Because adding the top limbs of two encoded numbers would overflow too soon. If you set both to 2^63 for example, they overflow immediately. Might be fine for wraparound arithmetic, but not in general.
Setting both to 2^63 means your original 256-bit numbers were 2^255, thus the addition would overflow no matter what intermediate encoding you’re using.
Sure, then set one to 2^62 and the other to -2^62 (namely: 0b1100..00). It's overflow as far as unsigned arithmetic is concerned, but not in the case of signed arithmetic.
That said, when you're dealing with 256-bit integers, you're almost assuredly not working with signed arithmetic.
volemo|9 months ago
vitus|9 months ago
That said, when you're dealing with 256-bit integers, you're almost assuredly not working with signed arithmetic.