On 2018-12-01 09:42:38 [+0100], Arnd Bergmann wrote:
> You are right that you can't take (or release) a mutex from interrupt
> context. However, I don't think converting a spinlock to a semaphore
> is going to help here either.
you can acquire a semaphore with a try_lock from interrupts context
On Mon, Dec 3, 2018 at 12:18 PM Sebastian Andrzej Siewior
wrote:
>
> On 2018-12-01 09:42:38 [+0100], Arnd Bergmann wrote:
> > You are right that you can't take (or release) a mutex from interrupt
> > context. However, I don't think converting a spinlock to a semaphore
> > is going to help here
Commit-ID: 7d79adb87fa79e4a4c59424fd5b5a922861fc5f6
Gitweb: https://git.kernel.org/tip/7d79adb87fa79e4a4c59424fd5b5a922861fc5f6
Author: Sebastian Andrzej Siewior
AuthorDate: Thu, 29 Nov 2018 16:02:10 +0100
Committer: Borislav Petkov
CommitDate: Mon, 3 Dec 2018 19:37:27 +0100
x86/fpu:
On Mon, 3 Dec 2018 at 22:04, tip-bot for Sebastian Andrzej Siewior
wrote:
>
> Commit-ID: 7d79adb87fa79e4a4c59424fd5b5a922861fc5f6
> Gitweb:
> https://git.kernel.org/tip/7d79adb87fa79e4a4c59424fd5b5a922861fc5f6
> Author: Sebastian Andrzej Siewior
> AuthorDate: Thu, 29 Nov 2018 16:02:10
On Mon, Dec 03, 2018 at 10:12:19PM +0100, Ard Biesheuvel wrote:
> > + * Using the FPU in hardirq is not allowed.
>
> According to the documentation in x86/kernel/fpu/core.c, this is not
> true. So which one is accurate?
I think you mean the irq from user mode... Yap, we do allow that.
On Mon, Dec 3, 2018 at 8:26 AM Arnd Bergmann wrote:
>
> On Mon, Dec 3, 2018 at 12:18 PM Sebastian Andrzej Siewior
> wrote:
> >
> > On 2018-12-01 09:42:38 [+0100], Arnd Bergmann wrote:
> > > You are right that you can't take (or release) a mutex from interrupt
> > > context. However, I don't
