* x..xx01..1 Subtract 1, turns all the trailing zeros into 1's and
* the 1 just to their left into a 0; the remainder is
* untouched
- * 0..0011..1 The xor with x..xx10..0 clears that remainder, sets
- * bottom to all 1
+ * 0..0011..1 The xor with the original, x..xx10..0, clears that
+ * remainder, sets the bottom to all 1
* 0..0100..0 Add 1 to clear the word except for the bit in 's'
*
* Another method is to do 'word &= -word'; but it generates a compiler