How to return the result of a map? Are there any predefined collectors?

[Brian Goetz]

Or, since your problem is all dealing in integers anyway, you can avoid 
boxing entirely by using IntStream.toArray(), which will return an int[].

On 2/22/2013 5:16 PM, Brian Goetz wrote:
>> The result of a sequence of map operations is a Stream<T>. However I needed
>> to cast the functions explicitly as follows
>>
>> Stream<Integer> doubles =
>> Arrays.asList(1,2,3,4).stream().map((Function<Integer, Integer>) n -> n *
>> 2).map((Function<Integer, Integer>) n -> n + 1);
>>
>> Q1. Is there a way to avoid the cast Function<Integer, Integer> here?
>
> Yes -- just don't do it.  Without the cast, it will select the override
> map(ToIntFunction<Integer>), which will get you out of boxed land, and
> return an IntStream.
>
> You can do even better by:
>
>     Streams.intRange(0, 5)  // IntStream
>            .map(x -> x*2)   // IntStream
>            .map(x -> x+1)   // IntStream
>
> and there's no boxing at all.
>
>> Also I wanted to convert the stream into a List. I understood the way to do
>> so is using a collector. Google searches led me to believe there is a
>> function toList() which should help. However I couldn't find a method
>> "toList" on Stream. Neither could I easily locate it anywhere in the java
>> docs.
>
>     .collect(Collectors.toList())
>
> If you have an IntStream and you want to put it into a List<Integer>:
>
> List<Integer> list =
>     intStream.boxed()  // Stream<Integer>
>              .collect(Collectors.toList());
>
>> Q2. Are there any predefined collectors I could use ? eg. in this case to
>> convert a Stream<Int> into a List<Int>
>
> See above -- Collectors.toList() or .toCollection(ctor).
>
> Or, you can use the explicit form of Collect:
>
>     stream.collect(ArrayList::new, ArrayList::add, ArrayList::addAll);
>
>