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by bonzini 1206 days ago
Absolutely not, he needs to compute (63 - x) / 7, where x is computed from a 64-bit integers. There are no bigints involved.
1 comments

Well, 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.