1Z0-829 Practice Question: Handling Date, Time, Text, Numeric and Boolean Values
A developer needs to parse the string "2023-12-31T23:59:60Z" (a leap second) into a java.time.Instant. Which statement is true?
⚠ Common exam trap
Many exam-takers assume '60' seconds is always invalid and will cause an exception, but the Java Time API specifically accommodates leap seconds by converting them to the nearest valid nanosecond-adjusted Instant.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
It returns an Instant representing 2023-12-31T23:59:59Z with an added nanosecond.
Java.time.Instant.parse() handles leap seconds by converting them to the last valid second of the minute (23:59:59) and then adding a nanosecond to account for the extra second. This behavior is specified by the ISO-8601 standard and the Java Time API, which does not support a true 60th second but represents it as an Instant with a fractional second adjustment.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
It returns an Instant representing 2023-12-31T23:59:59Z, ignoring the leap second.
Why it's wrong here
Leap second is not ignored; it is handled.
- ✓
It returns an Instant representing 2023-12-31T23:59:59Z with an added nanosecond.
Why this is correct
The 60th second is treated as the last nanosecond of the minute.
- ✗
It throws a DateTimeParseException because 60 seconds is invalid.
Why it's wrong here
Instant.parse handles leap seconds.
- ✗
It returns null because the string is invalid.
Why it's wrong here
It returns an Instant, not null.
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