Why can't I map integers to strings when streaming from an array?
JavaFunctional ProgrammingJava 8Java StreamJava Problem Overview
This code works (taken in the Javadoc):
List<Integer> numbers = Arrays.asList(1, 2, 3, 4);
String commaSeparatedNumbers = numbers.stream()
.map(i -> i.toString())
.collect(Collectors.joining(", "));
This one can't be compiled:
int[] numbers = {1, 2, 3, 4};
String commaSeparatedNumbers = Arrays.stream(numbers)
.map((Integer i) -> i.toString())
.collect(Collectors.joining(", "));
IDEA tells me I have an "incompatible return type String in lambda expression".
Why ? And how to fix that ?
Java Solutions
Solution 1 - Java
Arrays.stream(int[])
creates an IntStream
, not a Stream<Integer>
. So you need to call mapToObj
instead of just map
, when mapping an int
to an object.
This should work as expected:
String commaSeparatedNumbers = Arrays.stream(numbers)
.mapToObj(i -> ((Integer) i).toString()) //i is an int, not an Integer
.collect(Collectors.joining(", "));
which you can also write:
String commaSeparatedNumbers = Arrays.stream(numbers)
.mapToObj(Integer::toString)
.collect(Collectors.joining(", "));
Solution 2 - Java
Arrays.stream(numbers)
creates an IntStream
under the hood and the map operation on an IntStream
requires an IntUnaryOperator
(i.e a function int -> int
). The mapping function you want to apply does not respect this contract and hence the compilation error.
You would need to call boxed()
before in order to get a Stream<Integer>
(this is what Arrays.asList(...).stream()
returns). Then just call map
as you did in the first snippet.
Note that if you need boxed()
followed by map
you probably want to use mapToObj
directly.
The advantage is that mapToObj
doesn't require to box each int
value to an Integer
object; depending on the mapping function you apply of course; so I would go with this option which is also shorter to write.
Solution 3 - Java
You can create an Integer Stream using Arrays.stream(int[]) , you can call mapToObj
like mapToObj(Integer::toString)
.
String csn = Arrays.stream(numbers) // your numbers array
.mapToObj(Integer::toString)
.collect(Collectors.joining(", "));
Hope this helps..
Solution 4 - Java
No boxing, AFAIK, and no explosion of little strings added to the heap:
public static void main(String[] args) {
IntStream stream = IntStream.of(1, 2, 3, 4, 5, 6);
String s = stream.collect(StringBuilder::new, (builder, n) -> builder.append(',').append(n), (x, y) -> x.append(',').append(y)).substring(1);
System.out.println(s);
}
Solution 5 - Java
If the purpose of this sample and question is to figure out how to map strings to a stream of ints (for example, using a stream of ints to access an index in an Array of strings), you can also use boxing, then casting to an int (which would then allow accessing the index of the array).
int[] numbers = {0, 1, 2, 3};
String commaSeparatedNumbers = Arrays.stream(numbers)
.boxed()
.map((Integer i) -> Integer.toString((int)i))
.collect(Collectors.joining(", "));
The .boxed() call converts your IntStream (a stream of primitive ints) to a Stream
Just thought I'd offer another possibility. In programming, there are always multiple ways of accomplishing a task. Know as many as you can, then choose the one that fits the best for the task at hand, keeping in mind performance issues, intuitiveness, clarity of code, your preferences in coding style, and the most self-documenting.
Happy coding!
Solution 6 - Java
In addition to :: operator:
String numString = numbers.stream()
.map(String::valueOf)
.collect(Collectors.joining(", "));
This might be important in newer Java-Versions.