[Solved] Consider this Kafka consumer code snippet:try { ConsumerRecords records = consumer.poll(Duration.ofMillis(500));... }What happens if the poll() method throws a TimeoutException and it is not caught? | Kafka

Kafka - with Java/Python

Consider this Kafka consumer code snippet:

try {
  ConsumerRecords records = consumer.poll(Duration.ofMillis(500));
  // process records
} catch (TimeoutException e) {
  System.out.println("Polling timed out");
}

What happens if the poll() method throws a TimeoutException and it is not caught?

AThe exception propagates and may cause the consumer thread to terminate unexpectedly

BThe poll() call silently retries until successful

CThe consumer automatically reconnects and continues polling

DThe poll() method returns an empty ConsumerRecords object

Step-by-Step Solution

Solution:

Step 1: Understand exception propagation
If poll() throws a TimeoutException and it is not caught, the exception propagates up the call stack.
Step 2: Consequences
Uncaught exceptions can cause the consumer thread to terminate unexpectedly, potentially stopping message consumption.
Step 3: Evaluate other options
Options B, C, and D describe behaviors that do not happen automatically without explicit handling.
Final Answer:
The exception propagates and may cause the consumer thread to terminate unexpectedly -> Option A
Quick Check:
Uncaught exceptions propagate and can stop the thread [OK]

Quick Trick: Uncaught exceptions stop consumer thread unless handled [OK]

Common Mistakes:

MISTAKES

Assuming poll retries automatically on exceptions
Believing consumer reconnects without explicit code
Thinking poll returns empty records on exceptions

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Consider this Kafka consumer code snippet:

Step 1: Understand exception propagation

Step 2: Consequences

Step 3: Evaluate other options

Final Answer:

Quick Check:

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