Message ID | 1460047697-72830-1-git-send-email-zengzhaoxiu@163.com |
---|---|

State | Not Applicable, archived |

Headers | show |

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

? 2016?04?08? 08:18, Boris Brezillon ??: > 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. -- To unsubscribe from this list: send the line "unsubscribe linux-samsung-soc" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html

On Fri, 8 Apr 2016 09:51:04 +0800 Zeng Zhaoxiu <zhaoxiu.zeng@gmail.com> wrote: > > > ? 2016?04?08? 08:18, Boris Brezillon ??: > > 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.

? 2016?04?08? 10:18, Boris Brezillon ??: > On Fri, 8 Apr 2016 09:51:04 +0800 > Zeng Zhaoxiu <zhaoxiu.zeng@gmail.com> wrote: > >> >> ? 2016?04?08? 08:18, Boris Brezillon ??: >>> 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. -- To unsubscribe from this list: send the line "unsubscribe linux-samsung-soc" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html

Hi Zeng, On Fri, 8 Apr 2016 13:37:22 +0800 Zeng Zhaoxiu <zhaoxiu.zeng@gmail.com> wrote: > ? 2016?04?08? 10:18, Boris Brezillon ??: > > On Fri, 8 Apr 2016 09:51:04 +0800 > > Zeng Zhaoxiu <zhaoxiu.zeng@gmail.com> wrote: > > > >> > >> ? 2016?04?08? 08:18, Boris Brezillon ??: > >>> 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. 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

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;

`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. --- drivers/mtd/nand/s3c2410.c | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-)`