IntStream rangeClosed unable to return value other than int [duplicate]

Question

Why is this getting error? I thought map can return any value.

var s = IntStream.rangeClosed(1, 5).map(String::valueOf).collect(Collectors.toList());

| Error: | incompatible types: bad return type in method reference | java.lang.String cannot be converted to int | var s = IntStream.rangeClosed(1, 5).map(String::valueOf).collect(Collectors.toList()); |
^-------------^

Answer 1

Use mapToObj:

var s = IntStream.rangeClosed(1, 5).mapToObj(String::valueOf).collect(Collectors.toList());

map of IntStream can only map an int value to another int value. It accepts an IntUnaryOperator (which accepts an int and returns an int) as the mapper function, and returns an IntStream.

On the other hand, mapToObj allows you to map an int value to any reference type, and thus transform the IntStream to a Stream<SomeReferenceType>. It accepts an IntFunction<? extends U> (which accepts an int and returns a reference type) as the mapper function, and returns a Stream<U>.

Answer 2

Use mapToObj instead :

IntStream.rangeClosed(1, 5).mapToObj(String::valueOf).collect(Collectors.toList());

Answer 3

While the aforementioned answers are correct and mapToObj is the idiomatic approach to proceed with, I think it's important to understand why the problem arises and thus in future cases, you'll know how to decipher the problem simply by going through the documentation.

it's a very important skill for a programmer to dig into the documentation when not understanding the workings of a specific method.

So, let's go through the relevant stream pipeline operations:

IntStream.rangeClosed returns an IntStream as per the documentation:

Returns a sequential ordered IntStream from startInclusive (inclusive) to endInclusive (inclusive) by an incremental step of 1.

invoking map on an IntStream is expected to return an IntStream as per the documentation:

Returns a stream consisting of the results of applying the given function to the elements of this stream.

As well as that it's important to note that the method declaration for map is as follows:

IntStream map(IntUnaryOperator mapper)

i.e. it takes a IntUnaryOperator which in fact represents an operation on a single int-valued operand that produces an int-valued result.

However, you're passing a function String::valueOf which consumes an int as we're dealing with an IntStream and returns a String thus not compliant with IntUnaryOperator and this is the cause of the problem.

Whenever you want to take a primitive stream specialization and perform some mapping function and in turn yield a Stream<R> as a result then mapToObj is the way to go.

mapToObj is declared as:

mapToObj(IntFunction<? extends U> mapper)

IntFunction represents a function that accepts an int-valued argument and produces a result and this result is of type R which means you'll have a Stream<R> after the mapToObj.

Answer 4

Alternatively, you could use IntStream.boxed as :

var s = IntStream.rangeClosed(1, 5) // IntStream
                 .boxed() // Stream<Integer>
                 .map(String::valueOf) // Stream<String>
                 .collect(Collectors.toList());

since the IntStream originally is a sequence of primitive int-values elements.

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Another variant of performing such an operation would be :

var s = IntStream.rangeClosed(1, 5)
                 .boxed()
                 .map(a -> Integer.toString(a))
                 .collect(Collectors.toList());