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. -- dm-devel mailing list dm-devel@xxxxxxxxxx https://www.redhat.com/mailman/listinfo/dm-devel