On 31/03/2023 09:31, Jani Nikula wrote:
> On Thu, 30 Mar 2023, Andrew Morton <akpm@xxxxxxxxxxxxxxxxxxxx> wrote:
>> On Thu, 30 Mar 2023 21:53:03 +0000 David Laight <David.Laight@xxxxxxxxxx> wrote:
>>
>>>> But wouldn't all these issues be addressed by simply doing
>>>>
>>>> #define is_power_of_2(n) (n != 0 && ((n & (n - 1)) == 0))
>>>>
>>>> ?
>>>>
>>>> (With suitable tweaks to avoid evaluating `n' more than once)
>>>
>>> I think you need to use the 'horrid tricks' from min() to get
>>> a constant expression from constant inputs.
>>
>> This
>>
>> --- a/include/linux/log2.h~a
>> +++ a/include/linux/log2.h
>> @@ -41,11 +41,11 @@ int __ilog2_u64(u64 n)
>> * *not* considered a power of two.
>> * Return: true if @n is a power of 2, otherwise false.
>> */
>> -static inline __attribute__((const))
>> -bool is_power_of_2(unsigned long n)
>> -{
>> - return (n != 0 && ((n & (n - 1)) == 0));
>> -}
>> +#define is_power_of_2(_n) \
>> + ({ \
>> + typeof(_n) n = (_n); \
>> + n != 0 && ((n & (n - 1)) == 0); \
>> + })
>>
>> /**
>> * __roundup_pow_of_two() - round up to nearest power of two
>> _
>>
>> worked for me in a simple test.
>>
>> --- a/fs/open.c~b
>> +++ a/fs/open.c
>> @@ -1564,3 +1564,10 @@ int stream_open(struct inode *inode, str
>> }
>>
>> EXPORT_SYMBOL(stream_open);
>> +
>> +#include <linux/log2.h>
>> +
>> +int foo(void)
>> +{
>> + return is_power_of_2(43);
>> +}
>> _
>>
>>
>> foo:
>> # fs/open.c:1573: }
>> xorl %eax, %eax #
>> ret
>>
>>
>> Is there some more tricky situation where it breaks?
>
> It doesn't work with BUILD_BUG_ON_ZERO().
Like most programming problems, you just need another layer of
indirection! The below works for me in all the cases I could think of
(including __uint128_t).
#define __IS_POWER_OF_2(n) (n != 0 && ((n & (n - 1)) == 0))
#define _IS_POWER_OF_2(n, unique_n) \
({ \
typeof(n) unique_n = (n); \
__IS_POWER_OF_2(unique_n); \
})
#define is_power_of_2(n) \
__builtin_choose_expr(__is_constexpr((n)), \
__IS_POWER_OF_2((n)), \
_IS_POWER_OF_2(n, __UNIQUE_ID(_n)))
Although Jani's original might be easier to understand.
Steve
