Oracle Exam 1Z0-809 Topic 9 Question 27 Discussion

Actual exam question for Oracle's 1Z0-809 exam

Question #: 27
Topic #: 9

Given the code fragment:

List listVal = Arrays.asList(''Joe'', ''Paul'', ''Alice'', ''Tom'');

System.out.println (

// line n1

);

Which code fragment, when inserted at line n1, enables the code to print the count of string elements whose length is greater than three?

AlistVal.stream().filter(x -> x.length()>3).count()

BlistVal.stream().map(x -> x.length()>3).count()

ClistVal.stream().peek(x -> x.length()>3).count().get()

DlistVal.stream().filter(x -> x.length()>3).mapToInt(x -> x).count()

Show Suggested Answer

Suggested Answer: D

by Johnetta at May 27, 2022, 02:35 PM

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Shenika

6 days ago

Option B doesn't seem right because the map() method is used to transform the elements, not to filter them. The count() method would return the number of elements that are true, which is not what we want.

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Janine

16 days ago

I think option A is the correct answer. The stream() method is used to create a stream of the list elements, and the filter() method is used to filter the elements based on the condition x.length() > 3, which gives us the count of the string elements whose length is greater than three.

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Venita

16 days ago

I'm not sure about the answer. Can someone explain why option A is correct?

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Elliott

17 days ago

I agree with Ressie. Option A makes sense as it filters the elements based on length and then counts them.

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Ressie

18 days ago

I think the answer is A) listVal.stream().filter(x -> x.length()>3).count(). Because it filters out the elements with length greater than three and then counts them.

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