ntfs-83/kext/ntfs_index.c

1972 	u8 bit;
2013 	/* Find the first zero bit in the index bitmap. */
2028 		/* Check the next bit(s). */
2040 			bit = ffs(~(unsigned long)*bmp) - 1;
2041 			if (bit < 8)
2048 	 * we extend it by 8 bytes and allocate the first bit of the extension.
2070 	 * Get the page containing the bit we are allocating.  Note this has to
2113 	/* Finally, allocate the bit in the page. */
2114 	bit = 0;
2119 	 * very bit we want to allocate so we test for this case and if it has
2120 	 * happened we allocate the next bit along which must be free or
2126 		bit++;
2129 	*bmp |= 1 << bit;
2131 	/* Set @bmp_pos to the allocated index bitmap bit. */
2132 	bmp_pos = (bmp_pos << 3) + bit;
2134 	 * If we are caching the last set bit in the bitmap in the index inode
2135 	 * and we allocated beyond the last set bit, update the cached value.
2139 	ntfs_debug("Allocated index bitmap bit 0x%llx.", bmp_pos);
2143 	 * If the allocated bit is outside the data size need to extend it.
2226 	/* Free the index bitmap bit that we allocated. */
2906 		 * which are initialized to zero.  We then set the first bit in
2934 		 * attributes thus we can set up our cache of the last set bit
4856 	 * Zero the bit in the index bitmap corresponding to the index block
4869 	/* If this is not the last set bit, we are done. */
4881 	 * bit.  If we do not know the old last set bit (@idx_ni->last_set_bit
4883 	 * cleared it just now, start at the old last set bit.
4887 	 * occurs we invalidate the last set bit stored in the inode.
4913 			unsigned bit, byte = *bmp;
4918 			 * Determine the last set bit in the byte.
4924 			 * So we do the "bit = fls(byte) - 1" by hand which is
4927 			bit = 0;
4930 				bit += 4;
4934 				bit += 2;
4937 				bit++;
4940 			 * @bit now contains the last set bit in the byte thus
4941 			 * we can determine the last set bit in the bitmap.
4944 					(bmp - bmp_start)) << 3) + bit;
4961 	 * This was the last set bit.  Check if we would save disk space by
5182 	 * of the last set bit.
5201 		  * Skip the index root as it does not have a bit in the
5423 	 * Breaking this case up into two means we incur a little bit of