Subject: Re: [PATCH] mtd: nand: s3c2410: fix bug in
 s3c2410_nand_correct_data()
To: Boris Brezillon <boris.brezillon@free-electrons.com>
References: <1460047697-72830-1-git-send-email-zengzhaoxiu@163.com>
 <20160408021817.274d4e59@bbrezillon> <57070E88.5020008@gmail.com>
 <20160408041802.5c47a4be@bbrezillon>
Cc: zengzhaoxiu@163.com, kgene@kernel.org, k.kozlowski@samsung.com,
        richard@nod.at, dwmw2@infradead.org, computersforpeace@gmail.com,
        linux-arm-kernel@lists.infradead.org,
        linux-samsung-soc@vger.kernel.org, linux-mtd@lists.infradead.org,
        linux-kernel@vger.kernel.org
From: Zeng Zhaoxiu <zhaoxiu.zeng@gmail.com>
Message-ID: <57074392.2030601@gmail.com>
Date: Fri, 8 Apr 2016 13:37:22 +0800
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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@gmail.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@163.com wrote:
>>>
>>>> From: Zeng Zhaoxiu <zhaoxiu.zeng@gmail.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.