1Z0-829 Working with Streams and Lambda Expressions Practice Question
Exhibit
Refer to the exhibit.
var result = Stream.of("1", "2", "3")
.map(Integer::parseInt)
.collect(Collectors.teeing(
Collectors.summingInt(i -> i),
Collectors.counting(),
(sum, count) -> sum / count
));
System.out.println(result);What is the output of the following code? ```java System.out.println(Stream.of(1, 2).map(i -> i * 2.0).count()); ```
⚠ Common exam trap
The trap here is that candidates may mistakenly think the `map` operation changes the return type of `count()` or that the output would be a floating-point number (2.0) because the mapping produces `Double` values, but `count()` always returns a `long` regardless of the stream's element type.
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
✓
2
The code is: ```java System.out.println(Stream.of(1, 2).map(i -> i * 2.0).count()); ``` This creates a stream of integers 1 and 2, maps each to a Double (2.0 and 4.0), and then calls count() which returns the number of elements in the stream, which is 2. count() returns a long, but when printed, it outputs 2 (not 2.0). Option C is correct because the stream has two elements.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
2.0
Why it's wrong here
The result is an integer (2), not a double.
- ✗
1
Why it's wrong here
Incorrect calculation.
- ✓
2
Why this is correct
6/3 = 2, printed as 2.
- ✗
Compilation error
Why it's wrong here
The code compiles; teeing is available since Java 12.
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