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Since start, end, and entryIndex will always fit in 48 bits, we can represent them either either `uint64` or `int64`. We're consolidating on `int64` to reduce the amount of casting we do (and the attendant `nolint:gosec // G115` comments).
Since start, end, and entryIndex will always fit in 48 bits, we can represent them either either `uint64` or `int64`. We're consolidating on `int64` to reduce the amount of casting we do (and the attendant `nolint:gosec // G115` comments).
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LGTM with one optional improvement
| if !s.ReadUint64(&start) { | ||
| return nil, errors.New("invalid start") | ||
| } | ||
| if start > math.MaxInt64 { |
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Should this (and below) instead compare against 2^48?
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Agreed. Per Appendix C.2, "The largest possible Merkle Tree in this protocol is 2^48-1", and I think we already reject anything past that in proof.Marshal() and treedb. So we want > trees.MaxSize here and below.
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de76877
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re-LGTM with same optional question
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Since start, end, and entryIndex will always fit in 48 bits, we can represent them either either
uint64orint64. We're consolidating onint64to reduce the amount of casting we do (and the attendantnolint:gosec // G115comments).