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fix: correct stride bug in BitVector::unmarshal_bytes #2

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fix: correct stride bug in BitVector::unmarshal_bytes #2

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b5c8aae
fix: correct stride bug in BitVector::unmarshal_bytes
meguruli Sep 23, 2025
0a133e5
refactor: optimize deserialization loop for better readability
meguruli Sep 24, 2025
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2 changes: 1 addition & 1 deletion bloom/internal/bitvec/bitvec.mbt
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Original file line number Diff line number Diff line change
Expand Up @@ -370,7 +370,7 @@ pub fn BitVector::unmarshal_bytes(bytes : Bytes) -> BitVector {
let words_needed = words_needed(length)
let data = Array::make(words_needed, 0UL)
for i in 0..<words_needed {
data[i] = bytes[12 + i:20 + i].to_uint64_le()
data[i] = bytes[12 + i*8:20 + i*8].to_uint64_le()
}
Comment on lines 373 to 376
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@gemini-code-assist gemini-code-assist bot Sep 23, 2025

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medium

While the fix is correct, this loop can be made more readable and slightly more efficient by avoiding the multiplication on each iteration. Using an offset variable that is incremented by the word size makes the logic clearer.

For example:

  let mut offset = 12
  for i in 0..<words_needed {
    data[i] = bytes[offset:offset + 8].to_uint64_le()
    offset += 8
  }

As a follow-up, you might consider defining constants for the magic numbers used in serialization (e.g., header size 12, word size 8) to improve maintainability.

{ length, tail_mask, data }
}
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9 changes: 9 additions & 0 deletions bloom/internal/bitvec/bitvec_test.mbt
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Expand Up @@ -270,3 +270,12 @@ test "serialization_roundtrip" {
let deserialized = BitVector::from_string(serialized)
assert_eq(original.to_string(), deserialized.to_string())
}

test "marshal_unmarshal_bytes_roundtrip_multiword" {
let original = BitVector::ones(128)
// Modify the second word so the bug cannot be masked by identical bytes
original.clear(64)
let serialized = original.marshal_bytes()
let deserialized = BitVector::unmarshal_bytes(serialized)
assert_eq(original.equal(deserialized), true)
}