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SubjectRe: [PATCH] mm,x86: fix span coverage in e820_all_mapped()
On 2013/12/11 5:06, Yinghai Lu wrote:

> On Tue, Dec 10, 2013 at 1:06 AM, Xishi Qiu <qiuxishi@huawei.com> wrote:
>> In the following case, e820_all_mapped() will return 1.
>> A < start < B-1 and B < end < C, it means <start, end> spans two regions.
>> <start, end>: [start - end]
>> e820 addr: ...[A - B-1][B - C]...
>
> should be [start, end) right?
> and
> [A, B),[B, C)
>

Hi Yinghai,

It is right, in this case the function will return 1.

>>
>> Signed-off-by: Xishi Qiu <qiuxishi@huawei.com>
>> ---
>> arch/x86/kernel/e820.c | 15 +++------------
>> 1 files changed, 3 insertions(+), 12 deletions(-)
>>
>> diff --git a/arch/x86/kernel/e820.c b/arch/x86/kernel/e820.c
>> index 174da5f..31ecab2 100644
>> --- a/arch/x86/kernel/e820.c
>> +++ b/arch/x86/kernel/e820.c
>> @@ -85,20 +85,11 @@ int __init e820_all_mapped(u64 start, u64 end, unsigned type)
>>
>> if (type && ei->type != type)
>> continue;
>> - /* is the region (part) in overlap with the current region ?*/
>> + /* is the region (part) in overlap with the current region ? */
>> if (ei->addr >= end || ei->addr + ei->size <= start)
>> continue;
>> -
>> - /* if the region is at the beginning of <start,end> we move
>> - * start to the end of the region since it's ok until there
>> - */
>> - if (ei->addr <= start)
>> - start = ei->addr + ei->size;
>
> so in your case new start will be B ?
>
> next run will be C
>
>> - /*
>> - * if start is now at or beyond end, we're done, full
>> - * coverage
>> - */
>> - if (start >= end)
>
>
>> + /* is the region full coverage of <start, end> ? */
>> + if (ei->addr <= start && ei->addr + ei->size >= end)
>> return 1;
>> }
>> return 0;
>
> also e820 should be sanitized already to have [A,C).
>

Yes, it should be sanitized already, but maybe someone will change the e820
to support some feature, so this function will be a potential bomb.

> or you are talking about [A,B), [B+1, C)
> first run start will be B, and next run with [B+1, ...), that will be
> skipped...
> will not return 1.
>
> so old code should be ok.
>

In this case, old code is right, but I discuss in another one that
you wrote above.

Thanks,
Xishi Qiu

> Thanks
>
> Yinghai
>
> .
>





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