11:03:44PM +0530, Rabin Vincent wrote:
> > On Sat, Aug 15, 2009 at 07:17:11PM +0100, Russell King - ARM Linux wrote:
> > > +#ifndef CONFIG_SPARSEMEM
> > > +int pfn_valid(unsigned long pfn)
> > > +{
> > > +	struct meminfo *mi = &meminfo;
> > > +	unsigned int mid, left = 0, right = mi->nr_banks;
> > > +
> > > +	while ((mid = (right - left) / 2) > 0) {
> > > +		struct membank *bank = &mi->bank[mid];
> > > +
> > > +		if (pfn < bank_pfn_start(bank))
> > > +			right = mid;
> > > +		else if (pfn >= bank_pfn_end(bank))
> > > +			left = mid + 1;
> > > +		else
> > > +			return 1;
> > > +	}
> > > +	return 0;
> > > +}
> > > +EXPORT_SYMBOL(pfn_valid);
> > > +#endif
> > 
> > The above function will ignore the first bank.  I don't think that was
> > intended.
> 
> Why do you say that?
> 
> Let's start with 4 membank entries - 0 1 2 3.
> 
> 0 covers 0-10, 1 covers 20-30, 2 covers 40-50 and 3 covers 60-70.
> 
> Let's run through the steps looking for 5:
> 
> Initally, left = 0, right = 4, so mid=2.  5 < 40?  Yes, so right becomes 2.
> 
> left = 0, right = 2, so mid=1.  5 < 20?  Yes, so right becomes 1.
> 
> left = 0, right = 1, so mid=0.  5 < 0?  No.  5 >= 10?  No.  so return 1.
> 
> This doesn't look to me like it ignores the first bank.  I think you're
> wrong.  Please provide me with your proof ASAP because right now it's
> pending for Linus.

When mid == 0, the loop exits because the condition you have is :

  while ((mid = (right - left) / 2) > 0) {
                                  ^^^^^^

Thus bank[0] will never be checked.

Rabin