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
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>>
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