Given: ``java IntStream stream = IntStream.of(1,2,3); IntFunction<Integer> inFu = x -> x * x;//line n1 IntStream newStream = stream.map(inFu.apply(10));//line n2…

Question

IntStream stream = IntStream.of(1,2,3);
IntFunction<Integer> inFu = x -> x * x;//line n1
IntStream newStream = stream.map(inFu.apply(10));//line n2
newStream.forEach(System.out::print);

Options

• AReplace line n1 with: IntFunction<UnaryOperator> inFu = x -> y -> x*y;
• BReplace line n1 with: IntFunction<IntUnaryOperator> inFu = x -> y -> x*y; Replace line n2 with: IntStream newStream = stream.map(inFu.applyAsInt(10));
• CReplace line n2 with: IntStream newStream = stream.map(inFu.applyAsInt(10));
• DReplace line n2 with: IntStream newStream = stream.map(inFu.apply(10));

• A
  12% (2)
• B
  82% (14)
• D
  6% (1)

Explanation

Option B works because it changes inFu from a function that produces a value into a curried function that produces another function. With IntFunction<IntUnaryOperator>, calling inFu.apply(10) returns y -> 10*y - an IntUnaryOperator - which is exactly what IntStream.map() requires. Without this change, the n2 fix alone cannot help because map needs a mapping function, not an integer result.

Option A fails because UnaryOperator (raw/unparameterized) is not an IntUnaryOperator - they are separate functional interfaces, and IntStream.map() specifically requires IntUnaryOperator. Also, n2 is unchanged, so the wrong type is still passed.

Option C fails because IntFunction<Integer> (n1 unchanged) has no applyAsInt method - that method belongs to IntUnaryOperator. This is a compile error on n2 itself before map even comes into play.

Option D fails because with the original IntFunction<Integer>, inFu.apply(10) evaluates to the integer 100, not a function. IntStream.map() can't accept a bare int/Integer where it expects an IntUnaryOperator.

Memory tip: IntStream.map() takes a transformer function (an IntUnaryOperator), never a pre-computed value. If you find yourself calling .apply(n) directly inside map(...), ask "am I passing a function or an already-computed result?" - passing a result means you applied one level too early and need currying.

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