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.z...@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
zengzhao...@163.com wrote:
From: Zeng Zhaoxiu <zhaoxiu.z...@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.
