From: Zeng Zhaoxiu <[email protected]>
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.
---
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)
return 1;
return -1;
--
2.5.5
Hi Zeng,
On Fri, 8 Apr 2016 00:48:17 +0800
[email protected] wrote:
> From: Zeng Zhaoxiu <[email protected]>
>
> 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
--
Boris Brezillon, Free Electrons
Embedded Linux and Kernel engineering
http://free-electrons.com
在 2016年04月08日 08:18, Boris Brezillon 写道:
> Hi Zeng,
>
> On Fri, 8 Apr 2016 00:48:17 +0800
> [email protected] wrote:
>
>> From: Zeng Zhaoxiu <[email protected]>
>>
>> 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.
On Fri, 8 Apr 2016 09:51:04 +0800
Zeng Zhaoxiu <[email protected]> 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
> > [email protected] wrote:
> >
> >> From: Zeng Zhaoxiu <[email protected]>
> >>
> >> 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.
--
Boris Brezillon, Free Electrons
Embedded Linux and Kernel engineering
http://free-electrons.com
在 2016年04月08日 10:18, Boris Brezillon 写道:
> On Fri, 8 Apr 2016 09:51:04 +0800
> Zeng Zhaoxiu <[email protected]> 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
>>> [email protected] wrote:
>>>
>>>> From: Zeng Zhaoxiu <[email protected]>
>>>>
>>>> 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.
Hi Zeng,
On Fri, 8 Apr 2016 13:37:22 +0800
Zeng Zhaoxiu <[email protected]> wrote:
> 在 2016年04月08日 10:18, Boris Brezillon 写道:
> > On Fri, 8 Apr 2016 09:51:04 +0800
> > Zeng Zhaoxiu <[email protected]> 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
> >>> [email protected] wrote:
> >>>
> >>>> From: Zeng Zhaoxiu <[email protected]>
> >>>>
> >>>> 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.
Yes, probably, but it may be faster on a few CPUs :). Anyway, not sure
you should bother optimizing this now, especially since this test is in
the ECC correction path, and I doubt it makes any difference (detecting
and correcting errors is what takes most of the time here).
> We can use this new function instead.
>
In the end, I don't care that much which solution you'll choose, since
it's driver specific code. Pick whatever implementation you prefer and
resend the patch with your SoB.
Thanks,
Boris
--
Boris Brezillon, Free Electrons
Embedded Linux and Kernel engineering
http://free-electrons.com
From: Zhaoxiu Zeng <[email protected]>
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.
Signed-off-by: Zhaoxiu Zeng <[email protected]>
---
drivers/mtd/nand/s3c2410.c | 3 ++-
1 file changed, 2 insertions(+), 1 deletion(-)
diff --git a/drivers/mtd/nand/s3c2410.c b/drivers/mtd/nand/s3c2410.c
index 9c9397b..86ffb73 100644
--- a/drivers/mtd/nand/s3c2410.c
+++ b/drivers/mtd/nand/s3c2410.c
@@ -542,7 +542,8 @@ 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)
+ /* equal to "(diff0 & ~(1 << __ffs(diff0)))" */
+ if ((diff0 & (diff0 - 1)) == 0)
return 1;
return -1;
--
2.5.0
On Tue, 12 Apr 2016 15:30:35 +0800
[email protected] wrote:
> From: Zhaoxiu Zeng <[email protected]>
>
> 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.
>
> Signed-off-by: Zhaoxiu Zeng <[email protected]>
Applied.
Thanks,
Boris
> ---
> drivers/mtd/nand/s3c2410.c | 3 ++-
> 1 file changed, 2 insertions(+), 1 deletion(-)
>
> diff --git a/drivers/mtd/nand/s3c2410.c b/drivers/mtd/nand/s3c2410.c
> index 9c9397b..86ffb73 100644
> --- a/drivers/mtd/nand/s3c2410.c
> +++ b/drivers/mtd/nand/s3c2410.c
> @@ -542,7 +542,8 @@ 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)
> + /* equal to "(diff0 & ~(1 << __ffs(diff0)))" */
> + if ((diff0 & (diff0 - 1)) == 0)
> return 1;
>
> return -1;
--
Boris Brezillon, Free Electrons
Embedded Linux and Kernel engineering
http://free-electrons.com