explain what lsb and msb do for 0 and negatives

S32-setting-library/Numeric.pod

@@ -500,13 +500,30 @@ Returns the least significant bit position containing a 1 bit, counting
 bit positions from least significant to most significant.  (In other
 words, it's the base 2 logarithm of number represented by that 1 bit.)
 
+This function returns C<Nil> on a 0 value, since there are no bits set.
+Negative integers are treated as 2's complement, so always have
+a lowest bit set somewhere, if only the sign bit.  Hence, a -32768
+returns an lsb of 15 regardless of whether it's stored in an C<int16>
+or an C<Int>.
+
 =item msb
 
  multi method msb ( Int $x: ) is export
 
 Returns the most significant bit position containing a 1 bit, that is,
 the base 2 logarithm of the top 1 bit.
 
+This function returns C<Nil> on a 0 value.  For negative values, the
+function is dependent on the type.  For native types, signed integers
+are treated as unsigned, so a negative number stored in C<int64> will
+always return 63.  Negative integers stored in an C<Int> notionally
+have an infinite number of 1 bits on top, which is a problem.
+Instead of returning C<+Inf>, which is relatively useless, we return
+the position of the first of that infinite supply of sign bits.
+So C<msb(-1)> returns 0, C<msb(-2)> returns 1, and C<msb(-32768)>
+returns 15, just as if we'd converted it from C<int16> to C<uint16>
+and examined that for its top bit.
+
 =back
 
 =head2 Rat