Re: [PATCH v2 1/3] platform/x86: intel_scu_ipc: Check status after timeout in busy_loop()

[Stephen Boyd]

Quoting Kuppuswamy Sathyanarayanan (2023-09-06 13:20:49)
> On 9/6/2023 1:14 PM, Stephen Boyd wrote:
> > Quoting Andy Shevchenko (2023-09-06 13:04:54)
> >> On Wed, Sep 06, 2023 at 11:09:41AM -0700, Stephen Boyd wrote:
> >>>               status = ipc_read_status(scu);
> >>>               if (!(status & IPC_STATUS_BUSY))
> >>
> >>> -                     return (status & IPC_STATUS_ERR) ? -EIO : 0;
> >>> +                     goto not_busy;
> >>
> >> Wouldn't simple 'break' suffice here?
> >
> > Yes, at the cost of reading the status again when it isn't busy, or
> > checking the busy bit after the loop breaks out and reading it once
> > again when it is busy. I suppose the compiler would figure that out and
> > optimize so that break would simply goto the return statement.
> >
> > The code could look like this without a goto.
> >
> >       do {
> >               status = ipc_read_status(scu);
> >               if (!(status & IPC_STATUS_BUSY))
> >                       break;
> >       } while (time_before(jiffies, end));
> >
> >       if (status & IPC_STATUS_BUSY)
> >               status = ipc_read_status(scu);
>
> IMO, you can remove the if condition and read again the status in all cases.
> It is more readable. But it is up to you.
>

I don't really care either way. Just let me know what makes the
maintainers happy here.