squah-confluent commented on code in PR #18020:
URL: https://github.com/apache/kafka/pull/18020#discussion_r1868494415
##########
group-coordinator/src/main/java/org/apache/kafka/coordinator/group/modern/consumer/ConsumerGroup.java:
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@@ -791,6 +819,60 @@ private void validateMemberEpoch(
}
}
+ /**
+ * Computes the subscription type based on the provided information.
+ *
+ * @param subscribedRegularExpressions The subscribed regular expression
count.
+ * @param subscribedTopicNames The subscribed topic name count.
+ * @param numberOfMembers The number of members in the group.
+ *
+ * @return The subscription type.
+ */
+ public static SubscriptionType subscriptionType(
+ Map<String, Integer> subscribedRegularExpressions,
+ Map<String, Integer> subscribedTopicNames,
+ int numberOfMembers
+ ) {
+ if (subscribedRegularExpressions.isEmpty()) {
+ // If the members do not use regular expressions, the subscription
is
+ // considered as homogeneous if all the members are subscribed to
the
+ // same topics. Otherwise, it is considered as heterogeneous.
+ for (int subscriberCount : subscribedTopicNames.values()) {
+ if (subscriberCount != numberOfMembers) {
+ return HETEROGENEOUS;
+ }
+ }
+ return HOMOGENEOUS;
+ } else {
+ int count =
subscribedRegularExpressions.values().iterator().next();
+ if (count == numberOfMembers) {
+ // If all the members are subscribed to a single regular
expressions
+ // and none of them are subscribed to topic names, the
subscription
+ // is considered as homogeneous. If some members are
subscribed to
+ // topic names too, the subscription is considered as
heterogeneous.
+ for (int subscriberCount : subscribedTopicNames.values()) {
+ if (subscriberCount != 1) {
+ return HETEROGENEOUS;
+ }
+ }
+ return HOMOGENEOUS;
Review Comment:
I'm a little confused about this. Won't we return `HOMOGENEOUS` when all
members are subscribed to the same regex, but one member is additionally
subscribed to a topic outside the scope of the regex?
eg. all members subscribe to regex "foo*" and one member also subscribes to
topic "bar".
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