Map.invoke & map.put change visibility question

Hi,
In the documentation is stated that changes done by a entryProcessor are visible only after the processor has finished execution. I could not find statement about the opposite direction - what happens if map.invoke() and map.put() are trying to change the same entry? Is there possibility the map.put method to be executed in parallel to the map.invoke, so that map.put changes value while map.invoke is being executed on it? I saw in the forum statement that all map operations are executed in single thread (when modifying the same entry)- does this mean that map.put is sort-of standard processor and is queued just like any other processors?
Best Regards,
Georgi

Hi Georgi,
all operations related to a single entry are executed in a single-threaded manner on the storage node, meaning either one or zero thread does anything on any single entry. It is not possible that a put would occur for an entry while an entry-processor is processing that entry.
And yes, it is a usable analog to think on a put operation as an entry-processor doing a setValue on the entry, or a lock/unlock operation as an entry-processor which changes the lock ownership (note: an acquired lock for an entry does not affect anything other than a lock operation for the same entry, which will not succeed until the earlier lock was released).
Best regards,
Robert

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