java 使用 Java8 生成器生成无限自然数序列
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Infinite sequence of Natural numbers with Java8 generator
提问by David Phillips
I defined naturalfor Infinite sequence (Stream) of Natural numbers with Java8 iterator.
我用 Java8定义natural了Stream自然数的无限序列 ( ) iterator。
IntStream natural = IntStream.iterate(0, i -> i + 1);
natural
.limit(10)
.forEach(System.out::println);
Now, I want to define it with Java8 generator.
现在，我想用 Java8 定义它generator。
static Stream generate(Supplier s)
静态流生成（供应商）
What would be the simplest way? Thanks.
什么是最简单的方法？谢谢。
采纳答案by Joffrey
Note: @assylias managed to do it with a lambda using AtomicInteger.He should probably have the accepted answer.
注意：@assylias 使用AtomicInteger. 他可能应该得到公认的答案。
I'm not sure you can do that with a lambda (because it is stateful), but with a plain Supplierthis would work:
我不确定你可以用 lambda 来做到这一点（因为它是有状态的），但如果是简单的，Supplier这会起作用：
IntSupplier generator = new IntSupplier() {
int current = 0;
public int getAsInt() {
return current++;
}
};
IntStream natural = IntStream.generate(generator);
However, I highly prefer your current solution, because this is the purpose of iterate(int seed, IntUnaryOperator f)IMHO:
但是，我非常喜欢您当前的解决方案，因为这是iterate(int seed, IntUnaryOperator f)恕我直言的目的：
IntStream natural = IntStream.iterate(0, i -> i + 1);
回答by assylias
With a generator you need to keep track of your current index. One way would be:
使用生成器，您需要跟踪当前索引。一种方法是：
IntStream natural = IntStream.generate(new AtomicInteger()::getAndIncrement);
Note: I use AtomicInteger as a mutable integer rather than for its thread safety: if you parallelise the stream the order will not be as expected.
注意：我使用 AtomicInteger 作为可变整数而不是为了它的线程安全：如果你并行化流，顺序将不会像预期的那样。
回答by David Phillips
This is built into IntStream:
这内置于IntStream：
IntStream.range(0, Integer.MAX_VALUE)
This returns all values up to (but not including) Integer.MAX_VALUE.
这将返回所有值（但不包括）Integer.MAX_VALUE。
