Currently, Streams provides no mechanism to compare two topologies. This is a common operation when users want to have tests verifying that they don't accidentally alter their topology. They would save the known-good topology and then add a unit test verifying the current code against that known-good state.
However, because there's no way to do this comparison properly, everyone is reduced to using the string format of the topology (from `Topology#describe().toString()`). The major drawback is that the string format is meant for human consumption. It is neither machine-parseable nor stable. So, these compatibility tests are doomed to fail when any minor, non-breaking, change is made either to the application, or to the library. This trains everyone to update the test whenever it fails, undermining its utility.
We should fix this problem, and provide both a mechanism to serialize the topology and to compare two topologies for compatibility. All in all, I think we need:
- a way to serialize/deserialize topology structure in a machine-parseable format that is future-compatible. Offhand, I'd recommend serializing the topology structure as JSON, and establishing a policy that attributes should only be added to the object graph, never removed. Note, it's out of scope to be able to actually run a deserialized topology; we only want to save and load the structure (not the logic) to facilitate comparisons.
- a method to verify the equality of two topologies... This method tells you that the two topologies are structurally identical. We can't know if the logic of any operator has changed, only if the structure of the graph is changed. We can consider whether other graph properties, like serdes, should be included.
- a method to verify the compatibility of two topologies... This method tells you that moving from topology A to topology B does not require an application reset. Note that this operation is not commutative: `A.compatibleWith(B)` does not imply `B.compatibleWith(A)`. We can discuss whether `A.compatibleWith(B) && B.compatibleWith(A)` implies `A.equals(B)` (I think not necessarily, because we may want "equality" to be stricter than "compatibility").