[Java] DictionaryEncoder doesn't crash when decoding index outside of Dictionary
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- Java
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説明
### Describe the bug, including details regarding any error messages, version, and platform.
### Background
When manually creating an encoded vector, with values referencing indices in a Dictionary, it is possible to have values equal to `valueCount` of the Dictionary vector i.e. one index out of bounds. This is interpreted as a null value in the Dictionary.
---
### Example
_Dictionary_
| Index | Original Value |
|-------|---------------|
| 0 | Foo |
_Encoded Vector - 1_
| Index | Encoded Value | Expected Decode Outcome | Actual Outcome |
|-------|----------------|-----------------------------------|-----------------|
| 0 | 0 | Return the original value | Returns the original value |
_Encoded Vector - 2_
| Index | Encoded Value | Expected Decode Outcome | Actual Outcome |
|-------|----------------|-----------------------------------|-----------------|
| 0 | 1 | Raise `IllegalArgumentException` | Returns null |
_Encoded Vector - 3_
| Index | Encoded Value | Expected Decode Outcome | Actual Outcome |
|-------|----------------|-----------------------------------|-----------------|
| 0 | 2 | Raise `IllegalArgumentException` | Raises `IllegalArgumentException`|
---
### Test to reproduce the error
```java
@Test
public void testReferencingIndexOutOfBounds() {
// Index at which the original value will be stored at in the dictionary
var encodedIndex = 0;
// The encoded value that references an index in the dictionary
var indexReferenced = 1;
try (final IntVector encodedVector = new IntVector("encodings", allocator);
final VarCharVector dictionaryVector = newVarCharVector("dict", allocator); ) {
var originalValue = "Foo";
dictionaryVector.allocateNew(1);
dictionaryVector.setValueCount(1);
dictionaryVector.set(encodedIndex, originalValue.getBytes(StandardCharsets.UTF_8));
encodedVector.allocateNew(1);
encodedVector.setValueCount(1);
encodedVector.set(0, indexReferenced);
Dictionary dictionary =
new Dictionary(dictionaryVector, new DictionaryEncoding(1L, false, null));
try (ValueVector decoded = DictionaryEncoder.decode(encodedVector, dictionary)) {
fail("There should be an exception when decoding index outside dictionary's range.");
} catch (Exception e) {
assertEquals("Provided dictionary does not contain value for index " + indexReferenced, e.getMessage());
}
}
}
```
コントリビューションガイド
調査の方向性
Start at DictionaryEncoder.decode and reproduce the behavior with the provided testReferencingIndexOutOfBounds test. Confirm that an encoded index equal to the dictionary value count raises IllegalArgumentException with the expected message, while valid indices still decode normally.
索引モデルが issue の本文から書いたものです。
評価
- 技術スタック
- java
- 領域
- data
- issue の種類
- バグ
- 難易度
- 2/5
- 見積もり時間
- 1〜3時間
- 活発さ
- 静か
- 明瞭さ
- 明確に書かれている
- 初心者へのやさしさ
- 74/100