On 25/06/2021 14:51, David Lloyd wrote:
Is this better than the current solution of `nativeOrder() ==
BIG_ENDIAN` other than reducing a few keystrokes?
I wouldn't expect many developers would need to be concerned with the
platform endianness, it's more likely going to be something advanced,
maybe a high performance library or something interaction with the
native code or the platform. So it would be just a convenience, you can
already can use ByteOrder.nativeOrder() and test if it returns BIG or
LITTLE_ENDIAN.
-Alan
