Is Go\'s gob Encoding Deterministic, and Under What Conditions?

Deterministic Nature of gob Encoding

Question:

Is the output of gob encoding deterministic, meaning that for equal Go objects, the encoded results will always match?

Answer:

Under certain circumstances, gob encoding is deterministic. This means that for two Go objects x and y, if x is equal to y, the output of gob.Encode(x) and gob.Encode(y) will be the same, as long as:

• No tricksiness is involved with interfaces or maps
• The objects consist only of structs and arrays

Non-Determinism with Maps:

However, when maps are involved in the encoded objects, gob encoding becomes non-deterministic. This is because the iteration order of maps is random, causing their serialization order to vary. Consequently, the encoded results for maps may differ even if the maps are considered equal.

Self-Describing Streams:

Gob encoding streams are self-describing, meaning they contain type information for each data item. This implies that type descriptions are only transmitted once for each unique type. As a result, subsequent encoding of the same type will only reference the previous type specification.

This self-describing nature can lead to variations in the encoded output for the same values across multiple encoding instances. The first encoding will include both the type spec and the value, while subsequent encodings will only include the value and a reference to the previous type spec.

Conclusion:

Although the current implementation of gob encoding is deterministic for simple structs and arrays, it is essential to note that this behavior may change with future Go releases. Therefore, relying on the exact matching of gob-encoded results is inadvisable.

The above is the detailed content of Is Go\'s gob Encoding Deterministic, and Under What Conditions?. For more information, please follow other related articles on the PHP Chinese website!