On Wed, Jun 14, 2017 at 11:42:30AM +0100, Mark Rutland wrote: > On Wed, Jun 14, 2017 at 11:06:58AM +0100, Will Deacon wrote: > > Apologies, I misunderstood your algorithm (I thought step (a) was on one CPU > > and step (b) was on another). Still, I don't understand the need for the > > timeout. If you instead read back the flag immediately, wouldn't it still > > work? e.g. > > > > > > lock: > > Readl_relaxed flag > > if (locked) > > goto lock; > > > > Writel_relaxed unique ID to flag > > Readl flag > > if (locked by somebody else) > > goto lock; > > > > <critical section> > > > > unlock: > > Writel unlocked value to flag > > > > > > Given that we're dealing with iomem, I think it will work, but I could be > > missing something obvious. > > Don't we have the race below where both threads can enter the critical > section? > > // flag f initial zero (unlocked) > > // t1, flag 1 // t2, flag 2 > readl(f); // reads 0 l = readl(f); // reads 0 > > <thinks lock is free> <thinks lock is free> > > writel(1, f); > readl(f); // reads 1 > <thinks lock owned> > writel(2, f); > readl(f) // reads 2 > <thinks lock owned> > > <crticial section> <critical section>
Urgh, yeah, of course and *that's* what the udelay is trying to avoid, by "ensuring" that the <thinks lock is free> time and subsequent write propagation is all over before we re-read the flag. John -- how much space do you have on this device? Do you have, e.g. a byte for each CPU? Will
