If two updates are issued in rapid succession (within same processing task), this will trigger two events as expected.
to.update({position:0}) to.update({velocity:1})
However, since events are asynchronous, the old vector observed by the event listener will be the same in both events. For instance, this will cause the sequencer to be confused about the nature of the change, possibly leading to faulty behavior.
One possible solution is to bring the correct old state along with the event arguments.
If two updates are issued in rapid succession (within same processing task), this will trigger two events as expected.
to.update({position:0}) to.update({velocity:1})However, since events are asynchronous, the old vector observed by the event listener will be the same in both events. For instance, this will cause the sequencer to be confused about the nature of the change, possibly leading to faulty behavior.
One possible solution is to bring the correct old state along with the event arguments.