我是 Scala 语言的新手.

I'm new to the Scala language.

我需要 Long 类型的 Range.

I need Range for Long type.

我需要一个包含第 1 步的 [1, 2, 3 ... 10000000] 列表.如果我使用 until/to 我会因为使用 Long 而不是 Int 而出错.

I need a List of [1, 2, 3 ... 10000000] with step 1. If I use until/to I get an error because of using Long instead of Int.

我尝试编写一个简单的函数，它需要一个开始、一个结束和一个空列表，并生成一个 [start .. end] 列表.

I try to write simple function which expects a start, an end and and an empty List and generates a List of [start .. end].

这是我的功能:

def range_l(start : Long, end : Long, list : List[Long]) : List[Long] = {
    if (start == end){
        val add_to_list = start :: list
        return add_to_list
    }
    else {
        val add_to_list = start :: list
        range_l(start + 1, end, add_to_list)
    }
}

如果我这样称呼它:range_l(1L, 1000000L, List()) 我在以下行中得到 OutOfMemory 错误:add_to_list = start ::列表

If I call it like: range_l(1L, 1000000L, List()) i get OutOfMemory error in the following line: add_to_list = start :: list

你有什么建议吗?如何获得 Range[Long] 或如何优化功能.如何避免 OutOfMemory?

What can you advice me? How can I get Range[Long] or how can I optimize the function. How can I avoid OutOfMemory?

谢谢.

推荐答案

您可以使用以下语法创建这样的范围:

You can create such a range by using the following syntax:

val range = 1L to 10000000L

'L' 是强制性的，以告知编译器文字是长整数而不是整数.

The 'L' is mandatory to inform the compiler that the litterals are longs and not ints.

然后您可以在实例 range 上使用几乎所有 List 方法.它不应该填满你的记忆，因为中间值是在需要时生成的.该范围可以传递给任何期望 Traversable[Long]、Seq[Long]、Iterable[Long] 等的方法.

You can then use almost all List methods on the instance range. It should not fill you memory because the intermediate values are generated when needed. The range can be passed to any method expecting a Traversable[Long], a Seq[Long], an Iterable[Long], etc.

但是，如果你真的需要一个 List 只需调用 range.toList (并增加堆大小以容纳所有列表元素)...

However, if you really need a List just call range.toList (and increase the heap size to accommodate all the list elements)...