Convert Split to record #21583
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Convert Split to record #21583
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Previously Split.equals did not invoke ConnectorSplit.equals, so it maybe didn't matter. |
sopel39
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Apr 17, 2024
Since the records have generates equals and hashCode methods, we need to switch away from using Set<Split> in the scheduler implementation which seems incorrect anyway as it suggests that we want to remove duplicates but hashCode and equals are not part of the ConnectorSplit contract.
wendigo
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findepi
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Apr 19, 2024
| @@ -92,7 +92,7 @@ private enum State | |||
| private final BooleanSupplier anySourceTaskBlocked; | |||
| private final PartitionIdAllocator partitionIdAllocator; | |||
| private final Map<InternalNode, RemoteTask> scheduledTasks; | |||
| private final Set<Split> pendingSplits = new HashSet<>(); | |||
| private final List<Split> pendingSplits = new ArrayList<>(); | |||
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see the line below
splitAssignment.values().forEach(pendingSplits::remove); // AbstractSet.removeAll performs terribly here.
now that pendingSplits is a list, this will be List.remove, which is a slow thing.
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is the removeAll better?
colebow
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Since the records have generates equals and hashCode methods, we need to switch away from using Set in the scheduler implementation which seems incorrect anyway as it suggests that we want to remove duplicates but hashCode and equals are not part of the ConnectorSplit contract.
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