linux-kernel - Re: [PATCH v2] PCI: endpoint: Skip odd BAR when skipping 64bit BAR

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Message-ID: <CABEDWGyQHh=8fmZFAK6acecDSF_-pfpa9NCZOQ--WDhZiZPihw@mail.gmail.com>
Date:   Fri, 31 May 2019 09:52:31 -0700
From:   Alan Mikhak <alan.mikhak@...ive.com>
To:     Kishon Vijay Abraham I <kishon@...com>
Cc:     linux-pci@...r.kernel.org, linux-kernel@...r.kernel.org,
        lorenzo.pieralisi@....com, linux-riscv@...ts.infradead.org,
        Palmer Dabbelt <palmer@...ive.com>,
        Paul Walmsley <paul.walmsley@...ive.com>,
        Bjorn Helgaas <bhelgaas@...gle.com>,
        Gustavo Pimentel <gustavo.pimentel@...opsys.com>,
        wen.yang99@....com.cn, kjlu@....edu
Subject: Re: [PATCH v2] PCI: endpoint: Skip odd BAR when skipping 64bit BAR

On Thu, May 30, 2019 at 9:37 PM Kishon Vijay Abraham I <kishon@...com> wrote:
>
> Hi Alan,
>
> On 25/05/19 12:20 AM, Alan Mikhak wrote:
> > Hi Kishon,
> >
> > Yes. This change is still applicable even when the platform specifies
> > that it only supports 64-bit BARs by setting the bar_fixed_64bit
> > member of epc_features.
> >
> > The issue being fixed is this: If the 'continue' statement is executed
> > within the loop, the loop index 'bar' needs to advanced by two, not
> > one, when the BAR is 64-bit. Otherwise the next loop iteration will be
> > on an odd BAR which doesn't exist.
>
> IIUC you are fixing the case where the BAR is "reserved" (specified in
> epc_features) and is also a 64-bit BAR?

Correct. If BAR0 is specified in epc_features as reserved and also
64-bit, the loop would skip BAR0 by executing the 'continue'
statement. In this scenario, BAR1 doesn't exist since the 64-bit BAR0
spans the two 32-bit BAR0 and BAR1. The loop index 'bar' would be
advanced by 2 in this case so on the next iteration the loop would
process BAR2.

> If 2 consecutive BARs are marked as reserved in reserved_bar of epc_features,
> the result should be the same right?

If both BAR0 and BAR2 are reserved and also 64-bit, the loop would
check BAR0 on its first iteration and skip BAR0 and BAR1, check BAR2
on its second iteration and skip BAR2 and BAR3, then check BAR4 on its
third iteration. If BAR4 is also 64-bit and reserved, the loop would
skip BAR4 and BAR5 and that would be the final iteration of the loop.

Regards,
Alan