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How to filter a stream of integers to a list?

I am trying to process a stream Integers

and collect integers that match a predicate (via a function compare()

) into a list. Here's a rough outline of the code I wrote.

private List<Integer> process() {
    Z z = f(-1);
    return IntStream.range(0, 10)
        .filter(i -> compare(z, f(i)))
        .collect(Collectors.toCollection(ArrayList::new)); // Error on this line
}

private boolean compare(Z z1, Z z2) { ... }
private Z f(int i) { ... }

Unfortunately my solution doesn't compile and I can't figure out the compiler error for the highlighted line:

Collection method (Provider <R>, ObjIntConsumer <R>, BiConsumer <R, R>) in IntStream type is not applicable for arguments (Collector <Object, capture # 1-of ?, Collection <Object →)

Any suggestions?

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2 answers


IntStream

does not contain a method collect

that takes one type argument Collector

. Stream

does. Therefore, you must convert yours IntStream

to Stream

objects.

You can IntStream

in Stream<Integer>

or use mapToObj

to achieve the same.

For example:

return IntStream.range(0, 10)
    .filter(i -> compare(z, f(i)))
    .boxed()
    .collect(Collectors.toCollection(ArrayList::new));


boxed()

will return a stream consisting of the elements of this stream, each packed into an integer.

or

return IntStream.range(0, 10)
    .filter(i -> compare(z, f(i)))
    .mapToObj(Integer::valueOf)
    .collect(Collectors.toCollection(ArrayList::new));
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Or you can specify yourself Supplier, Accumulator and Combiner

:



 IntStream.range(0, 10)
            .filter(i -> compare(z, f(i)))  
            .collect(ArrayList::new, ArrayList::add, ArrayList::addAll);
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