Given:
Which action will enable the code to compile?
Exam 1z0-819 All QuestionsBrowse all questions from this exam
Question 174
Correct Answer: C
To ensure the code compiles, we need to correctly match the types expected by the `IntStream.map` method. This method accepts an `IntUnaryOperator`, which is a functional interface representing an operation on a single `int` operand that produces an `int` result. The line `IntFunction<IntUnaryOperator> func = x -> y -> x * y;` correctly defines a function `func` that takes an `int` and returns an `IntUnaryOperator`. This operator effectively multiplies the given operand by `x`, where `x` is specified when invoking `func.apply(10)`. Therefore, option C is the correct choice.
Discussion
ASPushkinOption: C
answer : looks like C and D(same) "func" should have input : int and output: IntUnaryOperator Correct answer : IntFunction<IntUnaryOperator> func = x->y->x*y; IntUnaryOperator oper = y->x*y; where x and y are type of int by definition
KiraguJohn
What is the difference between C and D?
d7bb0b2Option: C
NO ANSWER CORRECT THE ONLY wait to this is ok is: IntFunction<IntUnaryOperator> func = x -> y -> x * y; then instream.map(func.apply(...)).....
d7bb0b2
Instream.map() method only aceppt IntUnaryOperator not IntFunction
ASPushkin
B) is not right 1) Function is raw type 2) func.apply(10) has just one input parameter
ASPushkin
IntFunction is a raw type and it equals to IntFunction<Object> so right answer is IntFunction<IntUnaryOperator> func = x-> y->x*y;
[Removed]
There's so much missing in this question it's almost impossible to even try to solve this. ALSO C & D are the exact same.
[Removed]
It's definitely not A as the BiFunction's apply method takes 2 parameters. I'm guessing the correct answer has to be something like this IntFunction<IntFunction<Integer>>
d7bb0b2
IntFunction<IntFunction<Integer>> can not resolve the problem because map of Instream only aceppt IntUnaryOperator not IntFunction