2012/5/23 Matthew Wilcox [off-list ref]:

On Wed, May 23, 2012 at 09:12:18PM +0900, Akinobu Mita wrote:

quoted

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.

quoted

{
      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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