我正在尝试实现一个在其实现中使用其自身的另一个实例的流。该流前面有一些常量元素(使用 IntStream.concat),因此只要串联流惰性地创建非常量部分,这应该就可以工作。我认为使用StreamSupport.intStream 重载采用供应商 https://docs.oracle.com/javase/8/docs/api/java/util/stream/StreamSupport.html#intStream-java.util.function.Supplier-int-boolean-与 IntStream.concat (其中“创建一个延迟连接的流” https://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html#concat-java.util.stream.IntStream-java.util.stream.IntStream-)应该足够懒,只在需要元素时才创建第二个分割器,但即使创建流(不评估它)也会溢出堆栈。如何延迟连接流?
我正在尝试从以下位置移植流式素数筛这个答案 https://stackoverflow.com/a/10733621/3614835进入Java。该筛子使用其自身的另一个实例(ps = postponed_sieve()
在 Python 代码中)。如果我破坏最初的四个常量元素(yield 2; yield 3; yield 5; yield 7;
)到他们自己的流中,很容易将生成器实现为分割器:
/**
* based on https://stackoverflow.com/a/10733621/3614835
*/
static class PrimeSpliterator extends Spliterators.AbstractIntSpliterator {
private static final int CHARACTERISTICS = Spliterator.DISTINCT | Spliterator.IMMUTABLE | Spliterator.NONNULL | Spliterator.ORDERED | Spliterator.SORTED;
private final Map<Integer, Supplier<IntStream>> sieve = new HashMap<>();
private final PrimitiveIterator.OfInt postponedSieve = primes().iterator();
private int p, q, c = 9;
private Supplier<IntStream> s;
PrimeSpliterator() {
super(105097564 /* according to Wolfram Alpha */ - 4 /* in prefix */,
CHARACTERISTICS);
//p = next(ps) and next(ps) (that's Pythonic?)
postponedSieve.nextInt();
this.p = postponedSieve.nextInt();
this.q = p*p;
}
@Override
public boolean tryAdvance(IntConsumer action) {
for (; c > 0 /* overflow */; c += 2) {
Supplier<IntStream> maybeS = sieve.remove(c);
if (maybeS != null)
s = maybeS;
else if (c < q) {
action.accept(c);
return true; //continue
} else {
s = () -> IntStream.iterate(q+2*p, x -> x + 2*p);
p = postponedSieve.nextInt();
q = p*p;
}
int m = s.get().filter(x -> !sieve.containsKey(x)).findFirst().getAsInt();
sieve.put(m, s);
}
return false;
}
}
我第一次尝试 primes() 方法返回一个 IntStream,它将常量流与新的 PrimeSpliterator 连接起来:
public static IntStream primes() {
return IntStream.concat(IntStream.of(2, 3, 5, 7),
StreamSupport.intStream(new PrimeSpliterator()));
}
调用 primes() 会导致 StackOverflowError,因为 primes() 始终实例化 PrimeSpliterator,但 PrimeSpliterator 的字段初始值设定项始终调用 primes()。然而,StreamSupport.intStream 的重载需要一个供应商,这应该允许延迟创建 PrimeSpliterator:
public static IntStream primes() {
return IntStream.concat(IntStream.of(2, 3, 5, 7),
StreamSupport.intStream(PrimeSpliterator::new, PrimeSpliterator.CHARACTERISTICS, false));
}
然而,我却得到了一个具有不同回溯的 StackOverflowError (在重复时进行了修剪)。请注意,递归完全发生在对 primes() 的调用中——终端操作 iterator() 永远不会在返回的流上调用。
Exception in thread "main" java.lang.StackOverflowError
at java.util.stream.StreamSpliterators$DelegatingSpliterator$OfInt.<init>(StreamSpliterators.java:582)
at java.util.stream.IntPipeline.lazySpliterator(IntPipeline.java:155)
at java.util.stream.IntPipeline$Head.lazySpliterator(IntPipeline.java:514)
at java.util.stream.AbstractPipeline.spliterator(AbstractPipeline.java:352)
at java.util.stream.IntPipeline.spliterator(IntPipeline.java:181)
at java.util.stream.IntStream.concat(IntStream.java:851)
at com.jeffreybosboom.projecteuler.util.Primes.primes(Primes.java:22)
at com.jeffreybosboom.projecteuler.util.Primes$PrimeSpliterator.<init>(Primes.java:32)
at com.jeffreybosboom.projecteuler.util.Primes$$Lambda$1/834600351.get(Unknown Source)
at java.util.stream.StreamSpliterators$DelegatingSpliterator.get(StreamSpliterators.java:513)
at java.util.stream.StreamSpliterators$DelegatingSpliterator.estimateSize(StreamSpliterators.java:536)
at java.util.stream.Streams$ConcatSpliterator.<init>(Streams.java:713)
at java.util.stream.Streams$ConcatSpliterator$OfPrimitive.<init>(Streams.java:789)
at java.util.stream.Streams$ConcatSpliterator$OfPrimitive.<init>(Streams.java:785)
at java.util.stream.Streams$ConcatSpliterator$OfInt.<init>(Streams.java:819)
at java.util.stream.IntStream.concat(IntStream.java:851)
at com.jeffreybosboom.projecteuler.util.Primes.primes(Primes.java:22)
at com.jeffreybosboom.projecteuler.util.Primes$PrimeSpliterator.<init>(Primes.java:32)
at com.jeffreybosboom.projecteuler.util.Primes$$Lambda$1/834600351.get(Unknown Source)
at java.util.stream.StreamSpliterators$DelegatingSpliterator.get(StreamSpliterators.java:513)
at java.util.stream.StreamSpliterators$DelegatingSpliterator.estimateSize(StreamSpliterators.java:536)
at java.util.stream.Streams$ConcatSpliterator.<init>(Streams.java:713)
at java.util.stream.Streams$ConcatSpliterator$OfPrimitive.<init>(Streams.java:789)
at java.util.stream.Streams$ConcatSpliterator$OfPrimitive.<init>(Streams.java:785)
at java.util.stream.Streams$ConcatSpliterator$OfInt.<init>(Streams.java:819)
at java.util.stream.IntStream.concat(IntStream.java:851)
at com.jeffreybosboom.projecteuler.util.Primes.primes(Primes.java:22)
如何足够懒惰地连接流以允许流在其实现中使用自身的另一个副本?