是否有可能创建一个由Java 8中递归定义的惰性（更好的无限）集合？ [英] Is it possible to create a lazy (better infinite) collection defined by recursion in Java 8?
问题描述
我可以创建一个递归闭包：
I can create a recursive closure:
static IntUnaryOperator fibo;
fibo =
(i) ->
i<2 ? 1 : fibo.applyAsInt(i-1)+ fibo.applyAsInt(i-2);
但是，这只是一个例子。相反，如果我创建了一个延迟/无限列表/流，递归可以以相当不错的方式使用：没有成员必须被计算多次。
But of course, it has sense only as an example. On the contrary, if I created a lazy/infinite list/stream, the recursion could be used in quite nice way: no member have to be calculated more than once.
我想到了以下结构：
IntStream fi;
fi=IntStream.iterate(0, i -> fi[i-1]+fi[i-2]);
但这样做不会工作 - 我无法通过索引从流中获取项另一个问题是，如果我以后去沿着流，它将被消耗，我不能反复使用它。如果我将流复制到List，它不再懒了。
But that way it won't work - I can't get an item from the stream by index.The other problem is that if I'll later go along the stream, it will be consumed and I can't use it repeatedly. If I copy the stream to List, it is not lazy anymore.
因此，我需要一些结构，我可以通过索引寻址。 fibo（i）。
As a result, I need some construction that I can address by index. As fibo(i).
编辑。显然，解决方案不能是流，对于流不能使用两次。我不想在每次调用F（i）时重复所有计算。
Edit. Obviously, the solution cannot be a stream, for the stream cannot be used twice. I don't want to repeat all calculations on every call to F(i).
推荐答案
解决方案将创建为class FunctionalSequence 用于表示由具有整数参数的lambda函数定义的惰性，无限序列对象。该函数可以是迭代的或不迭代的。对于迭代情况， FunctionalSequence 类将有一个方法 initialize 用于设置起始值。
The solution will be created as a class FunctionalSequence for representation of a lazy, infinite sequence of objects, defined by a lambda function with integer argument. The function can be iterative or not. For the iterative case the FunctionalSequence class will have a method initialize for setting the start values.
这种类的对象的声明如下：
The declaration of an object of such class will look so:
FunctionalSequence<BigInteger> fiboSequence = new FunctionalSequence<>();
fiboSequence.
initialize(Stream.of(BigInteger.ONE,BigInteger.ONE)).
setSequenceFunction(
(i) ->
fiboSequence.get(i-2).add(fiboSequence.get(i-1))
);
注意，和问题中的递归lambda例子一样，我们不能声明对象，在一个运算符。一个运算符用于声明，另一个用于定义。
Notice, as in the recursive lambda example in the question, we cannot declare the object and define it recursively in one operator. One operator for declaration, another for definition.
FunctionalSequence 类别定义：
import java.util.Iterator;
import java.util.LinkedList;
import java.util.stream.Stream;
public class FunctionalSequence<T> implements Iterable<T>{
LinkedList<CountedFlighweight<T>> realList = new LinkedList<>();
StackOverflowingFunction<Integer, T> calculate = null;
public FunctionalSequence<T> initialize(Stream<T> start){
start.forEachOrdered((T value) ->
{
realList.add(new CountedFlighweight<>());
realList.getLast().set(value);
});
return this;
}
public FunctionalSequence<T> setSequenceFunction(StackOverflowingFunction<Integer, T> calculate){
this.calculate = calculate;
return this;
}
@Override
public Iterator<T> iterator() {
return new SequenceIterator();
}
public T get(int currentIndex) throws StackOverflowError{
if(currentIndex < 0) return null;
while (currentIndex >= realList.size()){
realList.add(new CountedFlighweight<T>());
}
try {
return (T) realList.get(currentIndex).get(calculate, currentIndex);
} catch (Exception e) {
return null;
}
}
public class SequenceIterator implements Iterator<T>{
int currentIndex;
@Override
public boolean hasNext() {
return true;
}
@Override
public T next() {
T result = null;
if (currentIndex == realList.size()){
realList.add(new CountedFlighweight<T>());
}
// here the StackOverflowError catching is a pure formality, by next() we would never cause StackOverflow
try {
result = realList.get(currentIndex).get(calculate, currentIndex);
} catch (StackOverflowError e) {
}
currentIndex++;
return result;
}
}
/**
* if known is false, the value of reference is irrelevant
* if known is true, and reference is not null, reference contains the data
* if known is true, and reference is null, that means, that the appropriate data are corrupted in any way
* calculation on corrupted data should result in corrupted data.
