So from what I understand, it works like this by design and it's not possible to test my code with the current coder implementation. Is that correct?
On Wed, 8 Jul 2020 at 21:41, Jeff Klukas <[email protected]> wrote: > On Wed, Jul 8, 2020 at 1:38 PM Kirill Zhdanovich <[email protected]> > wrote: > >> So it's correct implementation of TableRow that encode(decode(a)) != a? >> > > A TableRow can contain fields of any map implementation. It makes sense to > me that once a TableRow object is serialized and deserialized, that the > coder must make a choice about a concrete Map implementation to use. > > So, no I would not expect that a decoded TableRow would contain exactly > the same objects as before encoding. But I _would_ expect that > encode(decode(a)).equals(a) in the sense that Map.equals() can determine > two maps of different types to be equal as long as both maps contain > entries that are equal to one another. > > > -- Best Regards, Kirill
