linux-kernel - Re: [PATCH] mtd: nand: s3c2410: fix bug in s3c2410_nand_correct

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Message-ID: <57074392.2030601@gmail.com>
Date:	Fri, 8 Apr 2016 13:37:22 +0800
From:	Zeng Zhaoxiu <zhaoxiu.zeng@...il.com>
To:	Boris Brezillon <boris.brezillon@...e-electrons.com>
Cc:	zengzhaoxiu@....com, kgene@...nel.org, k.kozlowski@...sung.com,
	richard@....at, dwmw2@...radead.org, computersforpeace@...il.com,
	linux-arm-kernel@...ts.infradead.org,
	linux-samsung-soc@...r.kernel.org, linux-mtd@...ts.infradead.org,
	linux-kernel@...r.kernel.org
Subject: Re: [PATCH] mtd: nand: s3c2410: fix bug in
 s3c2410_nand_correct_data()

在 2016年04月08日 10:18, Boris Brezillon 写道:
> On Fri, 8 Apr 2016 09:51:04 +0800
> Zeng Zhaoxiu <zhaoxiu.zeng@...il.com> wrote:
>
>>
>> 在 2016年04月08日 08:18, Boris Brezillon 写道:
>>> Hi Zeng,
>>>
>>> On Fri,  8 Apr 2016 00:48:17 +0800
>>> zengzhaoxiu@....com wrote:
>>>
>>>> From: Zeng Zhaoxiu <zhaoxiu.zeng@...il.com>
>>>>
>>>> If there is only one bit difference in the ECC, the function should return 1.
>>>> The result of "diff0 & ~(1<<fls(diff0))" is equal to diff0, so the function
>>>> actually returns -1.
>>>>
>>>> Here, we can use the simple expression "(diff0 & (diff0 - 1)) == 0" to determine
>>>> whether the diff0 has only one 1-bit.
>>> Missing Signed-off-by here.
>>>
>>>> ---
>>>>    drivers/mtd/nand/s3c2410.c | 2 +-
>>>>    1 file changed, 1 insertion(+), 1 deletion(-)
>>>>
>>>> diff --git a/drivers/mtd/nand/s3c2410.c b/drivers/mtd/nand/s3c2410.c
>>>> index 9c9397b..c9698cf 100644
>>>> --- a/drivers/mtd/nand/s3c2410.c
>>>> +++ b/drivers/mtd/nand/s3c2410.c
>>>> @@ -542,7 +542,7 @@ static int s3c2410_nand_correct_data(struct mtd_info *mtd, u_char *dat,
>>>>    	diff0 |= (diff1 << 8);
>>>>    	diff0 |= (diff2 << 16);
>>>>    
>>>> -	if ((diff0 & ~(1<<fls(diff0))) == 0)
>>>> +	if ((diff0 & (diff0 - 1)) == 0)
>>> Or just
>>>
>>> 	if (hweight_long((unsigned long)diff0) == 1)
>>>
>>> which is doing exactly what the comment says.
>>>
>>> BTW, I don't understand why the current code is wrong? To me, it seems
>>> it's correctly detecting the case where only a single bit is different.
>>> What are you trying to fix exactly?
>>>
>>> Best Regards,
>>>
>>> Boris
>>>
>> For example, assuming diff0 is 1, then fls(diff0) is equal to 1, then "~(1 << fls(diff0))" is equal to 0xfffffffd,
>> then the result of "(diff0 & ~(1 << fls(diff0)))" is 1 , not we expected 0.
>>
>> __fls(diff0) and "(fls(diff0) - 1)" are all right, but fls(diff0) is wrong.
>>
> Indeed, I forgot that fls() was returning (position + 1). Anyway, I
> still think using hweight clarifies what you really want to test.
>

"(n & (n - 1))" is used in is_power_of_2() in incluse/linux/log2.h,
it's result is equal to "n & ~(1 << __ffs(n))".

"(diff & (diff - 1))" is simple and fast, although here is not performance critical.
To improve readability of this code, we should add a new function and use it.

/*
  *  Determine whether some value has more than one 1-bits
  */

static inline __attribute__((const))
bool more_than_1_bit_set(unsigned long n)
{
     return (n & (n - 1)) != 0;
}

OTOH, I found many determinations like "hweightN(n) > 1" distributed in kernel,
these determinations are slower than "(n & (n - 1)) != 0" on most CPUs.
We can use this new function instead.