3 ms·
Absolutely not, he needs to compute (63 - x) / 7, where x is computed from a 64-bit integers. There are no bigints involved.
by bonzini 4y ago
Absolutely not, he needs to compute (63 - x) / 7, where x is computed from a 64-bit integers. There are no bigints involved.
- vintermann 4y agoWell, there are variable length integers involved, since it's as part of a way of representing variable length integers. But it seems you're right that x is just a regular machine int in this context, as an implementation detail. I misunderstood.