store operation: single ancestor or multiple nodes?

[guykatz]

Hi all;
given this object graph
assumning I need to change, in one logical operation nodes: E,H & Q
what is the way I need to store things for this opetation:

1. to to the ancestor common parent and store it using store(A)
2. store all 3 changed nodes using storeAll(E,H,Q)
   will both work? if they both work will they perform the same?
   thanks.

[hg-ms]

Hi,
The storeAll(E,H,Q) is the recommended solution to update several objects at once. It will apply a Lazy-Storer to all three objects. The storeAll method will either persist all supplied instance or non and throw an exception in case of error.

Using store(A) would work too if using an Eager-Storer but this will store everything reachable by references from “A” and therefore cause a lot of unnecessary overhead. In your case that would write out the whole graph again starting at “A”