Re: portable way to get highest bit set?

Ben Bacarisse ha scritto:
> jak <[email protected]> writes:
> 
>> Ben Bacarisse ha scritto:
>>> jak <[email protected]> writes:
>>>
>>>> candycanearter07 ha scritto:
>>>>> Hi,
>>>>> What is the best/most portable way to get the highest bit set?
>>>>> ie. 011010001
>>>>> to  010000000
>>>>
>>>> Hi,
>>>> I don't think it's the best but the most portable could be the
>>>> mathematical approach:
>>>>
>>>> #include <stdio.h>
>>>> #include <math.h>
>>>>
>>>> int main()
>>>> {
>>>>       unsigned long val = 3000, ret;
>>> Test case: val = 0xFFFFFFFFFFFFFFFF
>>
>> This is due to the approximation of the floating point:
> 
> I know the reason.  I was just pointing out that you need to at least
> test the boundary cases!
> 
You were really kind to answer. I was also reading about 'feivatround'
and like you I found the 'short blanket'. I think there are less
complicated and, however, easily portable methods. Such as this:

#include <stdio.h>
#include <math.h>

unsigned long h_bit(unsigned long val)
{
     int i;
     for(i = -1; val; val /= 2, i++);
     return (unsigned long)pow(2., (double)i);
}

int main()
{
     unsigned long val = ~0UL, ret;

     ret = h_bit(val);
     printf("\nv: %lX r: %lX", val, ret);

     return 0;
}

Finally, the simplest things seem to be the best.
Thanks again.