* @author Pet
*
* @param <U>
*/
public class CountedFlighweight<U>{
private boolean known = false;
private U reference;
/**
* used for initial values setting
*/
private void set(U value){
reference = value;
known = true;
}
/**
* used for data retrieval or function counting and data saving if necessary
* @param calculate
* @param index
* @return
* @throws Exception
*/
public U get(StackOverflowingFunction<Integer, U> calculate, int index) throws StackOverflowError{
if (! known){
if(calculate == null) {
reference = null;
} else {
try {
reference = calculate.apply(index);
} catch (Exception e) {
reference = null;
}
}
}
known = true;
return reference;
}
}
@FunctionalInterface
public interface StackOverflowingFunction <K, U> {
public U apply(K index) throws StackOverflowError;
}
}
由于递归函数很容易满足StackOverflowError，我们应该组织递归，以便在这种情况下，整个递归序列将回滚，没有任何变化真正满足和抛出异常。
As the recursive function could easily meet the StackOverflowError, we should organize the recursion so that in that case the whole recursive sequence will roll back without any changes really met and throw the exception.
使用FunctionalSequence可能如下：
The use of the FunctionalSequence could look so:
// by iterator:
int index=0;
Iterator<BigInteger> iterator = fiboSequence.iterator();
while(index++<10){
System.out.println(iterator.next());
}
或者：
static private void tryFibo(FunctionalSequence<BigInteger> fiboSequence, int i){
long startTime = System.nanoTime();
long endTime;
try {
fiboSequence.get(i);
endTime = System.nanoTime();
System.out.println("repeated timing for f("+i+")=" + (endTime-startTime)/1000000.+" ns");
} catch (StackOverflowError e) {
endTime = System.nanoTime();
//e.printStackTrace();
System.out.println("failed counting f("+i+"), time=" + (endTime-startTime)/1000000.+" ns");
}
}
最后一个函数可以以下列方式使用：
The last function can be used in the following way:
tryFibo(fiboSequence, 1100);
tryFibo(fiboSequence, 100);
tryFibo(fiboSequence, 100);
tryFibo(fiboSequence, 200);
tryFibo(fiboSequence, 1100);
tryFibo(fiboSequence, 2100);
tryFibo(fiboSequence, 2100);
tryFibo(fiboSequence, 1100);
tryFibo(fiboSequence, 100);
tryFibo(fiboSequence, 100);
tryFibo(fiboSequence, 200);
tryFibo(fiboSequence, 1100);
这里是结果（堆栈限制为256K为测试的需要）：
Here are the results (the stack was limited to 256K for the needs of testing):
1
1
2
3
5
8
13
21
34
55
failed counting f(1100), time=3.555689 ns
repeated timing for f(100)=0.213156 ns
repeated timing for f(100)=0.002444 ns
repeated timing for f(200)=0.266933 ns
repeated timing for f(1100)=5.457956 ns
repeated timing for f(2100)=3.016445 ns
repeated timing for f(2100)=0.001467 ns
repeated timing for f(1100)=0.005378 ns
repeated timing for f(100)=0.002934 ns
repeated timing for f(100)=0.002445 ns
repeated timing for f(200)=0.002445 ns
repeated timing for f(1100)=0.003911 ns
看，对于同一索引的f（i）的可重复调用几乎没有时间 - 没有迭代。由于StackOverflowError，我们不能一次到达f（1100）。但是在我们达到f（200）之后，f（1100）变得可达。我们做到了！
Look, the repeatable call of the f(i) for the same index takes practically no time - no iterations were made. We cannot reach f(1100) at once because of the StackOverflowError. But after we have reached once f(200), f(1100) becomes reachable. We made it!
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