On 14/06/2017 12:01, Will Deacon wrote:
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?
Hi Will,
To be clear, the agents in our case are the kernel and UEFI. Within the
kernel, we use a kernel spinlock to lock the same djtag between threads,
for these reasons:
- kernel has a native spinlock
- we are limited in locking values, as the lock flag is only a 8b field
in v2 hw (called module select)
Thanks,
John
Will
.
