2012/5/23 Matthew Wilcox <matt...@wil.cx>:
> On Wed, May 23, 2012 at 09:12:18PM +0900, Akinobu Mita wrote:
>> size_t memweight(const void *ptr, size_t bytes)
>
> Why should this return size_t instead of unsigned long?
I just use the same type as the bytes argument without mature
consideration. If unsigned long is better than size_t, I'll
change the return type.
>> {
>> size_t w = 0;
>> size_t longs;
>> const unsigned char *bitmap = ptr;
>>
>> for (; bytes > 0 && ((unsigned long)bitmap) % sizeof(long);
>> bytes--, bitmap++)
>> w += hweight8(*bitmap);
>>
>> longs = bytes / sizeof(long);
>> BUG_ON(longs >= INT_MAX / BITS_PER_LONG);
>> w += bitmap_weight((unsigned long *)bitmap, longs * BITS_PER_LONG);
>> bytes -= longs * sizeof(long);
>> bitmap += longs * sizeof(long);
>>
>> for (; bytes > 0; bytes--, bitmap++)
>> w += hweight8(*bitmap);
>>
>> return w;
>> }
>
> bitmap_weight copes with a bitmask that isn't a multiple of BITS_PER_LONG
> in size already. So I think this can be done as:
>
> unsigned long memweight(const void *s, size_t n)
> {
> const unsigned char *ptr = s;
> unsigned long r = 0;
>
> while (n > 0 && (unsigned long)ptr % sizeof(long)) {
> r += hweight8(*ptr);
> n--;
> ptr++;
> }
>
> BUG_ON(n >= INT_MAX / 8)
>
> return r + bitmap_weight((unsigned long *)ptr, n * 8);
> }
This works perfectly on little-endian machines. But it doesn't work
on big-endian machines, if the bottom edge of memory area is not
aligned on long word boundary.
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