我有另一个快速的解决方案来实现这个(实际上是不干净的，但你知道的):

public static void main(String[] args) {
    System.out.println(StreamUtil.iterate(1, o -> o + 1).terminateOn(15)
            .map(o -> o.toString()).collect(Collectors.joining(", ")));
}

static interface TerminatedStream<T> {
    Stream<T> terminateOn(T e);
}

static class StreamUtil {
    static <T> TerminatedStream<T> iterate(T seed, UnaryOperator<T> op) {
        return new TerminatedStream<T>() {
            public Stream<T> terminateOn(T e) {
                Builder<T> builder = Stream.<T> builder().add(seed);
                T current = seed;
                while (!current.equals(e)) {
                    current = op.apply(current);
                    builder.add(current);
                }
                return builder.build();
            }
        };
    }
}

这是在int上做的一个版本-正如问题中所问的那样。

用法:

StreamUtil.takeWhile(IntStream.iterate(1, n -> n + 1), n -> n < 10);

下面是StreamUtil的代码:

import java.util.PrimitiveIterator;
import java.util.Spliterators;
import java.util.function.IntConsumer;
import java.util.function.IntPredicate;
import java.util.stream.IntStream;
import java.util.stream.StreamSupport;

public class StreamUtil
{
    public static IntStream takeWhile(IntStream stream, IntPredicate predicate)
    {
        return StreamSupport.intStream(new PredicateIntSpliterator(stream, predicate), false);
    }

    private static class PredicateIntSpliterator extends Spliterators.AbstractIntSpliterator
    {
        private final PrimitiveIterator.OfInt iterator;
        private final IntPredicate predicate;

        public PredicateIntSpliterator(IntStream stream, IntPredicate predicate)
        {
            super(Long.MAX_VALUE, IMMUTABLE);
            this.iterator = stream.iterator();
            this.predicate = predicate;
        }

        @Override
        public boolean tryAdvance(IntConsumer action)
        {
            if (iterator.hasNext()) {
                int value = iterator.nextInt();
                if (predicate.test(value)) {
                    action.accept(value);
                    return true;
                }
            }

            return false;
        }
    }
}

作为@StuartMarks回答的后续。我的StreamEx库具有takeWhile操作，该操作与当前JDK-9实现兼容。当在JDK-9下运行时，它只会委托给JDK实现(通过MethodHandle。invokeExact非常快)。在JDK-8下运行时，将使用“polyfill”实现。所以使用我的库可以像这样解决问题:

IntStreamEx.iterate(1, n -> n + 1)
           .takeWhile(n -> n < 10)
           .forEach(System.out::println);

甚至我也有类似的需求——调用web服务，如果失败，重试3次。如果在多次尝试后仍然失败，请发送电子邮件通知。在谷歌搜索了很多之后，anyMatch()成为了救星。我的示例代码如下。在下面的例子中，如果webServiceCall方法在第一次迭代本身中返回true，则stream不会在我们调用anyMatch()时进一步迭代。我相信，这就是你想要的。

import java.util.stream.IntStream;

import io.netty.util.internal.ThreadLocalRandom;

class TrialStreamMatch {

public static void main(String[] args) {        
    if(!IntStream.range(1,3).anyMatch(integ -> webServiceCall(integ))){
         //Code for sending email notifications
    }
}

public static boolean webServiceCall(int i){
    //For time being, I have written a code for generating boolean randomly
    //This whole piece needs to be replaced by actual web-service client code
    boolean bool = ThreadLocalRandom.current().nextBoolean();
    System.out.println("Iteration index :: "+i+" bool :: "+bool);

    //Return success status -- true or false
    return bool;
}

下面是我使用Java流库的尝试。

        IntStream.iterate(0, i -> i + 1)
        .filter(n -> {
                if (n < 10) {
                    System.out.println(n);
                    return false;
                } else {
                    return true;
                }
            })
        .findAny();

takeWhile是质子包库提供的函数之一。

Stream<Integer> infiniteInts = Stream.iterate(0, i -> i + 1);
Stream<Integer> finiteInts = StreamUtils.takeWhile(infiniteInts, i -> i < 10);

assertThat(finiteInts.collect(Collectors.toList()),
           hasSize(10));