Re: [PATCH 01/10] string: introduce memweight

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2012/5/23 Matthew Wilcox <matthew@xxxxxx>:
> 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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