felipecrv commented on code in PR #41299:
URL: https://github.com/apache/arrow/pull/41299#discussion_r1583502818
##########
docs/source/format/CanonicalExtensions.rst:
##########
@@ -251,6 +251,17 @@ Variable shape tensor
Values inside each **data** tensor element are stored in
row-major/C-contiguous
order according to the corresponding **shape**.
+UUID
+====
+
+* Extension name: `arrow.uuid`.
+
+* The storage type of the extension is ``FixedSizeBinary`` with a length of 16
bytes.
+
+.. note::
+ A specific UUID version is not required or guaranteed. This extension
represents
+ UUIDs as FixedSizeBinary(16) and does not interpret the bytes in any way.
Review Comment:
You should specify what byte (0 or 15) is most-significant byte.
For instance, Java treats the 0-th byte as the MSB [1] which is consistent
with the convention that big-endian the network byte order [2]:
```java
private UUID(byte[] data) {
long msb = 0;
long lsb = 0;
assert data.length == 16 : "data must be 16 bytes in length";
for (int i=0; i<8; i++)
msb = (msb << 8) | (data[i] & 0xff);
for (int i=8; i<16; i++)
lsb = (lsb << 8) | (data[i] & 0xff);
this.mostSigBits = msb;
this.leastSigBits = lsb;
}
```
When porting this code to C/C++, be careful with the fact that Java's
integer types are big-endian (inheritance from SPARC), unlike most
architectures we use today that are little-endian.
An advantage of putting the MSB at byte 0 is that when you parse an UUID
string you read the string from the MSB to the LSB and write the UUID data from
0 to 15.
[1]
https://github.com/openjdk/jdk/blob/819f3d6fc70ff6fe54ac5f9033c17c3dd4326aa5/src/java.base/share/classes/java/util/UUID.java#L116-L126
[2]
https://stackoverflow.com/questions/13514614/why-is-network-byte-order-defined-to-be-big-endian
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