转换嵌套循环的规范方式是在流上使用 flatMap ，例如

  IntStream.range（0，5）.flatMap（i-> IntStream.range（i，10））
 .forEach （的System.out ::的println）; 
 

您任务中棘手的部分是增加2，因为它在流API中没有直接的等效。有两种可能：

1. 使用 IntStream.iterate（y，x-> x + 2 ）来定义起始值和增量。然后你必须通过 limit 来修改无限流： .limit（（11-y）/ 2）$ c
$ b

   因此，您的循环的结果代码如下所示：

     IntStream.range（0,5）
    .flatMap（y-> IntStream.iterate（y，x-> x + 2）.limit（（11-y）/ 2）
    .map（x  - > x + y））。forEach（System.out :: println）; 
    




2. 使用 IntStream.range（0，（11-y） / 2）创建一个所需数量的升序流 int s并用 .map（t- > y + t * 2），让它为$ c>循环产生期望的内部值。

   
   
   

   然后，循环的结果代码如下所示：
   
   

     IntStream.range（0，5） 
    .flatMap（y-> IntStream.range（0，（11-y）/ 2）.map（t-> y + t * 2）.map（x  - > x + y）） 
    .forEach（System.out :: println）; 
    

I was playing around with Java 8. I had some trouble converting this for loop into Java 8 Stream.

for (int y = 0; y < 5; y ++) {
    for (int x = y; x < 10; x += 2) {
        System.out.println(x+y);
    }
}

Please help!

The canonical way of converting nested loops is to use flatMap on a stream, e.g.

IntStream.range(0, 5).flatMap(i->IntStream.range(i, 10))
  .forEach(System.out::println);

The tricky part on your task is the increment by two as this has no direct equivalent in the stream API. There are two possibilities:

1. Use IntStream.iterate(y, x->x+2) to define start value and increment. Then you have to modify the infinite stream by limiting the number of elements: .limit((11-y)/2).

   So the resulting code for your loop would look like:

   IntStream.range(0, 5)
       .flatMap(y->IntStream.iterate(y, x->x+2).limit((11-y)/2)
       .map(x -> x+y)).forEach(System.out::println);
   

2. Use IntStream.range(0, (11-y)/2) to create an stream of the desired number of ascending ints and modify it with .map(t->y+t*2) to have it produce the desired values of your inner for loop.

   Then, the resulting code for your loop would look like:

   IntStream.range(0, 5)
       .flatMap(y->IntStream.range(0, (11-y)/2).map(t->y+t*2).map(x -> x+y))
       .forEach(System.out::println);