假设您要流式传递迭代器的元素;让我们使用Scanner的具体示例，它实现Iterator<String>。
给定Iterator，说：
// Scanner implements Iterator<String>
Iterator<String> scanner = new Scanner(System.in);
从中创建Stream<String>的选项是笨重的：
StreamSupport.stream(
Spliterators.spliteratorUnknownSize(scanner, Spliterator.ORDERED), false);
或稍微简洁，但是迟钝：
StreamSupport.stream(
((Iterable<String>) () -> new Scanner(System.in)).spliterator(), false);
在JDK的某个地方是否有一个返回Stream<T>给定Iterator<T>的工厂方法？
答案 0 :(得分：8)
我只想使用Image: ic_stage_2.png
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entropy: 0.977354
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。是的，这是一个无限的流，但是来自Stream.generate(iterator::next)的扫描仪也是如此，除非你知道你要求多少行，在这种情况下它很容易使用
System.in这可以防止迭代器必须迭代两次（一次作为迭代器，一次作为流）。
如果你需要一个大小的流而不知道它有多大，但是不想让你的代码混乱，只需创建一个明确采用Stream.generate(iterator::next).limit(numLines);
的实用工具方法：
Iterable<T>最终比在一条线上把它全部塞进来更容易辨认，并且
然后，您只需拨打public static <T> Stream<T> stream(Iterable<T> it){
return StreamSupport.stream(it.spliterator(), false);
}
或stream(()->iterator);
您也可以轻松添加第二种便捷方法：
stream(()->new Scanner(System.in));让您拨打public static <T> Stream<T> stream(Iterator<T> it){
return stream(()->it);
}
或stream(iterator);
虽然，如果我要将stream(new Scanner(System.in));读入流中，我可能会使用Reader而不是扫描仪：
System.in