extra two bits. We need a sharper analysis in this case. The major
laziness was in the "the same is true of ah+al" clause: ah+al can't actually
have shift+1 digits + 1 bit unless bsize is odd and asize == bsize. In that
-case, we actually have (2*shift+1)*2 - shift = 3*shift + 2 allocated digits
-remaining, and that's obviously plenty to hold 2*shift + 2 digits + 2 bits.
+case, we actually have (2*shift+1)*2 - shift = 3*shift+2 allocated digits
+remaining, and that's obviously plenty to hold 2*shift+2 digits + 2 bits.
Else (bsize is odd and asize < bsize) ah and al each have at most shift digits,
so ah+al has at most shift digits + 1 bit, and (ah+al)*(bh+bl) has at most
-2*shift+1 digits + 2 bits, and again 2*shift+2 digits + 2 bits is
-enough to hold it.
+2*shift+1 digits + 2 bits, and again 2*shift+2 digits is enough to hold it.
*/
/* b has at least twice the digits of a, and a is big enough that Karatsuba